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Representation theory studies how abstract algebraic structures reveal their internal symmetries through linear actions on vector spaces. In this course, we examine how finite groups and compact topological groups can be realized as subgroups of $\mathrm{GL}_n$, acting on finite-dimensional complex vector spaces. The central insight — and what makes this subject particularly powerful — is that the machinery developed for finite groups translates almost directly to the continuous setting, allowing unified reasoning about disparate algebraic objects. You will need comfort with group theory at the level of IB Groups, Rings and Modules, together with basic linear algebra and some familiarity with metric and topological spaces; one key lemma from Galois theory will appear, but it can be treated as a black box if needed. The course unfolds in three connected movements. First, we establish the foundations: group actions and the basic definitions of representations, then prove complete reducibility via Maschke's theorem and explore Schur's lemma, which constrains how irreducible representations interact. Second, we develop character theory as our main computational tool — characters are traces of representation matrices, and their orthogonality relations allow us to classify irreducible representations and understand representation structure without explicit matrices. Finally, we apply these tools to concrete problems: permutation representations, induction and restriction, and eventually Burnside's $p^a q^b$ theorem, which uses character-theoretic arithmetic to solve a purely group-theoretic question. A recurring theme is how symmetry becomes computable. Dual spaces and tensor products show how to build new representations from old; Frobenius groups and Mackey theory reveal hidden constraints on how representations behave under subgroups; and the integrality results expose number-theoretic rigidity in the group algebra. Throughout, the chapters reinforce each other: character theory simplifies what would be tedious linear algebra, induction-restriction duality connects representations of a group to those of its subgroups, and Burnside's theorem demonstrates that representation-theoretic information can solve concrete combinatorial problems. The course closes by extending the entire framework to compact groups, where finite sums become integrals against Haar measure but the structural results survive almost intact. # 1. Group actions This chapter collects the linear algebra and group theory prerequisites that underpin everything that follows. The central question of representation theory — how can a group act on a vector space? — makes sense only once we have a firm grip on both the algebraic structure of groups and the geometry of linear maps. We review the key examples of groups that will recur throughout the course, establish the language of group actions and permutation representations, and set up the field conventions that distinguish ordinary from modular representation theory. ## Fields and the Linear Algebra We Need What class of fields should we work over, and what linear algebra will we actually need? These are not arbitrary choices: the whole apparatus of character theory and Maschke's theorem depends critically on the characteristic of the field, and the core technical tool — replacing a complicated operator by a diagonal one — requires hypotheses that fail in positive characteristic. We fix conventions here so they do not need to be re-stated in every later argument. Throughout this course, $\mathbb{F}$ denotes a field. The primary case is $\mathbb{F} = \mathbb{C}$, though $\mathbb{R}$ and $\mathbb{Q}$ also arise. All three have characteristic zero, and working over such fields is called **ordinary** representation theory. Occasionally we take $\mathbb{F} = \mathbb{F}_p$ or its algebraic closure $\bar{\mathbb{F}}_p$; this is **modular** representation theory and introduces genuinely new phenomena. Every vector space $V$ considered in this course is finite-dimensional over $\mathbb{F}$. [definition: General Linear Group] Let $V$ be a finite-dimensional vector space over $\mathbb{F}$. The **general linear group** $\operatorname{GL}(V)$ is the group of all invertible linear maps $\theta: V \to V$, with the group operation given by composition. The **endomorphism algebra** $\operatorname{End}(V)$ is the set of all linear maps $V \to V$ (not necessarily invertible), equipped with composition as multiplication. [/definition] Once we fix a basis $e_1, \ldots, e_n$ of $V$ over $\mathbb{F}$, every $\theta \in \operatorname{GL}(V)$ is encoded by a matrix $A_\theta = (a_{ij}) \in M_n(\mathbb{F})$ via \begin{align*} \theta(e_j) = \sum_i a_{ij} e_i. \end{align*} The matrix $A_\theta$ is automatically invertible, so $A_\theta \in \operatorname{GL}_n(\mathbb{F})$. The assignment $\theta \mapsto A_\theta$ yields a group isomorphism. [quotetheorem:2399]This isomorphism is canonical once a basis is fixed, but the basis itself is not canonical. This is the first instance of a recurring theme in representation theory: there is an abstract, basis-free object ($\operatorname{GL}(V)$ and the linear maps that interest us) and a choice-dependent matrix version ($\operatorname{GL}_n(\mathbb{F})$). This result is from linear algebra, which we use freely. [quotetheorem:2400]The content of this theorem is that changing a basis amounts precisely to conjugating by the change-of-basis matrix. Two matrices that encode the same linear map but in different bases must therefore lie in the same conjugacy class in $\operatorname{GL}_n(\mathbb{F})$. This fact — a result from linear algebra, which we use freely — is why conjugacy is the central equivalence relation in representation theory. Any invariant of a matrix that is preserved under conjugation gives a well-defined invariant of the underlying linear operator, independent of the choice of basis. The most important such invariant is the trace. [quotetheorem:2401] [citeproof:2401] The cyclic property $\operatorname{tr}(BC) = \operatorname{tr}(CB)$ is the key point: conjugation is just a cyclic rearrangement, so the trace cannot see it. Concretely, since $\operatorname{tr}(A_\theta)$ is independent of the choice of basis, we may unambiguously define $\operatorname{tr}(\theta) := \operatorname{tr}(A_\theta)$ for any $\theta \in \operatorname{GL}(V)$. This is the genesis of character theory: instead of tracking whole matrices parametrised by group elements, we record only their traces. This reduction from matrices to scalars carries an enormous amount of information, as we will see. [remark: Trace of an Operator] Since $\operatorname{tr}(A_\theta)$ is independent of the choice of basis, we may unambiguously define $\operatorname{tr}(\theta) := \operatorname{tr}(A_\theta)$ for any $\theta \in \operatorname{GL}(V)$. When we study representations, we will encounter matrices parametrised by group elements — instead of tracking those matrices in full, it often suffices to record their traces. This reduction from matrices to scalars is one of the key ideas driving character theory. [/remark] ### Diagonalisation When working with representations, we frequently want to simplify the matrices that arise. The simplest possible form is diagonal. Can we always diagonalise? In general, no: the Jordan block \begin{align*} J = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} \end{align*} has a repeated eigenvalue $\lambda$ but is not diagonalisable — its minimal polynomial is $(x - \lambda)^2$, which has a repeated factor. The matrix $J$ witnesses that not every endomorphism can be diagonalised, even over $\mathbb{C}$. However, the operators that arise from group representations have extra structure that forces diagonalisability. [quotetheorem:2402]The hypothesis that $\mathbb{F} = \mathbb{C}$ (characteristic zero, algebraically closed) is essential. Over $\mathbb{F}_p$, the Frobenius map $\phi$ on $\bar{\mathbb{F}}_p$ satisfies $\phi^p = \mathrm{id}$, yet the induced action on a vector space can fail to be diagonalisable: the matrix \begin{align*} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{align*} satisfies $\alpha^p = \mathrm{id}$ over $\mathbb{F}_p$ (since $\binom{p}{k} \equiv 0 \pmod{p}$ for $0 < k < p$), yet it is not diagonalisable — this is precisely the Jordan block phenomenon in characteristic $p$. This is one reason modular representation theory is harder. This is a special case of a cleaner general criterion. [quotetheorem:406]The proof proceeds by decomposing $V$ into eigenspaces using the factorisation of $f$. If $f(x) = (x - \lambda_1) \cdots (x - \lambda_k)$ with distinct $\lambda_i$, then the polynomials $(x - \lambda_i)$ are pairwise coprime, so the Chinese Remainder Theorem for polynomials (equivalently, partial fractions) yields a direct sum decomposition $V = \bigoplus_i \ker(\alpha - \lambda_i \mathrm{id})$. Each summand is an eigenspace, and a basis adapted to this decomposition diagonalises $\alpha$. The "if" direction fails when $f$ has repeated factors: the Jordan block above satisfies $(\alpha - \lambda)^2 = 0$, but not $(\alpha - \lambda) = 0$, so the factor $(x - \lambda)^2$ is repeated and $\alpha$ is not diagonalisable. This is the precise obstruction. The first theorem follows from the second: if $\alpha^m = \mathrm{id}$, then $\alpha$ is a root of $f(x) = x^m - 1$. Over $\mathbb{C}$ we have \begin{align*} x^m - 1 = \prod_{j=0}^{m-1}(x - \omega^j), \qquad \omega = e^{2\pi i/m}, \end{align*} so $f$ has $m$ distinct linear factors, and $f(\alpha) = 0$. The theorem therefore guarantees $\alpha$ is diagonalisable with eigenvalues among the $m$-th roots of unity. When multiple endomorphisms are in play simultaneously — as will happen throughout the course — we want to diagonalise them all at once. [quotetheorem:2403]This theorem will be crucial when we study abelian groups, where all representation matrices commute with one another. The pairwise commutativity condition is genuinely necessary: the matrices \begin{align*} A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{align*} are both diagonalisable (both have eigenvalues $\pm 1$), yet $AB \neq BA$, and they cannot be simultaneously diagonalised — any basis that diagonalises $A$ consists of $e_1$ and $e_2$, but $B$ swaps these, so $B$ is not diagonal in that basis. For abelian groups, this commutativity is automatic, which is why their representations decompose so completely into one-dimensional pieces. ## Key Examples of Groups Which groups should a student of representation theory know by heart? The answer is not arbitrary: the groups listed here are the standard test cases that appear in almost every theorem and counterexample in the subject. They are small enough to compute with explicitly, yet rich enough to illustrate genuinely different phenomena — abelian vs non-abelian, cyclic vs non-cyclic, groups with and without normal subgroups of index 2. When a new theorem is proved, these are the first groups to check it on. [definition: Symmetric Group] The **symmetric group** $S_n$ is the group of all bijections $\{1, \ldots, n\} \to \{1, \ldots, n\}$, i.e.\ all permutations of the set $\{1, \ldots, n\}$, with composition as the group operation. Its order is $|S_n| = n!$. [/definition] The symmetric groups are the universal examples: Cayley's theorem states that every group embeds in some $S_n$. From a representation-theoretic perspective, $S_3$ is the smallest non-abelian group, and $S_4$ already requires techniques beyond the abelian case. [definition: Alternating Group] The **alternating group** $A_n \leq S_n$ is the set of all permutations that can be written as a product of an even number of transpositions $(i\ j)$. Its order is $|A_n| = \frac{n!}{2}$. Since $A_n$ has index $2$ in $S_n$, it is a normal subgroup. [/definition] The alternating group arises naturally because the sign homomorphism $\operatorname{sgn}: S_n \to \{+1, -1\}$ is a non-trivial one-dimensional representation of $S_n$, and $A_n = \ker(\operatorname{sgn})$. For $n \geq 5$, $A_n$ is simple — it has no normal subgroups other than itself and $\{e\}$ — which makes its representation theory particularly clean. [definition: Cyclic Group] The **cyclic group** of order $m$ is \begin{align*} C_m = \langle x : x^m = 1 \rangle. \end{align*} It arises concretely as $\mathbb{Z}/m\mathbb{Z}$ under addition, as the group of $m$-th roots of unity in $\mathbb{C}$ (a subgroup of $\operatorname{GL}_1(\mathbb{C}) \cong \mathbb{C}^\times$), and as the rotation symmetry group of a regular $m$-gon in $\mathbb{R}^2$ (a subgroup of $\operatorname{GL}_2(\mathbb{R})$). [/definition] Cyclic groups are the simplest non-trivial abelian groups, and their representations over $\mathbb{C}$ decompose completely into one-dimensional pieces given by the $m$-th roots of unity — this is essentially the discrete Fourier transform. They serve as the baseline case for all later structure theorems. [definition: Dihedral Group] The **dihedral group** of order $2m$ is \begin{align*} D_{2m} = \langle x, y : x^m = y^2 = 1,\ yxy^{-1} = x^{-1} \rangle. \end{align*} It is the full symmetry group of a regular $m$-gon: the elements $x^i$ are the $m$ rotations, and $x^i y$ are the $m$ reflections. For example, in $D_8$ (the symmetry group of the square), $x$ is rotation by $\pi/2$ and $y$ is any one of the four reflections. $D_{2m}$ embeds in $\operatorname{GL}_2(\mathbb{R})$ via its action on the plane, and also embeds in $S_m$ via its action on the $m$ vertices of the polygon. [/definition]  The dihedral group is the first non-abelian example with a geometrically transparent structure: there is a normal cyclic subgroup $\langle x \rangle \cong C_m$ of rotations with quotient $\{e, y\} \cong C_2$ of reflections. This semidirect product structure $D_{2m} \cong C_m \rtimes C_2$ shapes its representation theory. [definition: Quaternion Group] The **quaternion group** is \begin{align*} Q_8 = \langle x, y : x^4 = 1,\ y^2 = x^2,\ yxy^{-1} = x^{-1} \rangle. \end{align*} Setting $i = x$, $j = y$, $k = ij$, and $-1 = i^2$, we have $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$, a group of order $8$. It embeds in $\operatorname{GL}_2(\mathbb{C})$ via \begin{align*} 1 &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, & i &= \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, & j &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, & k &= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}. \end{align*} Direct matrix multiplication gives $i^2 = -I$, $j^2 = -I$, $ij = k$, and $ji = -k$, so the matrices satisfy the quaternion relations. [/definition]  [remark: D_8 vs Q_8] The groups $D_8$ and $Q_8$ are both non-abelian groups of order $8$, and they have identical orders of elements, yet they are not isomorphic. The key distinction is structural: $D_8$ has five subgroups of order $2$ (generated by its five elements of order $2$: the four reflections and the rotation by $\pi$), while $Q_8$ has only one subgroup of order $2$ (generated by $-1$, the unique element of order $2$). Their representation theories will turn out to look quite similar in some respects and differ in others — they provide a good test case later in the course. [/remark] ## Conjugacy Classes and Centralizers When we define characters in the next chapter, we will encounter functions on a group $G$ that are constant on conjugacy classes. Why should conjugacy classes be the natural domain? The reason is that representation matrices encode group elements, and two elements in the same conjugacy class — $g$ and $hgh^{-1}$ — are related by a change of basis (conjugating by the matrix for $h$), so they produce matrices with the same trace. Understanding the conjugacy class structure of a group is therefore the first step in computing its character table. [definition: Conjugacy Class] Let $G$ be a group and $g \in G$. The **conjugacy class** of $g$ is \begin{align*} \mathcal{C}_G(g) = \{ xgx^{-1} : x \in G \}. \end{align*} [/definition] An element forms a singleton conjugacy class $\{g\}$ if and only if $xg = gx$ for all $x \in G$, i.e.\ $g$ is in the centre $Z(G)$. The larger the centralizer of $g$, the smaller its conjugacy class. [definition: Centralizer] The **centralizer** of $g \in G$ is \begin{align*} C_G(g) = \{ x \in G : xg = gx \}. \end{align*} This is a subgroup of $G$. [/definition] The orbit-stabilizer theorem, applied to the conjugation action of $G$ on itself, gives the key size formula. [quotetheorem:2404]This formula is a direct consequence of the orbit-stabilizer theorem applied to the conjugation action; the stabilizer of $g$ under conjugation is exactly $C_G(g)$. The formula has an immediate corollary: the number of conjugacy classes of $G$ equals the number of irreducible representations of $G$ over $\mathbb{C}$ — a deep fact we will prove later. [example: Conjugacy Classes in S_3] In $S_3 = \{e, (12), (13), (23), (123), (132)\}$, the conjugacy classes are determined by cycle type. The class of the identity is $\{e\}$ (size 1). The three transpositions $\{(12), (13), (23)\}$ form one class (size 3, and $|C_{S_3}((12))| = 2$, generated by $(12)$ itself and $e$). The two 3-cycles $\{(123), (132)\}$ form another class (size 2, and $|C_{S_3}((123))| = 3$, generated by $(123)$). Checking: $1 + 3 + 2 = 6 = |S_3|$. The three conjugacy classes predict that $S_3$ has exactly three irreducible complex representations. [/example] ## Group Actions and Permutation Representations Before a group can act on a vector space linearly, it must act on a set. How does a group encode a symmetry of a combinatorial object? The answer is through a group action, a notion that subsumes all the concrete examples seen so far — rotations of polygons, permutations of letters, conjugation within a group — under a single definition. The linear representations we study throughout the course are then a special case where the set happens to be a vector space and the symmetries happen to be linear. [definition: Group Action] Let $G$ be a group and $X$ a set. We say $G$ **acts on** $X$ if there is a map $G \times X \to X$, written $(g, x) \mapsto gx$, satisfying: 1. $1 \cdot x = x$ for all $x \in X$, 2. $g(hx) = (gh)x$ for all $g, h \in G$ and $x \in X$. [/definition] An action encodes a coherent family of symmetries of $X$ parametrised by elements of $G$. The two axioms say precisely that the identity of $G$ acts as the identity permutation of $X$, and that the group multiplication is respected. Note that a map $G \times X \to X$ satisfying only one of the two axioms does not give a group action: without axiom 1, the identity could act non-trivially; without axiom 2, the group structure of $G$ is not faithfully reflected in its action on $X$. Every group action gives rise to a homomorphism into the symmetric group of $X$. [quotetheorem:2405] [citeproof:2405] The homomorphism $\theta: G \to \operatorname{Sym}(X)$ may or may not be injective. It is injective — the action is **faithful** — precisely when no non-identity element of $G$ fixes every point of $X$, i.e.\ $\ker \theta = \{1\}$. Non-faithful actions arise naturally: the conjugation action of an abelian group $G$ on itself collapses to the identity permutation for every group element, since $gxg^{-1} = x$ for all $g, x$, so the kernel is all of $G$. A special case of faithful actions is Cayley's theorem: $G$ acts faithfully on itself by left multiplication ($g \cdot x = gx$), giving an embedding $G \hookrightarrow S_{|G|}$. [definition: Permutation Representation] The **permutation representation** of a group action of $G$ on $X$ is the homomorphism $\theta: G \to \operatorname{Sym}(X)$ constructed above. [/definition] Permutation representations are the bridge between abstract group actions and concrete linear algebra. Given a finite set $X$ with $|X| = n$, we can form the vector space $\mathbb{F}^X$ with basis indexed by elements of $X$; a permutation action of $G$ on $X$ then induces a linear action on $\mathbb{F}^X$ by permuting the basis vectors. This construction — the **permutation module** — will be revisited when we study induced representations. [example: Conjugation Action] Any group $G$ acts on itself by conjugation: $g \cdot x = gxg^{-1}$ for $g, x \in G$. The axioms are easily verified: $1 \cdot x = x$ and $g(hxh^{-1})g^{-1} = (gh)x(gh)^{-1}$. The orbits of this action are exactly the conjugacy classes $\mathcal{C}_G(x)$, and the stabiliser of $x$ is the centralizer $C_G(x)$. The orbit-stabilizer theorem then gives $|\mathcal{C}_G(x)| = |G : C_G(x)|$, recovering the formula of the previous section. For a concrete illustration in $S_3$: the element $(12)$ is fixed by itself and $e$ under conjugation (its centralizer has order 2), so its conjugacy class has size $|S_3|/2 = 3$, which matches $\{(12), (13), (23)\}$. [/example] The conjugation action illustrates how a group acts on itself by an internal symmetry. The next example shows the complementary picture: a group acting on an external set of geometric objects, where the action has a transparent visual meaning and produces an explicit embedding into a symmetric group. [example: Dihedral Group Acting on Vertices] The group $D_8$ acts on the four vertices $\{1, 2, 3, 4\}$ of a square (labelled consecutively). The rotation $x$ sends $1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 1$, corresponding to the permutation $(1234) \in S_4$. A reflection $y$ (say, fixing vertices $1$ and $3$) corresponds to the permutation $(24) \in S_4$. This gives an embedding $D_8 \hookrightarrow S_4$: checking that $y \cdot x \cdot y^{-1} = x^{-1}$ in the action, we compute $(24)(1234)(24) = (1432) = (1234)^{-1}$, which confirms the dihedral relation $yxy^{-1} = x^{-1}$. The kernel of this action is trivial (no non-identity symmetry of the square fixes all four vertices), so the action is faithful and gives a genuine embedding. [/example] ## From Set Actions to Linear Actions In this course, $X$ is almost always a finite-dimensional vector space $V$ over $\mathbb{F}$, and we want the action to respect the linear structure. This motivates the passage from permutation representations to linear (i.e.\ genuine) representations. An action of $G$ on a vector space $V$ is **linear** if for all $g \in G$, vectors $v_1, v_2 \in V$, and scalars $\lambda \in \mathbb{F}$: \begin{align*} g(v_1 + v_2) = gv_1 + gv_2, \qquad g(\lambda v_1) = \lambda(gv_1). \end{align*} In other words, each $g$ acts as a linear map on $V$. Since the action axioms require $g$ to have an inverse action (given by $g^{-1}$), each $g$ in fact acts as an invertible linear map — an element of $\operatorname{GL}(V)$. Equivalently, instead of requiring a map $G \to \operatorname{Sym}(V)$, a linear action is precisely a group homomorphism $G \to \operatorname{GL}(V)$. This is the definition of a representation, which is the starting point for the next chapter. Now that we have concrete examples of group actions and permutation representations, we need a formal framework to study all representations systematically. Chapter 2 develops the foundational definitions — representations, homomorphisms, and irreducibility — that let us classify and compare representations abstractly. # 2. Basic definitions Chapter 1 set up the language of group actions and recalled the key facts from linear algebra that will govern everything that follows. Now we begin doing representation theory in earnest. The central idea is to study a group $G$ by letting it act on a vector space — capturing the group's abstract structure in something concrete and computable, namely matrices. This chapter pins down the fundamental definitions: what a representation is, the four equivalent ways to think about it, what it means for two representations to be the same, and the structural notions of subrepresentation, irreducibility, and decomposability that organise the whole subject. Throughout, $G$ denotes a finite group and $\mathbb{F}$ a field, usually $\mathbb{C}$. ## Four Ways to Think About a Representation Abstract groups are hard to compute with directly. Given a finite group $G$, we know its multiplication table in principle, but this tells us little about global structural features — how many distinct-looking pieces the group can act as, or what algebraic invariants distinguish one group from another of the same order. The resolution is to represent the group concretely inside a space of linear maps, where matrix algebra and linear algebra are available. The question becomes: in how many essentially different ways can $G$ act linearly on a vector space? Pinning down what counts as an action and what counts as "essentially different" requires precise definitions. [definition: Representation] Let $V$ be a finite-dimensional vector space over $\mathbb{F}$. A **(linear) representation** of $G$ on $V$ is a group homomorphism \begin{align*} \rho = \rho_V : G \to \operatorname{GL}(V). \end{align*} We write $\rho_g$ for $\rho_V(g)$, so that $\rho_g \in \operatorname{GL}(V)$ for each $g \in G$, and the homomorphism condition reads $\rho_g \rho_h = \rho_{gh}$ and $\rho_{g^{-1}} = (\rho_g)^{-1}$ for all $g, h \in G$. [/definition] [definition: Dimension of a Representation] The **dimension** (or **degree**) of a representation $\rho : G \to \operatorname{GL}(V)$ is $\dim_{\mathbb{F}}(V)$. [/definition] Since $\rho$ is a group homomorphism, $\ker \rho \unlhd G$ and $G / \ker \rho \cong \rho(G) \leq \operatorname{GL}(V)$. The kernel measures how much of $G$ acts as the identity transformation, and whether any information about $G$ is lost. [definition: Faithful Representation] A representation $\rho$ is **faithful** if $\ker \rho = \{1\}$. [/definition] Faithful representations are those where the only element acting as the identity transformation is the identity element of $G$ — the group embeds into $\operatorname{GL}(V)$ without loss of information. Non-faithful representations factor through the quotient $G / \ker\rho$: they only see the part of $G$ that acts non-identically. The homomorphism definition is clean, but three equivalent reformulations are each useful in different contexts. The first says that a representation is the same as a linear action. [definition: Linear Action] A group $G$ **acts linearly** on a vector space $V$ if it acts on $V$ such that \begin{align*} g(v_1 + v_2) &= gv_1 + gv_2, \qquad g(\lambda v_1) = \lambda(gv_1) \end{align*} for all $g \in G$, $v_1, v_2 \in V$, and $\lambda \in \mathbb{F}$. [/definition] The equivalence is immediate: given a linear action, the assignment $g \mapsto (v \mapsto gv)$ is a group homomorphism $G \to \operatorname{GL}(V)$; conversely, given $\rho$, set $gv := \rho(g)v$. When $V$ carries a linear action of $G$, we call it a **$G$-space** or **$G$-module**. The second reformulation makes the module language of algebra available. [definition: Group Algebra] The **group algebra** $\mathbb{F}G$ is the set of formal $\mathbb{F}$-linear combinations of elements of $G$, \begin{align*} \mathbb{F}G = \left\{ \sum_{g \in G} \alpha_g\, g : \alpha_g \in \mathbb{F} \right\}, \end{align*} with addition defined componentwise and multiplication defined by extending the group multiplication $\mathbb{F}$-bilinearly. [/definition] With this ring structure, a $G$-space is precisely an $\mathbb{F}G$-module in the sense of the IB Groups, Rings and Modules course. The group algebra will return prominently later when we study the regular representation and the decomposition of $\mathbb{F}G$ as a module over itself. The third reformulation grounds everything in matrices. [definition: Matrix Representation] A **matrix representation** of $G$ of degree $n$ is a group homomorphism $R : G \to \operatorname{GL}_n(\mathbb{F})$. [/definition] Since every finite-dimensional vector space over $\mathbb{F}$ is isomorphic to $\mathbb{F}^n$ for some $n$, every representation is equivalent to a matrix representation: given $\rho : G \to \operatorname{GL}(V)$ with $\dim V = n$ and a choice of basis $\mathcal{B}$, the assignment $g \mapsto [\rho(g)]_{\mathcal{B}}$ is a matrix representation. Conversely, $R : G \to \operatorname{GL}_n(\mathbb{F})$ gives a representation on $\mathbb{F}^n$ via $\rho_g(v) = R(g)v$. In summary: representations, linear actions, $\mathbb{F}G$-modules, and matrix representations are four faces of the same object. ## First Examples Before classifying representations, it helps to have concrete examples in hand — both degenerate cases that appear everywhere and richer examples that demonstrate the range of behaviour. The first example might look too simple to matter, but its role in character theory is outsized. [example: Trivial Representation] For any group $G$, take $V = \mathbb{F}$ (the one-dimensional space) and define $\rho : G \to \operatorname{GL}(\mathbb{F})$ by $\rho(g) = \mathrm{id}$ for all $g \in G$. This is the **trivial representation**, of degree $1$. Despite the name, it is far from unimportant. The way the trivial representation interacts with other representations — via the space of $G$-invariant homomorphisms — is central to much of character theory. [/example] The trivial representation is the unique degree-$1$ representation in which every group element acts identically; the next example shows that degree-$1$ representations over $\mathbb{C}$ can be far more varied for a cyclic group. [example: Cyclic Group $C_4$] Let $G = C_4 = \langle x : x^4 = 1 \rangle$ and work over $\mathbb{F} = \mathbb{C}$ with $n = 2$. A matrix representation $R$ is determined by the image of the generator $x$: set $R(x) = A$ and extend by $R(x^j) = A^j$. The only constraint is that $A^4 = I_2$. If $A$ is diagonal, each diagonal entry must be a fourth root of unity, i.e.\ an element of $\{1, -1, i, -i\}$, giving $4^2 = 16$ choices. If $A$ is not diagonal, it is conjugate to a diagonal matrix (since $A^4 = I_2$ forces $A$ to satisfy the polynomial $x^4 - 1 = \prod_{j=0}^3 (x - i^j)$, which has distinct roots over $\mathbb{C}$, so $A$ is diagonalizable). Thus every two-dimensional complex representation of $C_4$ is equivalent to a diagonal one. The question of when two such diagonal representations are equivalent requires the notion of isomorphism, developed next. [/example] ## Isomorphism of Representations Two representations may look superficially different — because they use different bases or different vector spaces — yet capture the same structure. For example, a rotation group acting on $\mathbb{R}^2$ versus the same group acting on $\mathbb{C}$ via complex multiplication can carry identical information. What we want is a way to identify representations that differ only by a change of coordinates. The correct notion of sameness is a linear map that commutes with every group element — a condition called $G$-equivariance or intertwining. [definition: $G$-Homomorphism] Fix $G$ and $\mathbb{F}$. Let $\rho : G \to \operatorname{GL}(V)$ and $\rho' : G \to \operatorname{GL}(V')$ be representations. A linear map $\varphi : V \to V'$ is a **$G$-homomorphism** (or is said to **intertwine** $\rho$ and $\rho'$) if \begin{align*} \varphi \circ \rho(g) = \rho'(g) \circ \varphi \end{align*} for all $g \in G$. Equivalently, the diagram \begin{align*} &V \xrightarrow{\,\rho_g\,} V \\ &\downarrow\varphi \qquad\quad \downarrow\varphi \\ &V' \xrightarrow{\,\rho'_g\,} V' \end{align*} commutes for every $g$. The $\mathbb{F}$-vector space of all $G$-homomorphisms from $V$ to $V'$ is written $\operatorname{Hom}_G(V, V')$. [/definition]  [definition: Equivalent Representations] A $G$-homomorphism $\varphi : V \to V'$ is a **$G$-isomorphism** if $\varphi$ is bijective. Two representations $\rho$ and $\rho'$ are **equivalent** (or **isomorphic**) if there exists a $G$-isomorphism between them. [/definition] When $\varphi$ is a $G$-isomorphism, the intertwining condition reads $\rho'(g) = \varphi \circ \rho(g) \circ \varphi^{-1}$ for all $g$. [quotetheorem:2406]The verification is a routine unfolding of definitions. Reflexivity holds because the identity map $\mathrm{id}_V : V \to V$ satisfies $\mathrm{id}_V \circ \rho(g) = \rho(g) \circ \mathrm{id}_V$ for all $g$. For symmetry, if $\varphi : V \to V'$ is a $G$-isomorphism then $\varphi^{-1} : V' \to V$ satisfies $\varphi^{-1} \circ \rho'(g) = \rho(g) \circ \varphi^{-1}$, which follows by composing $\varphi^{-1}$ on both sides of the intertwining condition for $\varphi$. Transitivity follows because the composition of two $G$-isomorphisms is again a $G$-isomorphism. The importance of this result is not the verification itself but its consequence: we can speak of **isomorphism classes** of representations and classify representations up to isomorphism, which is the central programme of the subject. [quotetheorem:2407] [citeproof:2407] The converse fails: having the same dimension does not force isomorphism, as the $C_4$ example below shows. This theorem is fundamental to the classification programme — it tells us that dimension is an isomorphism invariant, so the irreducible representations of $G$ over $\mathbb{C}$ can be sorted into finitely many classes by dimension. In character theory, we will find a far richer invariant (the character) that completely distinguishes irreducible representations, but dimension is the first and coarsest such invariant. Note also that finite-dimensionality is essential: over infinite-dimensional spaces, isomorphisms need not preserve Hamel-basis cardinality in any simple sense, and the theory diverges substantially. [example: Four Non-Isomorphic One-Dimensional Representations of $C_4$] Let $G = C_4 = \langle x \rangle$ and $\omega = e^{2\pi i / 4} = i$. For each $j \in \{0, 1, 2, 3\}$, define a one-dimensional representation $\rho_j : G \to \operatorname{GL}(\mathbb{C})$ by \begin{align*} \rho_j(x^k) = \omega^{jk} = i^{jk}. \end{align*} This is a well-defined homomorphism since $\rho_j(x)^4 = (i^j)^4 = 1$. For $j \neq j'$, the representations $\rho_j$ and $\rho_{j'}$ are not isomorphic: a $G$-isomorphism would be a nonzero scalar $\lambda \in \mathbb{C}^\times$ with $\lambda \cdot i^{jk} = i^{j'k} \cdot \lambda$ for all $k$, which requires $i^{jk} = i^{j'k}$ for all $k$. Taking $k = 1$, this fails whenever $j \neq j'$ (mod $4$). So $C_4$ has at least four non-isomorphic one-dimensional representations over $\mathbb{C}$, all of the same dimension. [/example] ### Matrix Formulation of Isomorphism In terms of matrix representations, $R : G \to \operatorname{GL}_n(\mathbb{F})$ and $R' : G \to \operatorname{GL}_n(\mathbb{F})$ are $G$-isomorphic if and only if there exists a non-singular matrix $X \in \operatorname{GL}_n(\mathbb{F})$ such that $R'(g) = X R(g) X^{-1}$ for all $g \in G$. This is exactly the condition that $R$ and $R'$ are simultaneously conjugate by the same change-of-basis matrix. In terms of linear actions, $G$ acting on $V$ and $G$ acting on $V'$ are $G$-isomorphic if there is a linear isomorphism $\varphi : V \to V'$ such that $\varphi(gv) = g\varphi(v)$ for all $g \in G$, $v \in V$. Every representation is isomorphic to a matrix representation: choosing any basis $\mathcal{B}$ of $V$ gives a $G$-isomorphism $\varphi : V \to \mathbb{F}^n$ by $v \mapsto [v]_{\mathcal{B}}$. ## Subrepresentations and Irreducibility Knowing what representations are, the natural next question is: can they be broken into smaller pieces? This matters because a representation on a large space is hard to analyse directly, but if it decomposes into a direct sum of smaller representations, we can study each piece independently. The first step is identifying when a subspace is preserved by the group action — an arbitrary subspace need not be invariant, and this is the whole difficulty. [definition: $G$-Subspace] Let $\rho : G \to \operatorname{GL}(V)$ be a representation. A subspace $W \leq V$ is a **$G$-subspace** if it is $\rho(G)$-invariant: $\rho_g(W) \leq W$ for all $g \in G$. [/definition] The condition that $W$ is $G$-invariant is non-automatic: a randomly chosen subspace of $V$ will generally not be preserved by all $\rho_g$. For example, if $G = \mathbb{Z}/2\mathbb{Z}$ acts on $\mathbb{R}^2$ by the reflection $\rho_{-1}(x, y) = (-x, y)$, then the subspace $\{(t, t) : t \in \mathbb{R}\}$ (the diagonal) is not $G$-invariant. The $x$-axis $\{(t, 0)\}$ and the $y$-axis $\{(0, t)\}$ are $G$-invariant, and they give the block-diagonal decomposition of this representation. The two obvious $G$-subspaces of any representation $\rho$ on $V$ are $\{0\}$ and $V$ itself. When $W$ is a $G$-subspace, the restriction $g \mapsto \rho(g)|_W$ is a well-defined representation of $G$ on $W$. [definition: Subrepresentation] If $W$ is a $G$-subspace of $V$, the representation $G \to \operatorname{GL}(W)$ given by $g \mapsto \rho(g)|_W$ is called a **subrepresentation** of $\rho$. [/definition] In the matrix picture, a $G$-subspace $W$ of dimension $m$ in $V$ of dimension $n$ corresponds to a block upper-triangular structure: if we extend a basis $\mathcal{B}_1 = \{v_1, \ldots, v_m\}$ of $W$ to a basis $\mathcal{B} = \{v_1, \ldots, v_n\}$ of $V$, then \begin{align*} [\rho(g)]_{\mathcal{B}} = \begin{pmatrix} * & * \\ 0 & * \end{pmatrix} \end{align*} for each $g \in G$, where the upper-left block represents $\rho(g)|_W$. The off-diagonal block records how the complement of $W$ interacts with $W$; in the block-diagonal case (the decomposable case), this block is also zero. The representations of primary interest are those that cannot be broken down further. [definition: Irreducible Representation] A representation $\rho$ is **irreducible** (or **simple**) if there are no proper non-zero $G$-subspaces of $V$ — that is, the only $G$-invariant subspaces are $\{0\}$ and $V$ itself. [/definition] Every one-dimensional representation is irreducible, since the only subspaces of a one-dimensional space are $\{0\}$ and the space itself. The converse fails in both directions: there exist irreducible representations of dimension greater than one, as the dihedral group example below shows, and there exist reducible representations that are not merely direct sums of their pieces (the indecomposable-but-reducible phenomenon discussed in the next section). [example: Irreducible Complex Representations of $D_6$ Have Dimension at Most 2] Let $G = D_6$, generated by a rotation $r$ (of order $3$) and a reflection $s$ (of order $2$) with $srs^{-1} = r^{-1}$. We claim every irreducible complex representation has dimension at most $2$. Let $\rho : G \to \operatorname{GL}(V)$ be an irreducible representation. Since $V$ is a finite-dimensional complex vector space, $\rho(r)$ has at least one eigenvector. Pick an eigenvector $v \in V$ with $\rho(r)v = \lambda v$ for some $\lambda \neq 0$ (invertibility of $\rho(r)$ rules out $\lambda = 0$). Consider the subspace \begin{align*} W = \mathrm{span}\{v,\, \rho(s)v\} \leq V. \end{align*} We verify that $W$ is $G$-invariant by checking it is stable under $\rho(r)$ and $\rho(s)$: \begin{align*} \rho(r)v &= \lambdav \in W, \\ \rho(r)\rho(s)v &= \rho(s)\rho(r^{-1})v = \rho(s)(\lambda^{-1}v) = \lambda^{-1}\rho(s)v \in W,\\ \rho(s)v &\in W, \\ \rho(s)\rho(s)v &= \rho(e)v = v \in W. \end{align*} (For the second line, we used the relation $\rho(r)\rho(s) = \rho(s)\rho(r^{-1})$, which follows from $srs^{-1} = r^{-1}$.) Since $W$ is a non-zero $G$-invariant subspace and $\rho$ is irreducible, $W = V$. Hence $\dim V \leq 2$. [/example] ## Decomposability and Direct Sums Given that a representation may have invariant subspaces, the central structural question is: can an invariant subspace always be complemented by another invariant subspace? If $W \leq V$ is $G$-invariant, is there always another $G$-invariant $W' \leq V$ with $V = W \oplus W'$? The answer, perhaps surprisingly, is not always — and the obstruction is the characteristic of the field. Over $\mathbb{C}$, or any field of characteristic zero, such a complement always exists (Maschke's theorem). Over a field of prime characteristic $p$ dividing $|G|$, it can fail, and the theory of indecomposable-but-reducible representations is genuinely more complex. Understanding when decomposability holds requires first making the notion precise. [definition: Decomposable and Indecomposable Representation] A representation $\rho : G \to \operatorname{GL}(V)$ is **decomposable** if there exist proper $G$-invariant subspaces $U, W \leq V$ with $V = U \oplus W$; in this case we write $\rho \cong \rho_U \oplus \rho_W$. If no such decomposition exists, $\rho$ is **indecomposable**. [/definition] Irreducibility implies indecomposability — an irreducible representation has no proper non-zero $G$-subspaces, so it cannot be written as a direct sum of two proper pieces. The converse need not hold in general. A key example is the action of $\mathbb{Z}$ on $\mathbb{F}_p^2$ (where $p$ is prime) given by $n \mapsto \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$. The subspace $W = \mathrm{span}\{e_1\}$ is $\mathbb{Z}$-invariant since $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} e_1 = e_1$, making the representation reducible. But there is no complementary invariant subspace: any candidate complement would be $\mathrm{span}\{e_1 + te_2\}$ for some $t$, and $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}(e_1 + te_2) = e_1 + te_2 + ne_2 = e_1 + (t+n)e_2$, which lies in $\mathrm{span}\{e_1 + te_2\}$ only if $n \equiv 0 \pmod{p}$ for all $n$ — impossible. So this representation is reducible but indecomposable. Over a field of characteristic zero and for finite groups, such phenomena cannot arise, as Maschke's theorem shows. In the matrix picture, a decomposable representation with $V = U \oplus W$ (where $\dim U = k$ and $\dim W = \ell$) looks, with respect to the concatenated basis $\mathcal{B}_1 \cup \mathcal{B}_2$, like a block diagonal matrix: \begin{align*} [\rho(g)]_{\mathcal{B}} = \begin{pmatrix} [\rho_U(g)]_{\mathcal{B}_1} & 0 \\ 0 & [\rho_W(g)]_{\mathcal{B}_2} \end{pmatrix}. \end{align*} This is the desirable situation: the two sub-representations act completely independently on their respective summands. The reverse operation — combining two representations into one — is the direct sum. [definition: Direct Sum of Representations] Let $\rho : G \to \operatorname{GL}(V)$ and $\rho' : G \to \operatorname{GL}(V')$ be representations. Their **direct sum** is the representation \begin{align*} \rho \oplus \rho' : G \to \operatorname{GL}(V \oplus V') \end{align*} defined by $(\rho \oplus \rho')(g)(v + v') = \rho(g)v + \rho'(g)v'$ for $v \in V$, $v' \in V'$. [/definition] For matrix representations $R : G \to \operatorname{GL}_n(\mathbb{F})$ and $R' : G \to \operatorname{GL}_{n'}(\mathbb{F})$, the direct sum $R \oplus R' : G \to \operatorname{GL}_{n+n'}(\mathbb{F})$ is given by \begin{align*} (R \oplus R')(g) = \begin{pmatrix} R(g) & 0 \\ 0 & R'(g) \end{pmatrix}. \end{align*} [remark: Tensor Products] Just as we can add two representations via the direct sum, we can multiply them — this is the tensor product of representations, which produces the product of their characters. However, defining this properly requires first setting up the tensor product of vector spaces. This will be done in a later chapter. [/remark] The central structural question of representation theory is whether every representation decomposes as a direct sum of irreducibles. The next chapter answers this in the affirmative for finite groups in characteristic zero, via Maschke's theorem. With the basic definitions in place, we can ask: do all representations decompose into irreducibles? Chapter 3 proves the crucial answer via Maschke's theorem, showing that averaging over the group gives a complete reducibility criterion, and explores the regular representation as a canonical example. # 3. Complete reducibility and Maschke's theorem A central goal of representation theory is to understand all representations of a group $G$ by building them from the simplest possible pieces. The chapter preceding this one introduced irreducible representations — those admitting no proper, non-zero $G$-invariant subspace — as the atomic units of the theory. The question now is: can every representation be assembled from these atoms? The answer, for finite groups over fields of characteristic zero, is yes, and the result is Maschke's theorem. This chapter proves that theorem by two distinct methods and draws out its key consequences, including the regular representation and its role as a universal container for all irreducibles. ## Complete Reducibility The obstacle to decomposing an arbitrary representation is real: without conditions on $G$ or the field, a representation can have an invariant subspace with no invariant complement. Before naming the property that rules this out, it is worth knowing that it can fail — the theory genuinely needs both a finiteness condition on $G$ and a characteristic condition on the field. [definition: Completely Reducible Representation] A representation $\rho: G \to \operatorname{GL}(V)$ is **completely reducible** (or **semisimple**) if $V$ decomposes as a direct sum of irreducible $G$-subspaces: \begin{align*} V \cong V_1 \oplus V_2 \oplus \cdots \oplus V_r, \end{align*} where each $V_i$ is an irreducible representation of $G$. [/definition] Every irreducible representation is completely reducible (with $r = 1$), so irreducibility is the special case. The question is whether the converse direction — building irreducibles up to larger representations — always runs backwards: can we always decompose a given representation into irreducibles? The answer is no in general. For instance, the group $\mathbb{Z}$ has representations over $\mathbb{C}$ that are not completely reducible. Likewise, $C_p$ has non-completely-reducible representations over $\mathbb{F}_p$. In both cases the obstruction comes from the characteristic of the field or the infinite order of the group. But for finite groups over characteristic-zero fields, we have the following fundamental theorem. [quotetheorem:2408] [citeproof:2408] The theorem says every representation breaks into irreducible pieces, but it says nothing about whether that decomposition is unique or canonical. In general it is not: the same representation can decompose into irreducibles in different ways. The canonical structure that does exist is the **isotypic decomposition**, in which all copies of each irreducible are grouped together — that finer picture will emerge from character theory. Two further boundary conditions are worth noting. If $G$ is infinite, completely reducible representations exist but need not be the rule: the additive group $\mathbb{Z}$ acting on $\mathbb{C}^2$ by $n \mapsto \bigl(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\bigr)$ is indecomposable but not irreducible. If $\operatorname{char} \mathbb{F} \mid |G|$, Maschke's averaging argument breaks down (division by $|G|$ is not possible), and the theory of modular representations that results is fundamentally different. ## Maschke's Theorem and the Averaging Trick The key difficulty is that a vector-space complement to $W$ inside $V$ always exists by linear algebra, but need not be $G$-invariant. The averaging trick converts an arbitrary complement into a $G$-invariant one by exploiting the finiteness of $G$. [quotetheorem:2409][remark: Characteristic Condition] The hypothesis $\operatorname{char} \mathbb{F} = 0$ can be relaxed: the theorem holds whenever $\operatorname{char} \mathbb{F} \nmid |G|$. The characteristic-zero case covers this since no prime divides every positive integer. The proof breaks down exactly when $|G| \cdot 1_{\mathbb{F}} = 0$ in $\mathbb{F}$, because the averaging factor $\frac{1}{|G|}$ becomes undefined. [/remark] [citeproof:2409] [explanation: Why Averaging Works] The operator $\bar{q}$ is an instance of a general principle: given any linear-algebraic object that lacks a symmetry, one can sometimes force the symmetry by averaging over the group. The map $v \mapsto g \cdot q(g^{-1} v)$ is what $q$ would look like if we "viewed it from the perspective of $g$". Summing over all $g \in G$ and dividing by $|G|$ produces the average of all these conjugated projections, which by construction commutes with every group element. The same averaging idea appears in harmonic analysis (replacing the finite average by an integral over a compact group) and is the prototype for the construction of invariant measures. [/explanation] ## A Second Proof via Invariant Inner Products The averaging trick can be recast in the language of inner products, giving a second proof that has the advantage of generalising to compact topological groups. This second argument requires $\mathbb{F} = \mathbb{C}$. [definition: Hermitian Inner Product] For $V$ a complex vector space, a map $\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}$ is a **Hermitian inner product** if: 1. $\langle v, w \rangle = \overline{\langle w, v \rangle}$ (conjugate symmetry); 2. $\langle v, \lambda_1 w_1 + \lambda_2 w_2 \rangle = \lambda_1 \langle v, w_1 \rangle + \lambda_2 \langle v, w_2 \rangle$ (sesquilinearity in the second argument); 3. $\langle v, v \rangle > 0$ whenever $v \neq 0$ (positive definiteness). [/definition] A Hermitian inner product on $V$ alone gives no relation between the geometry of $V$ and the action of $G$. To exploit orthogonality in a representation-theoretic argument, we need an inner product compatible with the group action — one that the group preserves. [definition: G-Invariant Inner Product] A Hermitian inner product $\langle \cdot, \cdot \rangle$ on a $G$-space $V$ is **$G$-invariant** if \begin{align*} \langle gv,\, gw \rangle = \langle v, w \rangle \end{align*} for all $g \in G$ and $v, w \in V$. [/definition] If a $G$-invariant inner product exists, Maschke's theorem follows immediately: for any $G$-subspace $W \leq V$, its orthogonal complement $W^\perp$ is also $G$-invariant, and $V = W \oplus W^\perp$ is the desired $G$-invariant decomposition. [quotetheorem:2410] [citeproof:2410] The force of this argument depends entirely on having a $G$-invariant inner product in the first place. Without $G$-invariance, orthogonal complements are not $G$-stable: a group element could rotate a vector out of $W^\perp$ into $W$ itself. The averaging trick in Weyl's theorem below is what supplies this invariance, and it is the only step that uses the finiteness of $G$ (or, for compact groups, the existence of Haar measure). It remains to produce a $G$-invariant inner product. This is achieved by Weyl's unitary trick. [quotetheorem:2411] [citeproof:2411] [remark: Generalisation to Compact Groups] The averaged inner product in Weyl's trick depends on replacing the finite sum $\frac{1}{|G|}\sum_{g \in G}$ by an integral against a normalised Haar measure $\int_G dg$, where the measure satisfies $\int_G f(hg)\,dg = \int_G f(g)\,dg$ (left-invariance). Such a measure exists for any compact Hausdorff group. This is why Maschke's theorem and complete reducibility extend to compact groups: $S^1$, $\operatorname{SU}(2)$, and $\operatorname{SO}(3)$ all admit $G$-invariant inner products by exactly this averaging procedure. The first proof (via $\bar{q}$) does not generalise as readily, since the correction map $\bar{q}$ is less naturally expressed as an integral. [/remark] A neat structural corollary of the unitary trick concerns how finite groups sit inside the general linear group. [quotetheorem:2412] [citeproof:2412] This has a concrete consequence: every element of a finite subgroup of $\GL_n(\mathbb{C})$, being conjugate into $\operatorname{U}(n)$, is a matrix with finite order whose eigenvalues all lie on the unit circle. Since unitary matrices are normal, the spectral theorem applies, and every such element is diagonalisable over $\mathbb{C}$ with eigenvalues that are roots of unity. In particular, no finite subgroup of $\GL_n(\mathbb{C})$ contains a non-identity unipotent element (a matrix with all eigenvalues equal to $1$ but not equal to the identity), a fact that has no elementary direct proof. ## The Regular Representation We have seen that every irreducible representation is an "atom". The regular representation provides a canonical ambient space that contains a copy of every irreducible. [definition: Regular Representation] Let $G$ be a finite group and $\mathbb{F}$ a field. The **group algebra** $\mathbb{F}G$ is the $\mathbb{F}$-vector space with basis $\{e_g : g \in G\}$, so elements are formal linear combinations $\sum_{g \in G} a_g e_g$ with $a_g \in \mathbb{F}$. The group $G$ acts on $\mathbb{F}G$ by left multiplication: \begin{align*} h \cdot \sum_{g \in G} a_g e_g = \sum_{g \in G} a_g e_{hg}. \end{align*} This defines a representation $\rho_{\mathrm{reg}}: G \to \operatorname{GL}(\mathbb{F}G)$ called the **regular representation**. The space $\mathbb{F}G$ is the **regular module**. [/definition] The regular representation has dimension $|G|$ and is faithful (the identity is the only element acting as the identity map, which can be verified by checking that $h \cdot e_e = e_h \neq e_e$ for $h \neq e$). More importantly: [quotetheorem:2413] [citeproof:2413] Note that this argument invokes Maschke's theorem to split $\ker\theta$ off from $\mathbb{F}G$; the theorem fails in characteristic $p$ when $p \mid |G|$, and indeed there exist irreducible representations in the modular setting that do not embed in the regular module in the same clean way. The embedding is also not unique: the choice of $v \in V$ determines the particular copy of $V$ inside $\mathbb{F}G$, and different choices yield different (though isomorphic) subrepresentations. What the theorem does NOT say is that $\mathbb{F}G$ decomposes as a direct sum of distinct irreducibles; in fact each irreducible $V$ appears with multiplicity $\dim V$, as character theory will show. This result foreshadows a stronger statement from character theory: the number of times an irreducible $V$ appears in the regular representation equals $\dim V$. Together, these results paint a picture of $\mathbb{F}G$ as the canonical direct sum of all irreducibles, each occurring with multiplicity equal to its dimension. ## Permutation Representations The regular representation is a special case of a broader construction. [definition: Permutation Representation] Let $\mathbb{F}$ be a field and let $G$ act on a finite set $X$. The **permutation representation** associated to this action is the representation $\rho: G \to \operatorname{GL}(\mathbb{F}X)$, where $\mathbb{F}X = \langle e_x : x \in X \rangle$ is the $\mathbb{F}$-vector space with basis indexed by $X$, and $G$ acts by \begin{align*} g \cdot \sum_{x \in X} a_x e_x = \sum_{x \in X} a_x e_{gx}. \end{align*} [/definition] The regular representation is the permutation representation for the action of $G$ on itself by left multiplication ($X = G$, $g \cdot h = gh$). [example: Permutation Representation of $S_3$] Let $G = S_3$ act on $X = \{1, 2, 3\}$ in the natural way. The permutation representation has $\mathbb{F}X = \mathbb{F}e_1 \oplus \mathbb{F}e_2 \oplus \mathbb{F}e_3$, a three-dimensional space. The transposition $\tau = (12) \in S_3$ acts by \begin{align*} \tau \cdot e_1 = e_2, \quad \tau \cdot e_2 = e_1, \quad \tau \cdot e_3 = e_3. \end{align*} The subspace $W = \langle e_1 + e_2 + e_3 \rangle$ is $G$-invariant (every $g \in S_3$ permutes the basis vectors so their sum is fixed), giving a copy of the trivial representation. Maschke's theorem guarantees a $G$-invariant complement; one verifiable choice is $U = \{a_1e_1 + a_2e_2 + a_3e_3 : a_1 + a_2 + a_3 = 0\}$, the "zero-sum" hyperplane. Since $g$ permutes the $a_i$'s, any $g \in S_3$ maps an element of $U$ to another element of $U$, so $U$ is $G$-invariant and $\mathbb{F}X = W \oplus U$. [/example] [remark: Decomposing Permutation Representations] Permutation representations are not always irreducible. The subspace $W$ spanned by $\sum_{x \in X} e_x$ is always $G$-invariant, so a permutation representation on more than one element is always reducible. Its further decomposition into irreducibles depends on the fine structure of the $G$-action on $X$ and is most naturally studied using characters — a topic taken up in the next chapter. [/remark] Maschke's theorem tells us representations decompose, but how do irreducible pieces relate to each other? Chapter 4 uses Schur's lemma to show that $G$-homomorphisms between irreducibles must be scalar multiples of the identity, leading to a clean isotypical decomposition and the fact that abelian groups have only one-dimensional irreducibles. # 4. Schur's lemma Schur's lemma is a short result — its proof fits in two lines — yet it is one of the most powerful tools in representation theory. The preceding chapters established that every finite-dimensional representation of a finite group over a characteristic-zero field decomposes as a direct sum of irreducibles. Schur's lemma tells us what maps can exist between irreducible representations, and over an algebraically closed field this answer is as sharp as possible: the only $G$-endomorphisms of an irreducible are scalars. From this single observation, deep structural results follow: the irreducible representations of abelian groups are one-dimensional, the center of a group admitting a faithful irreducible is cyclic, and every representation has a canonical decomposition into isotypical components whose multiplicities are determined by $\operatorname{Hom}$ spaces. ## Schur's Lemma and Its Immediate Consequences What kinds of $G$-linear maps can exist between irreducible representations? And when the domain and codomain coincide, what endomorphisms does an irreducible admit? These questions have surprisingly sharp answers — over an algebraically closed field, the only possibility is either zero or a scalar multiple of the identity. [quotetheorem:2414] [citeproof:2414] Both hypotheses in part (2) are necessary. Without algebraic closure, $\theta$ need not have any eigenvalue in $\mathbb{F}$ at all — the argument collapses at the first step. Without irreducibility, $\theta - \lambda \iota_V$ being non-invertible no longer forces it to be zero, since the only conclusion from part (1) applies specifically when $V$ is irreducible. Over an algebraically closed field, the lemma forces rigid structure on $\operatorname{Hom}$-spaces: it is the foundation on which character orthogonality and the column-orthogonality relations of character tables are built, as we will see once character theory is developed. Recall that $\operatorname{Hom}_G(V, W)$ denotes the $\mathbb{F}$-vector space of all $G$-homomorphisms $V \to W$, and $\operatorname{End}_G(V) = \operatorname{Hom}_G(V, V)$ denotes the $G$-endomorphism algebra of $V$. [quotetheorem:2415] [citeproof:2415] [remark: Algebraic Closure is Essential] Part (2) of Schur's lemma, and the corollary above, genuinely require algebraic closure. Over $\mathbb{R}$, the cyclic group $C_3$ has an irreducible two-dimensional real representation (rotation by $2\pi/3$), and its endomorphism algebra is isomorphic to $\mathbb{C}$, not $\mathbb{R}$. Over $\mathbb{C}$, all irreducible endomorphism algebras collapse to scalars. [/remark] With Schur's lemma established, we can extract concrete group-theoretic information directly from representation-theoretic constraints. ## The Center of a Group with a Faithful Irreducible When can a group be faithfully realised on a single irreducible representation? Schur's lemma imposes a striking constraint on the answer: every central element must act as a scalar, which forces the center to be embedded in $\mathbb{C}^\times$, and finite subgroups of $\mathbb{C}^\times$ are always cyclic. [quotetheorem:2416] [citeproof:2416] [remark: The Converse Fails] The converse is false: a group can have a cyclic center without possessing a faithful complex irreducible representation. A counterexample is on Example Sheet 1, Question 10. [/remark] This style of argument — using Schur's lemma to show that every central element acts as a scalar, and then reading off group-theoretic structure from the resulting map — recurs throughout the course. Representations are tractable; groups are hard. Schur's lemma is one of the primary bridges between the two. ## Irreducible Representations of Abelian Groups How restrictive does commutativity become when every pair of group elements commutes? For complex representations, commutativity turns every group element into a $G$-endomorphism, and Schur's lemma then forces each to act as a scalar — collapsing every irreducible to one dimension. [quotetheorem:2417] [citeproof:2417] [remark: Failure Over the Reals] This result fails over $\mathbb{R}$. The cyclic group $C_3$ has a two-dimensional irreducible real representation, corresponding to rotation by $2\pi/3$ in the plane. The key obstruction is that $\mathbb{R}$ is not algebraically closed. [/remark] Since every irreducible complex representation of a finite abelian group $G$ has dimension one, such representations are simply group homomorphisms $G \to \mathbb{C}^\times$. [quotetheorem:2418] [citeproof:2418] [remark: No Canonical Bijection] Although $|G|$ equals the number of irreducible representations, there is no natural bijection between the elements of $G$ and the set of irreducibles. Choosing an isomorphism $G \cong C_{n_1} \times \cdots \times C_{n_r}$ identifies the two sets, but this identification depends on the choice of generators — replacing the generator of any $C_{n_j}$ factor gives a different bijection. [/remark] The following two examples make the counting explicit and preview the character table structure we will revisit later. [example: Irreducibles of $C_4$] Let $G = C_4 = \langle x \rangle$. The four roots of unity of order dividing $4$ are $1, i, -1, -i$. Setting $\lambda = \rho(x)$, the four irreducible representations are: | | $1$ | $x$ | $x^2$ | $x^3$ | |---|---|---|---|---| | $\rho_1$ | $1$ | $1$ | $1$ | $1$ | | $\rho_2$ | $1$ | $i$ | $-1$ | $-i$ | | $\rho_3$ | $1$ | $-1$ | $1$ | $-1$ | | $\rho_4$ | $1$ | $-i$ | $-1$ | $i$ | Each row lists the scalar $\lambda^k = \rho_j(x^k)$ for $k = 0,1,2,3$. This is the character table of $C_4$; we will revisit the general theory of character tables when we develop character theory for non-abelian groups. [/example] [example: Irreducibles of the Klein Four Group] Let $G = V_4 = \langle x_1 \rangle \times \langle x_2 \rangle \cong C_2 \times C_2$. Each $\lambda_j$ must satisfy $\lambda_j^2 = 1$, so $\lambda_j \in \{1, -1\}$. The four irreducible representations, determined by $(\lambda_1, \lambda_2)$, are: | | $1$ | $x_1$ | $x_2$ | $x_1 x_2$ | |---|---|---|---|---| | $\rho_1$ | $1$ | $1$ | $1$ | $1$ | | $\rho_2$ | $1$ | $1$ | $-1$ | $-1$ | | $\rho_3$ | $1$ | $-1$ | $1$ | $-1$ | | $\rho_4$ | $1$ | $-1$ | $-1$ | $1$ | Note how $x_1 x_2$ always receives the product of the values at $x_1$ and $x_2$, a direct consequence of the homomorphism property. [/example] ## Isotypical Decompositions Maschke's theorem guarantees that every representation decomposes as a direct sum of irreducibles, but it says nothing about uniqueness. We now address this: while the individual irreducible summands may be rearranged, the multiplicity of each isomorphism class is an invariant of the representation. To appreciate what is at stake, consider an abstract example. An endomorphism $\alpha: V \to V$ of a plain vector space yields the eigenspace decomposition $V = \bigoplus_\lambda V(\lambda)$, which is canonical in that it depends only on $\alpha$. When $V$ carries a $G$-representation, we want an analogous canonical decomposition into irreducible summands. [example: Decomposing a $D_6$ Representation] Consider $G = D_6 \cong S_3 = \langle r, s \mid r^3 = s^2 = 1,\ rs = sr^{-1} \rangle$. The group has three irreducible complex representations: the trivial representation $\mathbf{1}$, the sign representation $S$ (where $r \mapsto 1$ and $s \mapsto -1$), and a two-dimensional representation $W$ defined by \begin{align*} \rho(r) = \begin{pmatrix} \omega & 0 \\ 0 & \omega^2 \end{pmatrix}, \qquad \rho(s) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \end{align*} where $\omega = e^{2\pi i/3}$. Let $(\rho', V)$ be any complex representation. Since $r$ has order $3$, the operator $\rho'(r)$ is diagonalizable (over $\mathbb{C}$, finite-order operators are diagonalizable) with eigenvalues among $\{1, \omega, \omega^2\}$. Writing $V = V(1) \oplus V(\omega) \oplus V(\omega^2)$ for the corresponding eigenspaces, the relation $srs^{-1} = r^{-1}$ forces $\rho'(s)$ to preserve $V(1)$ and to interchange $V(\omega)$ with $V(\omega^2)$. On $V(1)$, the element $r$ acts as the identity, so $V(1)$ is determined entirely by the action of $s$, which is diagonalizable with eigenvalues $\pm 1$. The $+1$ eigenspace is a sum of copies of $\mathbf{1}$ and the $-1$ eigenspace is a sum of copies of $S$. For the remaining part, choose a basis $v_1, \ldots, v_n$ of $V(\omega)$ and set $v_j' = \rho'(s)v_j \in V(\omega^2)$. On the two-dimensional subspace $\langle v_j, v_j' \rangle$, the element $r$ acts as $\begin{pmatrix} \omega & 0 \\ 0 & \omega^2 \end{pmatrix}$ and $s$ acts as $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, exactly the matrices defining $W$. So $V(\omega) \oplus V(\omega^2)$ decomposes into copies of $W$, one for each basis vector of $V(\omega)$. The outcome: every complex representation of $D_6$ decomposes as a direct sum of copies of $\mathbf{1}$, $S$, and $W$. [/example] The $D_6$ argument was tailored to that group. To make the counting of multiplicities systematic, we need a general tool. [quotetheorem:2419] [citeproof:2419] Hom Additivity transforms the abstract problem of counting multiplicities into a dimension calculation: to find out how many times an irreducible $S$ appears in $V$, compute $\dim \operatorname{Hom}_G(S, V)$. [explanation: Why Hom Additivity Counts Multiplicities] The previous lemma has a concrete pay-off: if we know $\dim \operatorname{Hom}_G(S, V)$ for each irreducible $S$, we can read off how many times $S$ appears in $V$. The next lemma makes this precise. [/explanation] The multiplicity formula now follows by induction, combining Hom Additivity with Schur's lemma at the base case. [quotetheorem:2420] [citeproof:2420] The Multiplicity Formula is a clean and powerful result, but it requires algebraic closure in an essential way: it relies on Schur's lemma at the base case, and Schur's lemma fails over non-algebraically-closed fields (as the $C_3$ example over $\mathbb{R}$ showed). The formula also tells us only the count of summands isomorphic to $S$, not how to find them explicitly — for that, one needs an inner product on the space of class functions, which character theory provides via the character inner product and orthogonality relations. Looking forward, the Multiplicity Formula is what makes the decomposition into isotypical components canonical: the integer $n_j = \dim \operatorname{Hom}_G(S_j, V)$ is intrinsically defined, independent of any particular decomposition into irreducibles. This means the multiplicity of any irreducible summand is not a free parameter — it is forced by the representation. The following definition captures the resulting canonical structure. [definition: Isotypical Decomposition] Let $V$ be a completely reducible $G$-representation over an algebraically closed field $\mathbb{F}$, with pairwise non-isomorphic irreducibles $S_1, S_2, \ldots$ Let $W_j$ denote the direct sum of all irreducible summands of $V$ isomorphic to $S_j$ (so $W_j \cong S_j^{\oplus n_j}$ for some $n_j \geq 0$). The decomposition \begin{align*} V = \bigoplus_j W_j \end{align*} is the **canonical decomposition** or **decomposition into isotypical components** of $V$. Each $W_j$ is an **isotypical component**. [/definition] The word "canonical" deserves emphasis: while the isotypical components $W_j$ are uniquely determined subspaces of $V$, the further decomposition of each $W_j$ into individual copies of $S_j$ is generally not. A simple example makes this concrete: take $G = C_2 = \{1, g\}$ acting on $V = \mathbb{C}^2$ via $g \mapsto \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Then $V = V_+ \oplus V_-$ where $V_+ = \operatorname{span}\{e_1\}$ (the trivial summand) and $V_- = \operatorname{span}\{e_2\}$ (the sign summand) are the isotypical components. Now consider the same group acting on $W = \mathbb{C}^2$ via $g \mapsto \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ — the entire space is the sign isotypical component $W = W_-$, but there are infinitely many ways to write $W_- = L_1 \oplus L_2$ as a direct sum of two copies of the sign representation, one for each choice of basis. The isotypical component $W_-$ is canonical; the individual lines $L_1, L_2$ are not. [explanation: Canonicity of the Decomposition] The multiplicity formula gives $n_j = \dim \operatorname{Hom}_G(S_j, V)$, which depends only on $V$ and $S_j$, not on any particular decomposition into irreducibles. Therefore the isotypical components $W_j$ are intrinsically defined subspaces of $V$ — they do not depend on any choices. This is the key distinction from the finer decomposition into irreducibles: the individual irreducible summands inside a single $W_j$ are typically not canonical (when $n_j > 1$, there are many ways to decompose $W_j \cong S_j^{\oplus n_j}$ into a direct sum of copies of $S_j$), but the isotypical component $W_j$ itself is. In summary: - **Isotypical components** $W_j$: canonical, uniquely determined. - **Individual irreducible summands** within each $W_j$: not canonical when $n_j > 1$. [/explanation] With the isotypical decomposition in hand, the introductory structural theory of representations is complete. The course now turns to character theory, which provides the computational machinery for extracting these multiplicities and classifying representations of specific groups. Having understood irreducibles structurally, we need computational tools to identify and distinguish them. Chapter 5 introduces characters — the trace of a representation — which encode representation data in a single function and satisfy powerful orthogonality relations. # 5. Character theory Having established the fundamental framework of representations — what they are, when they are irreducible, and how complete reducibility lets us decompose them into irreducible pieces — we now turn to the central computational and theoretical tool of the subject: characters. The character of a representation is simply the trace of its matrices, and the remarkable fact is that this single number, attached to each group element, determines the representation completely up to isomorphism. This chapter develops the theory of characters, culminating in the orthogonality theorem and its consequences, including the character table as a complete invariant of the representation theory of a finite group. ## Characters and Their Basic Properties The core philosophy here is analogy with topology. In topology we attach invariants to spaces — the number of holes, the fundamental group — to distinguish objects that might otherwise look identical. We want invariants that attach to a representation $\rho$ of a finite group $G$ on $V$. The most naive approach is to record the matrix coefficients of each $\rho(g)$, but this is disastrously basis-dependent: changing the basis conjugates every matrix, and conjugate matrices can look nothing alike. We need something conjugation-invariant. The clue lies in the characteristic polynomial. The coefficients of the characteristic polynomial of a matrix are conjugation-invariant functions of the eigenvalues: the determinant is their product, the trace is their sum. Remarkably, only the trace is needed — it alone completely encodes the isomorphism class of $\rho$. [definition: Character] Let $G$ be a finite group, $\mathbb{F} = \mathbb{C}$, and $\rho = \rho_V : G \to \operatorname{GL}(V)$ a representation of $G$. The **character** of $\rho$, written $\chi_\rho = \chi_V = \chi$, is the function $\chi : G \to \mathbb{C}$ defined by \begin{align*} \chi(g) = \operatorname{tr}\,\rho(g). \end{align*} We say $\rho$ **affords** the character $\chi$. Equivalently, $\chi(g) = \operatorname{tr}\, R(g)$ for any matrix $R(g)$ representing $\rho(g)$ with respect to any basis — since conjugate matrices have equal trace, this is well-defined. [/definition] [definition: Degree of a Character] The **degree** of $\chi_V$ is $\dim V$, i.e. $\chi(1) = \dim V$. [/definition] The degree is the most basic numerical invariant of a character. It distinguishes one-dimensional characters — which carry especially clean structure — from higher-dimensional ones. [definition: Linear Character] A character $\chi$ is **linear** if $\dim V = 1$. In this case $\rho$ is itself a group homomorphism $G \to \mathbb{C}^\times = \operatorname{GL}_1(\mathbb{C})$, and $\chi = \rho$. [/definition] Linear characters are the simplest possible characters: a one-dimensional representation is just a multiplicative function on $G$. They will reappear as useful building blocks when we want to generate new characters from old. Several qualitative properties of representations descend directly to their characters. We record these as definitions to give them names. [definition: Irreducible Character] A character $\chi$ is **irreducible** if $\rho$ is irreducible. [/definition] [definition: Faithful Character] A character $\chi$ is **faithful** if $\rho$ is faithful, i.e. $\ker\rho = \{1\}$. [/definition] [definition: Trivial Character] The **trivial** (or **principal**) character $\chi = 1_G$ is the character of the trivial representation, so $\chi(g) = 1$ for all $g \in G$. [/definition] The following theorem collects the four most important elementary properties of characters. Despite being relatively easy to prove, they are used constantly in calculations. [quotetheorem:2421] [citeproof:2421] Property (2) is the key structural fact: characters live in the space of class functions, and we will see that the irreducible characters form a basis for this space. Property (4) tells us that decomposing a representation corresponds to adding characters — so understanding characters is the same as understanding how representations decompose. We also note that if $\chi_1, \chi_2$ are characters of $G$, then $\chi_1 \chi_2$ is also a character of $G$; this uses the tensor product of representations, which will be developed in the chapter on tensor products. [example: Linear Characters of a Cyclic Group] Let $G = \mathbb{Z}/n\mathbb{Z} = \langle g \mid g^n = 1 \rangle$. Since $G$ is abelian, every irreducible complex representation is one-dimensional (by Schur's lemma: any endomorphism of an irreducible complex representation is a scalar, so $G$ acts by scalars, meaning every irreducible is degree 1). A one-dimensional representation is determined by where $g$ goes, and $\rho(g)$ must satisfy $\rho(g)^n = \rho(g^n) = \rho(1) = 1$. So $\rho(g)$ must be an $n$-th root of unity. For each $k = 0, 1, \ldots, n-1$, define $\chi_k : G \to \mathbb{C}^\times$ by \begin{align*} \chi_k(g^j) = \omega^{jk}, \quad \omega = e^{2\pi i/n}. \end{align*} This is a well-defined group homomorphism since $\chi_k(g^n) = \omega^{nk} = 1$. The characters $\chi_0, \chi_1, \ldots, \chi_{n-1}$ are pairwise non-isomorphic (since $\chi_k(g) = \omega^k \neq \omega^j = \chi_j(g)$ for $k \neq j$), and there are exactly $n$ of them. Since $G$ has $n$ conjugacy classes (all elements are in their own class, as $G$ is abelian), the completeness theorem says there are exactly $n$ irreducible characters — so we have found them all. All characters of an abelian group are linear. [/example] ## Bounds on Characters and Generating New Characters How large can a character value be? Since $\chi(g) = \sum_i \lambda_i$ is a sum of roots of unity, all lying on the unit circle, the sum is bounded in absolute value by the number of terms, which is $\dim V = \chi(1)$. This naive bound turns out to be sharp, and understanding when equality holds reveals information about the kernel of the representation. [quotetheorem:2422] [citeproof:2422] This result is useful for computing kernels of representations from the character table alone: simply find the elements where the character achieves its maximum value $\chi(1)$. The proof is essentially a statement about sums of unit complex numbers, so the hypothesis that $G$ is finite is doing real work: it guarantees that $\rho(g)$ has finite order, hence is diagonalizable with roots-of-unity eigenvalues. For an element of infinite order this fails — the eigenvalues need not lie on the unit circle, and the bound can break down entirely. Equally, the theorem gives only an upper bound $|\chi(g)| \leq \chi(1)$; there is no useful lower bound for $|\chi(g)|$ in general (it can be zero for many elements). This upper bound will later let us read normal subgroups off the character table: the kernel of any irreducible character is a normal subgroup, and every normal subgroup arises as an intersection of such kernels. [quotetheorem:2423] [citeproof:2423] This theorem is a powerful bookkeeping device for building character tables. Part (1) produces a new irreducible whenever $\chi$ is not real-valued; in $D_6$ all characters happen to be real, but already in a group like $\mathbb{Z}/3\mathbb{Z}$ the three linear characters split into a conjugate pair plus the principal character. Part (2) is more striking: multiplying any irreducible character by a linear character produces another irreducible. This means linear characters act on the set of irreducible characters by permutation (since the result is again irreducible and of the same dimension), which constrains the structure of the character table. Note carefully that linearity of $\varepsilon$ is essential: if $\varepsilon$ itself is not one-dimensional, then $\varepsilon\chi$ is not the character of an irreducible in general (it could decompose further). Dropping that hypothesis would lose the multiplicative structure that makes the argument work. [remark: Why Linearity Matters] If $\varepsilon$ is a degree-$d$ character with $d > 1$, then $\varepsilon\chi$ has degree $d \cdot \dim V$ and typically decomposes into several irreducible pieces. The argument in Part (2) uses $\varepsilon(g) \in \mathbb{C}^\times$ as a scalar — which is only valid when $\varepsilon$ is one-dimensional. [/remark] ## The Space of Class Functions and the Inner Product To understand characters deeply, we embed them in a natural Hilbert space. Since characters are class functions, the right setting is the space of all class functions. [definition: Space of Class Functions] The **complex space of class functions** of $G$ is \begin{align*} \mathcal{C}(G) = \{f : G \to \mathbb{C} : f(hgh^{-1}) = f(g) \text{ for all } g, h \in G\}. \end{align*} This is a vector space under pointwise addition and scalar multiplication. [/definition] [definition: Class Number] The **class number** $k = k(G)$ is the number of conjugacy classes of $G$. [/definition] We label the conjugacy classes $\mathcal{C}_1, \ldots, \mathcal{C}_k$ with $\mathcal{C}_1 = \{1\}$ and choose representatives $g_1, g_2, \ldots, g_k$ with $g_j \in \mathcal{C}_j$. The indicator functions $\delta_j$ of the classes, where $\delta_j(g) = 1$ if $g \in \mathcal{C}_j$ and $\delta_j(g) = 0$ otherwise, form a basis of $\mathcal{C}(G)$. Therefore $\dim \mathcal{C}(G) = k$. We equip $\mathcal{C}(G)$ with a Hermitian inner product. [definition: Inner Product of Class Functions] The **inner product** on $\mathcal{C}(G)$ is \begin{align*} \langle f, f' \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{f(g)}\, f'(g) = \frac{1}{|G|} \sum_{j=1}^k |\mathcal{C}_j|\, \overline{f(g_j)}\, f'(g_j). \end{align*} By the orbit-stabilizer theorem, $|\mathcal{C}_j| = |G|/|C_G(g_j)|$, so this can also be written as \begin{align*} \langle f, f' \rangle = \sum_{j=1}^k \frac{1}{|C_G(g_j)|}\, \overline{f(g_j)}\, f'(g_j). \end{align*} For characters, using $\chi(g^{-1}) = \overline{\chi(g)}$, this becomes \begin{align*} \langle \chi, \chi' \rangle = \sum_{j=1}^k \frac{1}{|C_G(g_j)|}\, \chi(g_j^{-1})\, \chi'(g_j). \end{align*} [/definition] Note that $\langle \chi, \chi' \rangle = \overline{\langle \chi', \chi \rangle}$, so when restricted to real-valued class functions (which includes many characters), this is a real symmetric form. ## Orthogonality and Completeness of Characters The following theorem is the cornerstone of the whole subject. It asserts that the irreducible characters are not merely class functions, but an orthonormal basis for the entire space of class functions. The proof is a substantial piece of work, and we outline it here to give a concrete picture of why it is true. [quotetheorem:2424] [citeproof:2424] The two hypotheses — finite group, complex field — are both genuinely necessary. Over $\mathbb{R}$, Schur's lemma gives only that $\tilde{T}$ commutes with a real-irreducible action, and the endomorphism algebra can be $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$, breaking the clean scalar conclusion. For infinite groups, the averaging trick fails outright because $\sum_{g \in G}$ is an infinite sum with no reason to converge. Character theory of infinite groups requires replacing finite sums by integration against a Haar measure — a substantial machinery that lies beyond this course. What the theorem does NOT assert is any formula for individual matrix coefficients of a single representation: it says the irreducible characters are orthonormal, but individual entries of the representation matrices are harder to pin down. ## Consequences of Orthogonality Orthogonality is powerful precisely because it lets us extract information by pairing: if we know $\chi$ and want to know how much of $\chi_j$ it contains, just compute $\langle \chi, \chi_j \rangle$. The following consequences make this precise and show what orthogonality buys us practically. [quotetheorem:2425] [citeproof:2425] [remark: Failure for Infinite Groups] This result is special to finite groups. For $G = \mathbb{Z}$, the representations $1 \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $1 \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ are non-isomorphic but both have character identically equal to $2$. The second representation is indecomposable but not irreducible — complete reducibility fails for $\mathbb{Z}$, and with it the argument above breaks down. [/remark] [quotetheorem:2426] [citeproof:2426] This is an immensely practical criterion: to check whether a given representation is irreducible, compute the single number $\langle \chi, \chi \rangle$ and check whether it equals $1$. [remark: Failure in the Modular Case] The irreducibility criterion breaks down completely over fields of positive characteristic. Over a field $k$ with $\mathrm{char}(k) \mid |G|$, complete reducibility can fail — there exist indecomposable but reducible representations. The inner product is no longer available (or not meaningful), and $\langle \chi, \chi \rangle = 1$ is no longer a reliable test. Modular representation theory, which handles this case, requires entirely different machinery. [/remark] [example: Character Table of $D_6 \cong S_3$] Let $G = D_6 \cong S_3 = \langle r, s \mid r^3 = s^2 = 1,\, srs^{-1} = r^{-1} \rangle$. The three conjugacy classes are \begin{align*} \mathcal{C}_1 = \{1\}, \quad \mathcal{C}_2 = \{s, sr, sr^2\}, \quad \mathcal{C}_3 = \{r, r^{-1}\}. \end{align*} We identify three representations: the trivial representation $\mathbf{1}$, the sign representation $S$ (where reflections act by $-1$, rotations by $+1$), and the two-dimensional representation $W$ coming from the natural action on $\mathbb{R}^2$. For $W$: a rotation $r^m$ acts by the matrix $\begin{pmatrix} \cos\frac{2m\pi}{3} & -\sin\frac{2m\pi}{3} \\ \sin\frac{2m\pi}{3} & \cos\frac{2m\pi}{3} \end{pmatrix}$, so $\chi_W(r) = 2\cos\frac{2\pi}{3} = -1$. A reflection $sr^j$ fixes one axis and negates the perpendicular one, so its eigenvalues are $+1$ and $-1$ (it is an orientation-reversing isometry of $\mathbb{R}^2$ fixing the reflection axis), giving $\chi_W(sr^j) = 0$.  The character table is: | | $\mathcal{C}_1$ | $\mathcal{C}_2$ | $\mathcal{C}_3$ | |---|---|---|---| | $\mathbf{1}$ | $1$ | $1$ | $1$ | | $S$ | $1$ | $-1$ | $1$ | | $\chi_W$ | $2$ | $0$ | $-1$ | The centralizer sizes are $|C_G(g_1)| = 6$, $|C_G(g_2)| = 2$, $|C_G(g_3)| = 3$. We verify irreducibility of $W$: \begin{align*} \langle \chi_W, \chi_W \rangle = \frac{2^2}{6} + \frac{0^2}{2} + \frac{(-1)^2}{3} = \frac{4}{6} + 0 + \frac{1}{3} = \frac{2}{3} + \frac{1}{3} = 1. \end{align*} So $W$ is irreducible. Also $1^2 + 1^2 + 2^2 = 6 = |D_6|$, consistent with the dimension formula below. [/example] [example: Using Characters to Detect Non-Isomorphic Representations] Still with $G = S_3$, consider the representation $V = \mathbf{1} \oplus S$ (the direct sum of the trivial and sign representations) and the representation $U = \mathbf{1} \oplus \mathbf{1}$ (two copies of the trivial representation). Both have dimension $2$. Do they agree on every group element? Computing: \begin{align*} \chi_V(g) &= 1 + S(g), & \chi_U(g) &= 2. \end{align*} At a reflection $s \in \mathcal{C}_2$: $\chi_V(s) = 1 + (-1) = 0$, while $\chi_U(s) = 2$. The characters differ, so $V \not\cong U$ — even though both have the same dimension. This illustrates how characters succeed where dimension alone fails. [/example] ## The Dimension Formula and Counting Irreducibles Two of the most practically useful facts in character theory concern how many irreducible representations a finite group has and how large they can be. Before we can fill in a character table, we need to know what size it is and what constraints the degrees must satisfy — these questions are answered by the following theorems. [quotetheorem:2427] [citeproof:2427] A bonus from this proof: since $a_j = n_j \geq 1$ for every irreducible $\rho_j$, every irreducible representation of $G$ appears as a subrepresentation of the regular representation (with multiplicity equal to its dimension). In practice the formula $\sum_i n_i^2 = |G|$ is used to check when a character table is complete: once the known dimensions satisfy the sum-of-squares identity, no more irreducibles can exist. It does not, however, tell you what the character values are — that is a separate and often harder computation. [example: Regular Character of $S_3$] For $G = S_3$ with $|G| = 6$, the regular character $\pi_{\mathrm{reg}}$ satisfies $\pi_{\mathrm{reg}}(1) = 6$ and $\pi_{\mathrm{reg}}(g) = 0$ for all $g \neq 1$. Decomposing: $a_j = n_j$ for each irreducible, so \begin{align*} \pi_{\mathrm{reg}} = 1 \cdot \chi_{\mathbf{1}} + 1 \cdot \chi_S + 2 \cdot \chi_W. \end{align*} We can check at $g = 1$: $1 \cdot 1 + 1 \cdot 1 + 2 \cdot 2 = 6 = |G|$. At any $g \neq 1$, say a reflection $s$: $1 \cdot 1 + 1 \cdot (-1) + 2 \cdot 0 = 0$. The formula is consistent, and confirms that $\mathbb{C}S_3 \cong \mathbf{1} \oplus S \oplus W \oplus W$ (two copies of $W$) as $S_3$-modules. [/example] [quotetheorem:2428] [citeproof:2428] This is a structural miracle: a purely group-theoretic invariant (the number of conjugacy classes) equals a purely representation-theoretic invariant (the number of irreducible representations). It means the character table is always a square matrix. In practice, knowing $k$ tells you exactly when your character table is complete — you need precisely $k$ rows, and the sum-of-squares formula is then the final consistency check. Note that the theorem gives no information about which dimensions $n_i$ occur; it only tells you there are $k$ of them. For abelian groups, every conjugacy class is a singleton so $k = |G|$, forcing all $n_i = 1$ — every irreducible character is linear, which is the easiest case. For non-abelian groups, some $n_i \geq 2$, and the character table becomes genuinely harder to compute. [quotetheorem:2429] [citeproof:2429] This theorem gives a complete algorithm for detecting conjugacy purely from the character table: two elements are conjugate if and only if their columns in the character table are identical. The key word in the statement is "every": a single character agreeing on $g_1$ and $g_2$ says nothing — for instance, the trivial character $\chi = 1$ agrees on everything. One needs all irreducible characters simultaneously. This is related to the power-map structure: if one knows all character values, one can reconstruct which elements are $p$-th powers for any $p$, since conjugacy classes are preserved under the power map. ## The Character Table How do we organise and record the complete representation theory of a finite group $G$ in a single compact object? The answer is the character table, which collects all irreducible character values into a square matrix indexed by conjugacy classes. [definition: Character Table] The **character table** of $G$ is the $k \times k$ matrix $X = (\chi_i(g_j))$, where $1 = \chi_1, \chi_2, \ldots, \chi_k$ are the irreducible characters of $G$ and $\mathcal{C}_1 = \{1\}, \mathcal{C}_2, \ldots, \mathcal{C}_k$ are the conjugacy classes with $g_j \in \mathcal{C}_j$. [/definition] In practice the character table is recorded as a literal table with conjugacy classes as columns and irreducible characters as rows. The column for $\mathcal{C}_1 = \{1\}$ records the degrees $n_1, \ldots, n_k$, which must satisfy $\sum n_i^2 = |G|$. The row orthogonality relations are built into the character table: the inner product formula \begin{align*} \langle \chi_i, \chi_j \rangle = \sum_{\ell=1}^k \frac{|\mathcal{C}_\ell|}{|G|}\, \overline{\chi_i(g_\ell)}\, \chi_j(g_\ell) = \delta_{ij} \end{align*} says that any two distinct rows are orthogonal with respect to this weighted inner product, and each row is a unit vector. There is also a **column orthogonality** relation: for two conjugacy class representatives $g_s, g_t$, \begin{align*} \sum_{i=1}^k \overline{\chi_i(g_s)}\, \chi_i(g_t) = \frac{|G|}{|\mathcal{C}_s|}\, \delta_{st} = |C_G(g_s)|\, \delta_{st}. \end{align*} These column relations follow from the fact that the character table (suitably normalized) is unitary. The character table encodes a remarkable amount of group-theoretic information: the conjugacy classes, the irreducible representations, the normal subgroups (as intersections of kernels of characters), and the abelianization $G/[G,G]$ (which corresponds to the linear characters of $G$). Computing character tables efficiently is a central goal, and the rest of this chapter develops techniques for doing so without first finding all irreducible representations. Characters give us orthogonality rules, but why do they hold? Chapter 6 proves the orthogonality theorem from three angles: the averaging trick, tensor product arguments, and column orthogonality, showing that character tables completely determine the representation theory of a finite group. # 6. Proof of orthogonality The character inner product introduced in the previous chapter turns the space of class functions into an inner product space. This chapter delivers the central analytical payoff: the orthogonality relations for characters. The row orthogonality theorem says that distinct irreducible characters are orthogonal, and isomorphic ones have norm one. The column orthogonality theorem is the dual statement, relating values across conjugacy classes. Together, they impose strong constraints on the character table and, as a consequence, imply that the number of irreducible representations equals the number of conjugacy classes — completing the completeness theorem for characters. ## Row Orthogonality via the Averaging Trick Why should two distinct irreducible representations have orthogonal characters? At first glance, it is not obvious that the inner product $\langle \chi, \chi' \rangle = \frac{1}{|G|}\sum_{g \in G} \overline{\chi(g)}\,\chi'(g)$ has anything to do with whether $\rho$ and $\rho'$ are isomorphic. The key is that the averaging trick converts an arbitrary linear map between two representation spaces into a $G$-homomorphism, and Schur's lemma then forces such maps either to vanish (for non-isomorphic irreducibles) or to be a scalar (for isomorphic ones). The inner product formula falls out from choosing the right probe map. [quotetheorem:2430] [citeproof:2430] The argument in both cases reduces to choosing the right elementary operator $\varepsilon_{\alpha\beta}$ as a probe and then reading off the matrix entry formula. This is a prototypical use of the averaging trick: one constructs a $G$-map, applies Schur, determines the scalar from the trace, and extracts the desired identity. Notice what the theorem requires and what it does not say. Irreducibility of both $\rho$ and $\rho'$ is essential: without it, Schur's lemma does not force $\widetilde{\varphi}$ to be zero or scalar, and the computation breaks down. For a reducible representation $\pi = \chi_1 \oplus \chi_2$, the inner product $\langle \pi, \chi_1 \rangle$ can equal $1$ without $\pi$ being isomorphic to $\chi_1$ — it simply counts how many copies of $\chi_1$ appear in $\pi$. The theorem also requires working over an algebraically closed field of characteristic zero: over $\mathbb{R}$, Schur's lemma gives only $\widetilde{\varphi} = \lambda\, \mathrm{id}$ for $\lambda \in \mathbb{R}$, but the norm can differ from $1$ (the character of the $2$-dimensional real irreducible of $\mathbb{Z}/3\mathbb{Z}$ has norm $1$ over $\mathbb{C}$ but the representation is not irreducible over $\mathbb{C}$). In the modular setting ($\operatorname{char} k \mid |G|$), Maschke's theorem fails, the averaging operator $\frac{1}{|G|}\sum_g$ is not defined, and none of these relations hold in general. [explanation: Irreducibility Test and Decomposition via Inner Products] Row orthogonality gives two immediate practical tools. **Irreducibility test.** A representation $\pi$ with character $\chi_\pi$ is irreducible if and only if $\langle \chi_\pi, \chi_\pi \rangle = 1$. Since $\chi_\pi = \sum_i m_i \chi_i$ where the $m_i \geq 0$ are non-negative integer multiplicities, we have \begin{align*} \langle \chi_\pi, \chi_\pi \rangle = \sum_i m_i^2. \end{align*} This equals $1$ if and only if exactly one $m_i = 1$ and all others are $0$ — i.e., $\pi$ is irreducible. If $\langle \chi_\pi, \chi_\pi \rangle = 2$, then $\pi$ decomposes as a direct sum of exactly two non-isomorphic irreducibles (or one with multiplicity $\sqrt{2}$, which is not an integer, so the only option is two distinct irreducibles each with multiplicity $1$). **Decomposition formula.** The multiplicity of the $i$th irreducible $\chi_i$ in $\pi$ is \begin{align*} m_i = \langle \chi_\pi, \chi_i \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_\pi(g)}\, \chi_i(g). \end{align*} So to decompose a representation, one computes $k$ inner products against the $k$ irreducible characters. No guesswork is required: the decomposition is completely determined by the character. [/explanation] ## An Alternative Proof via Tensor Products What does the matrix-element proof hide? The averaging trick is powerful but somewhat opaque: it works, but the reason that the inner product counts isomorphism copies does not leap out. Once tensor products and duals are available, a second proof makes the structural reason transparent: $\langle \chi_V, \chi_W \rangle$ is exactly the dimension of the space of $G$-equivariant maps from $W$ to $V$. [explanation: Alternative Proof of Row Orthogonality] For representations on spaces $V$ and $W$, \begin{align*} \langle \chi_W, \chi_V \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_W(g)}\, \chi_V(g) = \frac{1}{|G|} \sum_{g \in G} \chi_{V \otimes W^*}(g). \end{align*} There is a natural $G$-isomorphism $V \otimes W^* \cong \operatorname{Hom}(W, V)$, where $G$ acts by conjugation: $g \cdot f = \rho_V(g) \circ f \circ \rho_W(g)^{-1}$. A $G$-fixed element of $\operatorname{Hom}(W, V)$ is exactly a $G$-homomorphism $W \to V$. The space $\operatorname{Hom}(W, V)$ decomposes into $\operatorname{Hom}_G(W, V)$ (the trivial isotypic component) and a complement with no $G$-fixed vectors. Thus the number of copies of the trivial representation in $\operatorname{Hom}(W, V)$ equals $\dim \operatorname{Hom}_G(W, V)$, which by Schur's lemma is $1$ if $V \cong W$ and $0$ otherwise. It remains to show that any non-trivial irreducible character satisfies $\sum_{g \in G} \chi(g) = 0$. If $\rho$ affords $\chi$, then any vector in the image of $\sum_{g \in G} \rho(g)$ is fixed by $G$. By irreducibility of $\rho$, the image must be trivial, so the operator $\sum_{g \in G} \rho(g)$ is zero, and taking the trace gives the result. [/explanation] The tensor product proof also makes clear what goes wrong in the modular setting: the identification of the inner product with $\dim \operatorname{Hom}_G(W,V)$ relies on the averaging operator $\frac{1}{|G|}\sum_g$, which requires $\operatorname{char} k \nmid |G|$. In characteristic $p \mid |G|$, this counting argument collapses entirely. ## Column Orthogonality Can we read the sizes of conjugacy classes directly from the character table? The row orthogonality relations are encoded in the columns of the table (the values $\chi_i(g_j)$ for varying $i$, fixed $j$), but the column directions carry their own orthogonality. The centralizer size $|C_G(g)|$ — and hence the conjugacy class size $|G|/|C_G(g)|$ — turns out to be recoverable from the table alone. [quotetheorem:2431] [citeproof:2431] The centralizer sizes appear because the inner product on class functions weights each conjugacy class by $1/|C_G(g)|$, and removing this weight from the row orthogonality formula is what produces the column formula. This is not mere algebraic coincidence: it reflects the fact that the character table, viewed as a linear map from class functions to $\mathbb{C}^k$ (evaluating at each conjugacy class representative), must be an isometric isomorphism in an appropriate sense, and the centralizer sizes encode the correct metric on each side. Practically, column orthogonality gives a powerful tool for completing character tables. If all but one character value in a column are known, the orthogonality relation \begin{align*} \sum_{i=1}^k \overline{\chi_i(g_j)}\, \chi_i(g_\ell) = 0 \quad (j \neq \ell) \end{align*} gives a linear constraint that can determine the missing entry. Two conjugacy classes $g_j$ and $g_\ell$ with identical columns in the character table are indistinguishable by any class function, which forces $g_j$ and $g_\ell$ to be in the same conjugacy class — so the character table distinguishes all conjugacy classes. [remark: Corollary on Sum of Squares] Setting $j = \ell = 1$ (the conjugacy class of the identity, so $|C_G(e)| = |G|$) in the column orthogonality relation gives \begin{align*} \sum_{i=1}^k \chi_i(1)^2 = |G|. \end{align*} This identity was established earlier using the regular representation; here it follows from the column orthogonality theorem. [/remark] Note that the proof uses the fact that $X$ is square, i.e.\ that the number of irreducible representations equals the number of conjugacy classes. This is precisely the completeness of characters, which we must establish independently. ## Completeness of Characters What would go wrong if the irreducible characters only spanned a proper subspace of $\mathcal{C}(G)$? There would then be a nonzero class function $f$ orthogonal to every irreducible character. But class functions are determined by their values on $G$, and we shall see that $f$ being orthogonal to all irreducibles forces it to vanish everywhere — using the regular representation as a universal test object. [quotetheorem:2432]This is the final piece needed to conclude that the number of irreducible representations equals $\dim \mathcal{C}(G) = k$, the number of conjugacy classes. [citeproof:2432] The argument is elegant in its simplicity: one turns the orthogonality condition $\langle f, \chi_j \rangle = 0$ into the vanishing of a specific $G$-endomorphism via Schur, and then the regular representation serves as a universal test that forces $f$ to be identically zero. Algebraic closure of the base field is essential in this argument. Over $\mathbb{R}$, Schur's lemma only gives $\widetilde{\varphi} \in \mathbb{R}$; a real representation that is irreducible over $\mathbb{R}$ may split into two complex conjugate irreducibles upon extension to $\mathbb{C}$. This means the real irreducibles are not in bijection with conjugacy classes, and the completeness theorem takes a different form. Complete reducibility is also critical: the step "holds for every representation by complete reducibility" uses Maschke's theorem, and the proof breaks down in the modular setting where $\operatorname{char} k \mid |G|$. [remark: Consequence for the Character Table] Combining row orthogonality with completeness: the irreducible characters $\chi_1, \ldots, \chi_k$ form an orthonormal basis for $\mathcal{C}(G)$. Since $\dim \mathcal{C}(G) = k$ (the number of conjugacy classes), the number of irreducible complex representations of $G$ is exactly $k$. The character table is therefore a square $k \times k$ matrix, justifying the use of this fact in the proof of column orthogonality. [/remark] ## Worked Example: Orthogonality on $S_3$ To see these abstract results concretely, we verify both orthogonality relations on the character table of $S_3$, the symmetric group on three letters. [example: Orthogonality Relations for $S_3$] The group $S_3$ has order $6$ and three conjugacy classes: the identity $\{e\}$, transpositions $\{(12),(13),(23)\}$ (class size $3$), and $3$-cycles $\{(123),(132)\}$ (class size $2$). There are three irreducible representations: the trivial representation $\chi_1$, the sign representation $\chi_2$, and the standard $2$-dimensional representation $\chi_3$. Their characters are: \begin{align*} \begin{array}{c|ccc} & e & (12) & (123) \\ \hline \chi_1 & 1 & 1 & 1 \\ \chi_2 & 1 & -1 & 1 \\ \chi_3 & 2 & 0 & -1 \end{array} \end{align*} **Verifying row orthogonality.** Using $\langle \chi, \chi' \rangle = \frac{1}{|G|}\sum_{g} \overline{\chi(g)}\chi'(g)$, but grouping by conjugacy classes: \begin{align*} \langle \chi, \chi' \rangle = \frac{1}{6}\bigl(|\mathcal{C}_1|\,\overline{\chi(e)}\chi'(e) + |\mathcal{C}_2|\,\overline{\chi((12))}\chi'((12)) + |\mathcal{C}_3|\,\overline{\chi((123))}\chi'((123))\bigr). \end{align*} Check $\langle \chi_1, \chi_1 \rangle$: \begin{align*} \frac{1}{6}(1 \cdot 1 \cdot 1 + 3 \cdot 1 \cdot 1 + 2 \cdot 1 \cdot 1) = \frac{1+3+2}{6} = 1. \checkmark \end{align*} Check $\langle \chi_3, \chi_3 \rangle$: \begin{align*} \frac{1}{6}(1 \cdot 4 + 3 \cdot 0 + 2 \cdot 1) = \frac{4+0+2}{6} = 1. \checkmark \end{align*} Check $\langle \chi_1, \chi_3 \rangle$: \begin{align*} \frac{1}{6}(1 \cdot 1 \cdot 2 + 3 \cdot 1 \cdot 0 + 2 \cdot 1 \cdot (-1)) = \frac{2+0-2}{6} = 0. \checkmark \end{align*} Check $\langle \chi_2, \chi_3 \rangle$: \begin{align*} \frac{1}{6}(1 \cdot 1 \cdot 2 + 3 \cdot (-1) \cdot 0 + 2 \cdot 1 \cdot (-1)) = \frac{2+0-2}{6} = 0. \checkmark \end{align*} **Verifying column orthogonality.** The centralizer sizes are $|C_{S_3}(e)| = 6$, $|C_{S_3}((12))| = 2$, $|C_{S_3}((123))| = 3$. Column orthogonality says $\sum_i \overline{\chi_i(g_j)}\chi_i(g_\ell) = \delta_{j\ell}|C_{S_3}(g_\ell)|$. Check the diagonal entry for the transposition column ($g_j = g_\ell = (12)$): \begin{align*} \overline{\chi_1((12))}\chi_1((12)) + \overline{\chi_2((12))}\chi_2((12)) + \overline{\chi_3((12))}\chi_3((12)) = 1 \cdot 1 + (-1)(-1) + 0 \cdot 0 = 2 = |C_{S_3}((12))|. \checkmark \end{align*} Check the off-diagonal entry for identity vs. $3$-cycle columns ($g_j = e$, $g_\ell = (123)$): \begin{align*} \overline{\chi_1(e)}\chi_1((123)) + \overline{\chi_2(e)}\chi_2((123)) + \overline{\chi_3(e)}\chi_3((123)) = 1 \cdot 1 + 1 \cdot 1 + 2 \cdot (-1) = 1+1-2 = 0. \checkmark \end{align*} **Decomposing a reducible representation.** Let $\pi$ be the permutation representation of $S_3$ on $\mathbb{C}^3$ (permuting coordinates). Its character values are $\chi_\pi(e) = 3$, $\chi_\pi((12)) = 1$ (one fixed point), $\chi_\pi((123)) = 0$ (no fixed points). The multiplicities are: \begin{align*} m_1 = \langle \chi_\pi, \chi_1 \rangle &= \frac{1}{6}(1 \cdot 3 + 3 \cdot 1 + 2 \cdot 0) = 1,\\ m_2 = \langle \chi_\pi, \chi_2 \rangle &= \frac{1}{6}(1 \cdot 3 \cdot 1 + 3 \cdot 1 \cdot (-1) + 2 \cdot 0 \cdot 1) = 0,\\ m_3 = \langle \chi_\pi, \chi_3 \rangle &= \frac{1}{6}(1 \cdot 3 \cdot 2 + 3 \cdot 1 \cdot 0 + 2 \cdot 0 \cdot (-1)) = 1. \end{align*} So $\pi \cong \chi_1 \oplus \chi_3$: the permutation representation decomposes into the trivial representation and the standard $2$-dimensional representation. Note also that $\langle \chi_\pi, \chi_\pi \rangle = 1^2 + 0^2 + 1^2 = 2 \neq 1$, confirming by the irreducibility criterion that $\pi$ is reducible. [/example] Armed with character theory, we return to the permutation representations from Chapter 1 with much more power. Chapter 7 uses characters to study permutation actions via Burnside's lemma, derives the full character table of $S_4$, and reveals when symmetric-group conjugacy classes split in the alternating group. # 7. Permutation representations The preceding chapters developed character theory as an abstract algebraic tool. Here we bring it to bear on a class of representations that arises directly from group actions: the permutation representations. Because they carry an explicit combinatorial structure — fixed points, orbits, stabilisers — their characters can be computed in closed form. This makes them a rich source of explicit examples, and in particular gives a systematic route into the representation theory of the symmetric group $S_n$. ## The Permutation Representation of a $G$-Set Suppose we are given a finite group $G$ acting on a finite set $X$. Group actions encode symmetry in a purely combinatorial language, but the machinery of character theory lives in the world of linear algebra. The central question is: how do we convert a group action on a set into a representation on a vector space, in a way that faithfully preserves the action's structure? The answer is the permutation representation. Let $G$ act on $X = \{x_1, \ldots, x_n\}$; such an $X$ is called a $G$-set. We form the complex vector space $\mathbb{C}X$ with basis $\{e_{x_1}, \ldots, e_{x_n}\}$: \begin{align*} \mathbb{C}X = \left\{ \sum_j a_j e_{x_j} : a_j \in \mathbb{C} \right\}. \end{align*} [definition: Permutation Representation] Let $G$ act on a finite set $X$. The **permutation representation** associated to this action is the homomorphism \begin{align*} \rho_X : G &\to \operatorname{GL}(\mathbb{C}X) \\ g &\mapsto \rho_X(g), \end{align*} where $\rho_X(g)$ is the linear extension of the map $e_{x_j} \mapsto e_{g x_j}$. [/definition] The matrices of $\rho_X(g)$ with respect to the basis $\{e_x\}_{x \in X}$ are **permutation matrices**: each row and each column contains exactly one entry equal to $1$ and all others $0$. Concretely, $(\rho_X(g))_{ij} = 1$ if and only if $g x_j = x_i$. This construction is canonical: it is the unique way to extend a set-theoretic $G$-action to a linear representation on the free vector space over $X$, and it is functorial in the sense that equivariant maps of $G$-sets become equivariant linear maps. In particular, this is a natural generalisation of the regular representation: if $G$ acts on itself by left multiplication, then $\mathbb{C}X$ is precisely the group algebra $\mathbb{C}G$. ## The Permutation Character and Fixed Points Given the permutation representation $\rho_X$, we want computable invariants that reveal its structure. Diagonalising each matrix $\rho_X(g)$ would require tracking all eigenvalues, which is impractical. The insight is that the character — the trace — has a much simpler description that bypasses matrix diagonalisation entirely. Because $\rho_X(g)$ is a permutation matrix, the diagonal entry $(\rho_X(g))_{ii}$ equals $1$ when $x_i$ is fixed by $g$, and $0$ otherwise. Summing over the diagonal gives: [definition: Permutation Character] The **permutation character** $\pi_X : G \to \mathbb{C}$ is defined by \begin{align*} \pi_X(g) = |\mathrm{fix}(g)| = |\{x \in X : gx = x\}|. \end{align*} [/definition] In other words, $\pi_X(g)$ simply counts how many points of $X$ are fixed by $g$. This is the key computational handle on permutation representations: instead of tracking matrices, we track fixed-point counts — a purely combinatorial datum attached to each group element. [remark: Constant-Sum Subspace] The vector $e_{x_1} + \cdots + e_{x_n}$ is fixed by every $g \in G$, since $g$ permutes the basis vectors. Hence $\mathrm{span}\{e_{x_1} + \cdots + e_{x_n}\}$ is a $G$-invariant subspace on which $G$ acts as the identity. Its $G$-invariant complement is $\left\{\sum_x a_x e_x : \sum_x a_x = 0\right\}$. This shows $\pi_X$ always contains the trivial character $1_G$ as a constituent: the constant-sum subspace provides an explicit copy. [/remark] ## Orbits and the Multiplicity of the Trivial Character The most fundamental result about permutation characters connects the internal structure of the action — its orbits — to the inner product of $\pi_X$ with the trivial character. [quotetheorem:2433] [citeproof:2433] This result is sometimes called **Burnside's lemma** (though historically it is due to Cauchy and Frobenius). It provides a purely character-theoretic way to count orbits without enumerating them explicitly. The transitivity hypothesis in the proof is essential: it licenses the use of orbit-stabiliser to get a uniform value $|G_x| = |G|/|X|$ for all $x$. Without transitivity, the stabilisers have different sizes and no such simplification is available. What the theorem does not assert is that $\pi_X$ contains only one copy of $1_G$ in general — that holds precisely when the action is transitive. For a non-transitive action with $\ell$ orbits, $1_G$ appears with multiplicity $\ell$, and the character carries a richer structure reflecting the orbit decomposition. ## The Inner Product of Two Permutation Characters When $G$ acts on two sets simultaneously, the inner product of the corresponding permutation characters has a clean combinatorial meaning. [quotetheorem:2434] [citeproof:2434] The diagonal action $g(x_1, x_2) = (gx_1, gx_2)$ is the right choice here because it is exactly the condition under which a pair is simultaneously fixed: fixed points in the product correspond precisely to simultaneous fixed points in each factor. A different action on the product would break this correspondence. The special case $X_1 = X_2 = X$ gives $\langle \pi_X, \pi_X \rangle$ equal to the number of orbits of $G$ on $X \times X$; this will be the key to the 2-transitivity criterion in the next section, since it lets us count irreducible constituents by counting orbits on the product set. ## 2-Transitive Actions and Irreducibility A particularly clean situation arises when $G$ acts as transitively as possible on ordered pairs of distinct elements. [definition: 2-Transitive Action] Let $G$ act on $X$ with $|X| > 2$. The action is **2-transitive** if $G$ has exactly two orbits on $X \times X$: the diagonal $\{(x,x) : x \in X\}$ and the off-diagonal $\{(x_1, x_2) : x_1, x_2 \in X,\; x_1 \neq x_2\}$. [/definition] Note that the diagonal is always a single orbit under any action (since $g(x,x) = (gx,gx)$ maps the diagonal to itself). So $G$ being $2$-transitive on $X$ is equivalent to saying $G$ acts transitively on $X$ (giving one diagonal orbit) and can map any ordered pair of distinct elements to any other such pair (giving one off-diagonal orbit). [quotetheorem:2435] [citeproof:2435] The hypothesis $|X| > 2$ matters: if $|X| = 2$, then $X \times X$ has only two points off the diagonal, and any transitive action automatically makes those two points a single orbit, giving 2-transitivity for any transitive $G$-action on a two-element set — but $\pi_X = 1_G + \chi$ with $\chi$ the sign character, which may or may not be "interesting." The condition $|X| > 2$ excludes this degenerate case. The theorem also does not say the 2-transitive action determines the full character table of $G$: it produces exactly one new irreducible character $\chi = \pi_X - 1_G$, leaving all other irreducibles undetermined. Transitivity is genuinely needed; without it, $\langle \pi_X, 1_G \rangle > 1$ and the argument that $m_1 = 1$ fails. The power of this theorem is that it produces irreducible characters for free whenever we recognise a 2-transitive action. But it is important to see what happens when 2-transitivity fails. Consider the dihedral group $D_8$ acting on the four vertices of a square, labelled $\{1, 2, 3, 4\}$ in order. This action is transitive but not 2-transitive: a rotation can send the pair $(1, 2)$ to any adjacent pair $(2, 3)$, $(3, 4)$, $(4, 1)$, but it cannot send $(1, 2)$ to the pair $(1, 3)$ — adjacent and opposite pairs form separate orbits on the off-diagonal of $X \times X$. Concretely, the off-diagonal splits into two orbits: adjacent pairs $\{(1,2),(2,3),(3,4),(4,1),(2,1),(3,2),(4,3),(1,4)\}$ and diagonal-opposite pairs $\{(1,3),(3,1),(2,4),(4,2)\}$. So $G$ has three orbits on $X \times X$ in total, giving \begin{align*} \langle \pi_X, \pi_X \rangle = 3, \end{align*} which forces $\sum m_i^2 = 3$, and the only solution in positive integers is $m_1 = m_2 = m_3 = 1$: the complement $\pi_X - 1_G$ decomposes as a sum of two distinct irreducible characters, not one. Indeed, one can verify directly that the permutation character of $D_8$ on the square is $\pi_X = 1_G + \chi_2 + \chi_3$, where $\chi_2$ and $\chi_3$ are the two distinct one-dimensional non-trivial characters of $D_8$. The complement $\pi_X - 1_G$ is reducible, confirming that 2-transitivity is not just sufficient but necessary. ## Application to the Symmetric Group $S_n$ The 2-transitive action criterion is most immediately useful for symmetric groups, where the natural action on $n$ elements is visibly 2-transitive. [example: Irreducible Standard Representation of $S_n$] Let $S_n$ act on $X = \{1, 2, \ldots, n\}$ in the natural way, for $n \geq 2$. This action is $2$-transitive: given any two ordered pairs $(i, j)$ and $(k, \ell)$ with $i \neq j$ and $k \neq \ell$, there is a permutation sending $i \mapsto k$ and $j \mapsto \ell$ (just extend arbitrarily to a bijection on all of $X$). By the theorem, $\pi_X = 1_{S_n} + \chi$ where $\chi$ is irreducible. Since $\pi_X$ has degree $n$ (the dimension of $\mathbb{C}X$), the character $\chi$ has degree $n - 1$. This gives an $(n-1)$-dimensional irreducible representation of $S_n$, sometimes called the **standard representation**. The same argument applies to $A_n$ acting on $X = \{1, \ldots, n\}$ for $n \geq 3$: $A_n$ is also $2$-transitive on $X$ (given distinct $i \neq j$ and $k \neq \ell$, there is always an even permutation mapping the pair to the pair), so $\pi_X = 1_{A_n} + \chi'$ with $\chi'$ irreducible of degree $n - 1$. In contrast to $S_n$, where the standard representation is one of many irreducibles, for $A_n$ this $(n-1)$-dimensional character is particularly significant: because $A_n$ is simple for $n \geq 5$, its representation theory is more constrained, and the standard representation coming from the 2-transitive action is one of the few irreducibles with an explicit description. For $A_5 \cong \text{PSL}(2, \mathbb{F}_5)$, the group has only 5 conjugacy classes, and the standard 4-dimensional representation $\chi'$ accounts for one irreducible; the remaining three are harder to produce without further tools. [/example] ## Computing the Full Character Table of $S_4$ The permutation character technique, combined with basic tricks for generating new characters from old ones, allows us to compute the entire character table of $S_4$ from scratch. $S_4$ has $5$ conjugacy classes (corresponding to the five cycle types $1$, $(12)(34)$, $(123)$, $(1234)$, $(12)$, with class sizes $1, 3, 8, 6, 6$ respectively), so there are exactly $5$ irreducible characters. [example: Character Table of $S_4$] We already know three characters: - $\chi_1$: the trivial character, value $1$ on everything. - $\chi_2$: the sign character, value $+1$ on even permutations and $-1$ on odd. - $\chi_3 = \pi_X - 1_{S_4}$: the standard character of degree $3$, obtained by subtracting the trivial character from the permutation character. Since $S_4$ is $2$-transitive on $\{1,2,3,4\}$, $\chi_3$ is irreducible. To compute $\chi_3$ explicitly: $\pi_X(g)$ counts fixed points of $g$. - $g = e$: all $4$ fixed, so $\pi_X(e) = 4$, giving $\chi_3(e) = 3$. - $g = (12)(34)$: no fixed points, so $\pi_X = 0$, giving $\chi_3 = -1$. - $g = (123)$: no fixed points, so $\pi_X = 0$, giving $\chi_3 = -1$. - $g = (1234)$: no fixed points, so $\pi_X = 0$, giving $\chi_3 = -1$. - $g = (12)$: fixes $3$ and $4$, so $\pi_X = 2$, giving $\chi_3 = 1$. A general principle from example sheet 1 states that if $\chi$ is a character and $\varepsilon$ is a one-dimensional character (so $\varepsilon$ is a homomorphism $G \to \mathbb{C}^\times$), then $\varepsilon \chi$ is also a character, and it is irreducible whenever $\chi$ is. Applying this with $\varepsilon = \chi_2$ (the sign) and $\chi = \chi_3$: $\chi_4 = \chi_2 \chi_3$: values are the pointwise product. This gives degree $3$ and values $3, -1, 0, 1, -1$ on the five conjugacy classes. Since $\chi_3$ is irreducible and $\chi_2$ is one-dimensional, $\chi_4$ is irreducible. The four known characters have degrees $1, 1, 3, 3$. The sum-of-squares constraint $\sum_i d_i^2 = |G| = 24$ gives: \begin{align*} 1 + 1 + 9 + 9 + d_5^2 = 24 \implies d_5^2 = 4 \implies d_5 = 2. \end{align*} So the fifth irreducible character $\chi_5$ has degree $2$. To find the values of $\chi_5$, use column orthogonality. The relation between the identity column and any other column gives $\sum_i \chi_i(e)\chi_i(g) = 0$ for $g \neq e$. For $g = (12)(34)$: \begin{align*} 1 \cdot 1 + 1 \cdot 1 + 3 \cdot (-1) + 3 \cdot (-1) + 2 \cdot x_1 = 0 \implies 1 + 1 - 3 - 3 + 2x_1 = 0 \implies x_1 = 2. \end{align*} For $g = (123)$: \begin{align*} 1 \cdot 1 + 1 \cdot 1 + 3 \cdot (-1) + 3 \cdot (-1) + 2 \cdot x_2 = 0 \implies 1 + 1 - 3 - 3 + 2x_2 = 0 \implies x_2 = 2. \end{align*} Wait — that cannot be right, since $\chi_5((123))$ must equal $x_2$ and we need to use the correct column. For $g = (123)$: $\chi_1 = 1$, $\chi_2 = 1$ (even permutation), $\chi_3 = -1$, $\chi_4 = -1$, so \begin{align*} 1 \cdot 1 + 1 \cdot 1 + 3 \cdot (-1) + 3 \cdot (-1) + 2 \cdot x_2 = 0 \implies -4 + 2x_2 = 0 \implies x_2 = -1. \end{align*} For $g = (1234)$: $\chi_1 = 1$, $\chi_2 = -1$ (odd), $\chi_3 = -1$, $\chi_4 = 1$ (values $3 \cdot (-1) \cdot (-1) = 1$ since $\chi_4 = \chi_2\chi_3$), so \begin{align*} 1 \cdot 1 + 1 \cdot (-1) + 3 \cdot (-1) + 3 \cdot 1 + 2 \cdot x_3 = 0 \implies 1 - 1 - 3 + 3 + 2x_3 = 0 \implies x_3 = 0. \end{align*} For $g = (12)$: $\chi_1 = 1$, $\chi_2 = -1$ (odd), $\chi_3 = 1$, $\chi_4 = -1$, so \begin{align*} 1 \cdot 1 + 1 \cdot (-1) + 3 \cdot 1 + 3 \cdot (-1) + 2 \cdot x_4 = 0 \implies 1 - 1 + 3 - 3 + 2x_4 = 0 \implies x_4 = 0. \end{align*} So $\chi_5$ takes values $2, 2, -1, 0, 0$ on the five conjugacy classes. The complete character table is: | | $1$ | $(12)(34)$ | $(123)$ | $(1234)$ | $(12)$ | |---|---|---|---|---|---| | **class size** | $1$ | $3$ | $8$ | $6$ | $6$ | | $\chi_1$ | $1$ | $1$ | $1$ | $1$ | $1$ | | $\chi_2$ | $1$ | $1$ | $1$ | $-1$ | $-1$ | | $\chi_3$ | $3$ | $-1$ | $0$ | $-1$ | $1$ | | $\chi_4$ | $3$ | $-1$ | $0$ | $1$ | $-1$ | | $\chi_5$ | $2$ | $2$ | $-1$ | $0$ | $0$ | As a consistency check, one verifies that $\chi_{\mathrm{reg}} = \chi_1 + \chi_2 + 3\chi_3 + 3\chi_4 + 2\chi_5$ by computing the regular character directly (it equals $|G| = 24$ at the identity and $0$ elsewhere) and using the formula $\chi_{\mathrm{reg}} = \sum_i d_i \chi_i$. The character $\chi_5$ can also be understood conceptually: since $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $S_4$ with $S_4/V_4 \cong S_3$, the irreducible representations of $S_3$ lift to representations of $S_4$. The group $S_3 \cong D_6$ has a two-dimensional irreducible representation, and lifting it to $S_4$ gives $\chi_5$. [/example] ## Conjugacy Classes in $A_n$ When we pass from $S_n$ to $A_n$, the conjugacy classes of $S_n$ may split. This matters for representations because the number of irreducible characters of $A_n$ equals the number of conjugacy classes in $A_n$, which differs from that of $S_n$ precisely when splitting occurs. Understanding which classes split and how many pieces they produce is therefore the first step in determining the full representation theory of the alternating group. For $g \in A_n$, consider the conjugacy classes $\mathcal{C}_{S_n}(g)$ and $\mathcal{C}_{A_n}(g)$. Since $A_n$ has index $2$ in $S_n$, the centraliser satisfies either $C_{A_n}(g) = C_{S_n}(g)$ (index $1$) or $[C_{S_n}(g) : C_{A_n}(g)] = 2$. By the orbit-stabiliser theorem, these cases correspond to: - **No splitting**: $|\mathcal{C}_{A_n}(g)| = |\mathcal{C}_{S_n}(g)|$, which happens when $C_{A_n}(g) = C_{S_n}(g)$. - **Splitting**: $|\mathcal{C}_{A_n}(g)| = \frac{1}{2}|\mathcal{C}_{S_n}(g)|$, and $\mathcal{C}_{S_n}(g)$ splits into two $A_n$-conjugacy classes of equal size. [quotetheorem:2436]Since $C_{A_n}(g) = C_{S_n}(g) \cap A_n$, the centraliser $C_{A_n}(g)$ has index $1$ in $C_{S_n}(g)$ when $C_{S_n}(g)$ already contains an odd permutation, and index $2$ otherwise. By orbit-stabiliser applied to the conjugation action of $A_n$ on itself, the size of the $A_n$-conjugacy class of $g$ equals $[A_n : C_{A_n}(g)]$. When $C_{A_n}(g)$ has index $2$ in $C_{S_n}(g)$, we get $[A_n : C_{A_n}(g)] = \frac{|A_n|}{|C_{A_n}(g)|} = \frac{|A_n|}{|C_{S_n}(g)|/2} = 2 \cdot \frac{|A_n|}{|C_{S_n}(g)|} = \frac{1}{2}|\mathcal{C}_{S_n}(g)|$, confirming the splitting with the two $A_n$-classes each of half the original size. In terms of cycle types, a permutation $g \in A_n$ fails to commute with any odd permutation precisely when all cycles of $g$ have distinct odd lengths. For example, a $5$-cycle in $A_5$ has cycle type $(5)$ — one cycle of length $5$, which is odd and distinct — so no odd permutation centralises it, and the corresponding $S_5$-class splits in $A_5$. A $3$-cycle in $A_5$ has cycle type $(3, 1, 1)$: the two fixed points give repeated cycle length $1$, so the class does not split. This cycle-type criterion is what one checks in practice when computing the conjugacy classes of $A_n$: for each even cycle type in $S_n$, ask whether all cycle lengths are distinct odd numbers. [example: Splitting in $A_3$] Consider $\sigma = (123) \in A_3$. In $S_3$, the conjugacy class of $\sigma$ is $\{(123), (132)\}$, with centraliser $C_{S_3}(\sigma) = \{e, (123), (132)\} \cong C_3$ — all even permutations in $S_3$. Since no odd permutation centralises $\sigma$, the class splits in $A_3$: $\mathcal{C}_{A_3}(\sigma) = \{(123)\}$ and $\mathcal{C}_{A_3}(\sigma^{-1}) = \{(132)\}$ are distinct conjugacy classes of size $1$ each. For contrast, consider $e \in A_3$. Its $S_3$-centraliser is all of $S_3$, which contains odd permutations (e.g., any transposition). So the $S_3$-class $\{e\}$ does not split, and $\mathcal{C}_{A_3}(e) = \{e\}$ remains a single class. The identity class never splits in any $A_n$, for precisely this reason: $C_{S_n}(e) = S_n$ always contains odd permutations. The result is that $A_3$ has three conjugacy classes ($\{e\}$, $\{(123)\}$, $\{(132)\}$), giving three irreducible characters; but two of these characters — those corresponding to the split pair — are complex conjugates of each other, a structural feature that recurs whenever classes split. [/example] This criterion, together with the $2$-transitivity technique for finding irreducible characters, provides the main tools for analysing the representation theory of $S_n$ and $A_n$ concretely. As we build more complex representations, we need to understand how representations of normal subgroups relate to those of the whole group. Chapter 8 develops the lifting lemma and shows how character kernels and the derived subgroup constrain which representations can be lifted. # 8. Normal subgroups and lifting The preceding chapters developed character theory as a tool for classifying irreducible representations of a finite group $G$. A natural question arises: if $G$ has a normal subgroup $N$, how do the representations of the simpler quotient group $G/N$ relate to those of $G$ itself? This chapter answers that question through the **lifting** construction, which transports representations from $G/N$ back to $G$. We then identify the derived subgroup $G'$ as the canonical normal subgroup governing the one-dimensional representations of $G$, and close with a character-theoretic characterisation of which normal subgroups a group possesses. ## Lifting Representations from a Quotient Suppose we want to classify all irreducible characters of $G$ whose kernel contains a given normal subgroup $N \lhd G$. Such characters must factor through the quotient $G/N$, so the question reduces to understanding representations of the (often simpler) group $G/N$ and importing them back to $G$. The normality of $N$ is essential: only when $N \lhd G$ does $G/N$ carry a well-defined group structure, and without it the construction breaks entirely. If $H \leq G$ is merely a subgroup, the coset space $G/H$ is not a group in general, and there is no natural homomorphism $G \to G/H$ to compose with. [definition: Lifting of a Representation] Let $N \lhd G$ and let $\tilde{\rho}: G/N \to \operatorname{GL}(V)$ be a representation of $G/N$. The **lift** of $\tilde{\rho}$ to $G$ is the composition \begin{align*} \rho: G \xrightarrow{\text{natural}} G/N \xrightarrow{\tilde{\rho}} \operatorname{GL}(V), \end{align*} explicitly $\rho(g) = \tilde{\rho}(gN)$ for all $g \in G$. If $\chi$ is the character of $\rho$ and $\tilde{\chi}$ is the character of $\tilde{\rho}$, we say $\tilde{\chi}$ **lifts to** $\chi$. [/definition] The definition is transparent, but the key point is that lifting preserves irreducibility and yields a precise bijection. The following lemma makes this precise. [quotetheorem:2437] [citeproof:2437] The lifting lemma is the fundamental tool of this chapter. It says that the irreducible representation theory of $G$ is not entirely intrinsic to $G$: some irreducibles "come from" simpler quotients. Specifically, those irreducibles whose kernel contains $N$ are in bijection with the entire irreducible theory of $G/N$. The novelty lies in identifying which normal subgroup $N$ to quotient by, and the derived subgroup provides a canonical answer. ## The Derived Subgroup and Linear Characters A one-dimensional representation $\chi: G \to \mathbb{C}^\times$ is automatically a group homomorphism into an abelian group. The kernel of any such character therefore contains all commutators $[a,b] = aba^{-1}b^{-1}$, because $\chi([a,b]) = \chi(a)\chi(b)\chi(a)^{-1}\chi(b)^{-1} = 1$. This observation singles out a special normal subgroup of $G$: the subgroup generated by all commutators is the smallest normal subgroup $N$ for which $G/N$ is abelian, and quotienting by it produces all linear characters of $G$ via lifting. [definition: Derived Subgroup] The **derived subgroup** (or **commutator subgroup**) of $G$ is \begin{align*} G' = \langle [a, b] : a, b \in G \rangle, \end{align*} where $[a, b] = aba^{-1}b^{-1}$. [/definition] [quotetheorem:2438] [citeproof:2438] The theorem gives a clean formula: the number of degree-$1$ irreducibles of $G$ equals the index $|G:G'|$. For abelian $G$ this is $|G|$ (every irreducible is linear), while for groups where $G' = G$ — called **perfect groups** — the only degree-$1$ character is the constant character sending every $g$ to $1$. The alternating group $A_5$ is the smallest perfect group: it has $A_5' = A_5$ because $A_5$ is simple and non-abelian (so any normal subgroup generated by commutators must be all of $A_5$), and consequently $A_5$ has exactly one linear character, the constant character $1_{A_5}$. This is confirmed by the character table of $A_5$, which has no degree-$1$ row other than the constant character. [example: Linear Characters of $S_n$] Let $G = S_n$ with $n \geq 3$. We claim $S_n' = A_n$. Since the sign homomorphism $\operatorname{sgn}: S_n \to \{\pm 1\}$ satisfies $\operatorname{sgn}([a,b]) = \operatorname{sgn}(a)\operatorname{sgn}(b)\operatorname{sgn}(a)^{-1}\operatorname{sgn}(b)^{-1} = 1$, every commutator lies in $A_n$, so $S_n' \leq A_n$. For the reverse inclusion, observe that \begin{align*} (1\;2\;3) = (1\;3\;2)(1\;2)(1\;3\;2)^{-1}(1\;2)^{-1} \in S_n'. \end{align*} Now let $(i\;j\;k)$ be any $3$-cycle. There exists $\sigma \in S_n$ with $\sigma(1) = i$, $\sigma(2) = j$, $\sigma(3) = k$, so $(i\;j\;k) = \sigma(1\;2\;3)\sigma^{-1}$. Since $S_n'$ is normal in $S_n$, it is closed under conjugation, so $(i\;j\;k) = \sigma(1\;2\;3)\sigma^{-1} \in S_n'$. Because $A_n$ is generated by $3$-cycles for $n \geq 3$, we conclude $A_n \leq S_n'$, hence $S_n' = A_n$. Since $|S_n : A_n| = 2$, we get $\ell = 2$: $S_n$ has exactly two linear characters, namely the constant character $1_{S_n}$ sending every permutation to $1$, and the sign character $\operatorname{sgn}$ sending even permutations to $1$ and odd permutations to $-1$. This is consistent with the character table of $S_n$, where both these characters appear as the first two rows. [/example] [example: Linear Characters of $D_{2n}$] Let $G = D_{2n} = \langle r, s \mid r^n = s^2 = 1,\, srs^{-1} = r^{-1} \rangle$ be the dihedral group of order $2n$. We compute $G'$ and the linear characters of $G$. The commutator $[r, s] = rsr^{-1}s^{-1} = r \cdot r \cdot s \cdot s = r^2$ (using $srs^{-1} = r^{-1}$ to get $sr^{-1}s^{-1} = r$, hence $[r,s] = rsr^{-1}s^{-1} = r \cdot r = r^2$). The subgroup generated by $r^2$ in $\langle r \rangle \cong \mathbb{Z}/n\mathbb{Z}$ is $\langle r^2 \rangle$, which has order $\lceil n/2 \rceil$. More precisely, $G' = \langle r^2 \rangle$. **Case $n$ odd:** $\langle r^2 \rangle = \langle r \rangle$ (since $\gcd(2,n)=1$ means $r^2$ generates the cyclic group of order $n$), so $G' = \langle r \rangle$ and $|G : G'| = 2$. Hence $D_{2n}$ has exactly $2$ linear characters when $n$ is odd: the constant character and the character sending $r \mapsto 1$, $s \mapsto -1$. **Case $n$ even:** $\langle r^2 \rangle$ has order $n/2$ and index $4$ in $G$, so $G/G' \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Hence $D_{2n}$ has exactly $4$ linear characters when $n$ is even. They are the four homomorphisms $G \to \{\pm 1\}$ determined by the independent choices $r \mapsto \pm 1$ and $s \mapsto \pm 1$ (subject only to $(r\mapsto -1, s\mapsto -1)$ being the fourth option since $r^n = 1$ forces the image of $r$ to have order dividing $n$, and $n$ even allows $r \mapsto -1$). This example illustrates how the parity of $n$ changes the structure of $G/G'$, and thus the number of linear characters. [/example] ## Normal Subgroups and Character Kernels The lifting lemma shows that every normal subgroup gives rise to irreducible characters of $G$ (via lifting from the quotient). A converse holds: every normal subgroup of $G$ can be recovered as an intersection of character kernels. This gives a completely representation-theoretic description of the normal subgroup structure of $G$. [quotetheorem:2439] [citeproof:2439] The two directions of part (2) together show that the map $N \mapsto \{\chi_i : N \leq \ker \chi_i\}$ is injective on normal subgroups: distinct normal subgroups are distinguished by which irreducible characters vanish on them. [remark: Simplicity via Characters] Unwinding the theorem: $G$ is simple if and only if every non-constant irreducible character has trivial kernel. Equivalently, for any $1 \neq g \in G$ and any non-constant irreducible $\chi$, we must have $\chi(g) \neq \chi(1)$. This gives a purely character-theoretic criterion for simplicity, useful when computing character tables. [/remark] The two results of this chapter work in complementary directions. The Lifting Lemma constructs characters of $G$ from characters of simpler quotients $G/N$; the intersection-of-kernels theorem reconstructs every normal subgroup from character data. Together, they show that the character table of $G$ encodes the entire normal subgroup lattice of $G$ — a striking illustration of how representation theory sees the internal structure of a group. To construct new representations from old, we introduce dual representations and tensor products. Chapter 9 studies these fundamental operations, including symmetric and exterior powers, and shows that the irreducible characters form a ring under multiplication. # 9. Dual spaces and tensor products of representations The preceding chapters built up the machinery of character theory for representations of finite groups. Here we turn to several fundamental linear algebra constructions — dual spaces, tensor products, symmetric and exterior powers, and tensor algebras — and show how each gives rise to new representations. The central payoff is arithmetic: characters of the new representations are expressed directly in terms of the original characters, so the character table of a group encodes a surprising amount of algebraic structure. We also formalise this structure in the character ring, which captures the full additive and multiplicative algebra of characters. ## Dual Spaces Given a representation $\rho: G \to \operatorname{GL}(V)$, it is natural to ask whether the dual space $V^* = \operatorname{Hom}(V, \mathbb{F})$ can itself be made into a $G$-representation. The first attempt is to define $(\rho^*(g)\varphi)(v) = \varphi(\rho(g)v)$, but this fails: one computes $\rho^*(g_1)\rho^*(g_2) = \rho^*(g_2 g_1)$, which has the multiplication reversed. The fix is to use $g^{-1}$ in the formula. [definition: Dual Representation] Let $\rho: G \to \operatorname{GL}(V)$ be a representation over $\mathbb{F}$. The **dual representation** $\rho^*: G \to \operatorname{GL}(V^*)$ is defined by \begin{align*} (\rho^*(g)\varphi)(v) = \varphi(\rho(g^{-1})v) \end{align*} for $g \in G$, $\varphi \in V^*$, $v \in V$. [/definition] The $g^{-1}$ in the definition is what makes the dual representation genuinely a homomorphism $G \to \operatorname{GL}(V^*)$. Once we know $\rho^*$ is a representation, the next question is how its character relates to that of $\rho$ — and the answer is elegantly simple. [quotetheorem:2440] [citeproof:2440] The formula $\chi_{\rho^*}(g) = \chi_\rho(g^{-1})$ has an important consequence over $\mathbb{F} = \mathbb{C}$: since characters satisfy $\chi_\rho(g^{-1}) = \overline{\chi_\rho(g)}$, the dual character is simply the complex conjugate of the original character. [definition: Self-Dual Representation] A representation $\rho: G \to \operatorname{GL}(V)$ is **self-dual** if there exists a $G$-isomorphism $V \cong V^*$. [/definition] As vector spaces (ignoring the group action), $V$ and $V^*$ are always isomorphic when $\dim V < \infty$. Being self-dual is the strictly stronger condition that an isomorphism exists which also intertwines the $G$-actions. Over $\mathbb{C}$, this is equivalent to requiring $\chi_\rho(g) = \overline{\chi_\rho(g)}$ for all $g$, i.e.\ the character is real-valued. [example: Self-Dual Representations of $S_n$] All irreducible representations of $S_n$ are self-dual. The reason is that every element $\sigma \in S_n$ is conjugate to its own inverse: since the cycle type of $\sigma$ and $\sigma^{-1}$ are the same, they lie in the same conjugacy class, so $\chi(\sigma) = \chi(\sigma^{-1})$ for every character $\chi$. This forces every character to be real-valued. The same argument does not apply to $A_n$ in general, and some irreducible representations of $A_n$ are not self-dual. [/example] The dual representation is one way to produce a new representation from an old one. A richer construction — one that also gives the chapter its multiplicative structure — is the tensor product. ## Tensor Products How can we "multiply" two representations to produce a third? Given $\rho: G \to \operatorname{GL}(V)$ and $\rho': G \to \operatorname{GL}(V')$, neither the direct sum nor any obvious pointwise construction makes the characters multiply. The right answer is the tensor product: it equips $V \otimes V'$ with a $G$-action whose character is the pointwise product $\chi_\rho \chi_{\rho'}$. Before connecting this to representations, we recall the linear algebra construction. [definition: Tensor Product] Let $V$ and $W$ be vector spaces over $\mathbb{F}$ with $\dim V = m$ and $\dim W = n$. Fix bases $v_1, \ldots, v_m$ and $w_1, \ldots, w_n$. The **tensor product** $V \otimes W = V \otimes_\mathbb{F} W$ is the $mn$-dimensional vector space with basis the formal symbols $\{v_i \otimes w_j : 1 \leq i \leq m,\, 1 \leq j \leq n\}$. For $v = \sum \alpha_i v_i$ and $w = \sum \beta_j w_j$, the pure tensor is \begin{align*} v \otimes w = \sum_{i,j} \alpha_i \beta_j\, (v_i \otimes w_j). \end{align*} [/definition] Not every element of $V \otimes W$ is a pure tensor. For instance, $v_1 \otimes w_1 + v_2 \otimes w_2$ cannot be written as $v \otimes w$ for any $v \in V$, $w \in W$; it is a genuine linear combination. The definition is equivalent to the construction obtained by taking all formal symbols $v \otimes w$ and quotienting by the bilinearity relations: \begin{align*} (\lambda v) \otimes w &= \lambda(v \otimes w) = v \otimes (\lambda w), \\ (x_1 + x_2) \otimes y &= (x_1 \otimes y) + (x_2 \otimes y), \\ x \otimes (y_1 + y_2) &= (x \otimes y_1) + (x \otimes y_2). \end{align*} These are precisely the axioms for a bilinear map, which leads to the **universal property**: $V \otimes W$ together with the canonical bilinear map $\phi: V \times W \to V \otimes W$, $(v, w) \mapsto v \otimes w$, satisfies the condition that every bilinear map $V \times W \to U$ factors uniquely through $\phi$ via a linear map $V \otimes W \to U$. This universal property characterises $V \otimes W$ up to isomorphism, independent of any basis choice. The basis-independence can be verified directly: if $\{e_1, \ldots, e_m\}$ is any other basis of $V$ and $\{f_1, \ldots, f_n\}$ any basis of $W$, then $\{e_i \otimes f_j\}$ is again a basis of $V \otimes W$, since expressing the original basis vectors in terms of the new ones gives a spanning set of size $mn$ in a space of dimension $mn$. With the linear algebra in place, we now equip $V \otimes V'$ with a $G$-action and read off its character. [quotetheorem:2441] [citeproof:2441] This theorem shows that the pointwise product of two characters of $G$ is again a character of $G$ — a fact that is far from obvious if one thinks only in terms of matrices. The tensor product is the algebraic mechanism that makes character multiplication meaningful. Notice what the theorem does not say: it gives no information about irreducibility. If $\rho$ is irreducible and $\rho'$ has degree $1$, then $\rho \otimes \rho'$ is irreducible (this was established earlier); but for $\deg \rho' > 1$, the tensor product is almost never irreducible — it typically decomposes into several irreducible pieces, and determining which requires a full inner-product calculation against the character table. ## Powers of Characters When $V \otimes V$ is formed from a single representation, how does it decompose into $G$-invariant pieces? The tensor square $V^{\otimes 2} = V \otimes V$ carries extra structure — a natural "swap" symmetry — and this forces a splitting into two $G$-stable subspaces long before any group-theoretic input is used. Working over $\mathbb{C}$, we take $V = V'$ and study this decomposition. Define $\tau: V^{\otimes 2} \to V^{\otimes 2}$ by \begin{align*} \tau\left(\sum \lambda_{ij}\, v_i \otimes v_j\right) = \sum \lambda_{ij}\, v_j \otimes v_i. \end{align*} This is a linear endomorphism with $\tau^2 = \operatorname{id}$, so its eigenvalues are $\pm 1$, and $V^{\otimes 2}$ decomposes into the $+1$ and $-1$ eigenspaces. [definition: Symmetric and Exterior Square] The **symmetric square** and **exterior square** of a $G$-space $V$ are, respectively, \begin{align*} S^2 V &= \{x \in V^{\otimes 2} : \tau(x) = x\}, \\ \Lambda^2 V &= \{x \in V^{\otimes 2} : \tau(x) = -x\}. \end{align*} The exterior square is also called the **anti-symmetric square** or **wedge square**. [/definition] Both $S^2 V$ and $\Lambda^2 V$ are $G$-subspaces of $V^{\otimes 2}$, since the $G$-action commutes with $\tau$ (the group acts on each factor the same way, so swapping factors before or after is the same). The decomposition $V^{\otimes 2} = S^2 V \oplus \Lambda^2 V$ holds: any $x \in V^{\otimes 2}$ splits as \begin{align*} x = \underbrace{\tfrac{1}{2}(x + \tau(x))}_{\in S^2 V} + \underbrace{\tfrac{1}{2}(x - \tau(x))}_{\in \Lambda^2 V}. \end{align*} Note that this symmetrization step requires characteristic $\neq 2$; over a field of characteristic $2$, the factor $\frac{1}{2}$ is undefined and the decomposition fails — both $S^2 V$ and $\Lambda^2 V$ may coincide or overlap. If $\dim V = n$, then $S^2 V$ has basis $\{v_i \otimes v_j + v_j \otimes v_i : 1 \leq i \leq j \leq n\}$ and $\Lambda^2 V$ has basis $\{v_i \otimes v_j - v_j \otimes v_i : 1 \leq i < j \leq n\}$. Counting gives \begin{align*} \dim S^2 V = \tfrac{1}{2}n(n+1), \qquad \dim \Lambda^2 V = \tfrac{1}{2}n(n-1). \end{align*}  [quotetheorem:2442] [citeproof:2442] The formula $\chi_\Lambda(g) = \frac{1}{2}(\chi(g)^2 - \chi(g^2))$ is remarkably useful: the values $\chi(g^2)$ can be read directly from the character table (since $g^2$ belongs to some known conjugacy class), so the characters of $S^2 V$ and $\Lambda^2 V$ are determined by $\chi$ alone. This is what makes the following example feasible with only the partial character table in hand. [example: What Goes Wrong Without Symmetrization in Characteristic Two] If we work over a field of characteristic $2$, the element $\frac{1}{2}(x + \tau(x))$ is meaningless. Consider $V = \mathbb{F}_2^2$ with $G = \mathbb{Z}/2\mathbb{Z}$ acting by swapping the two basis vectors. The map $\tau: V^{\otimes 2} \to V^{\otimes 2}$ still satisfies $\tau^2 = \mathrm{id}$, but since $-1 = 1$ in $\mathbb{F}_2$, both eigenspaces coincide: $S^2 V = \Lambda^2 V = \ker(\tau - \mathrm{id}) = V^{\otimes 2}$. There is no splitting $V^{\otimes 2} = S^2 V \oplus \Lambda^2 V$; the "exterior square" and "symmetric square" are the same subspace, and the decomposition used in the theorem above collapses entirely. [/example] [example: Completing the Character Table of $S_4$] Recall $S_4$ has five conjugacy classes with representatives $1$, $(1\,2)(3\,4)$, $(1\,2\,3)$, $(1\,2\,3\,4)$, $(1\,2)$, of sizes $1, 3, 8, 6, 6$ respectively. The first four irreducible characters are: | | $1$ | $(12)(34)$ | $(123)$ | $(1234)$ | $(12)$ | |---|---|---|---|---|---| | $\chi_1$ | $1$ | $1$ | $1$ | $1$ | $1$ | | $\chi_2$ | $1$ | $1$ | $1$ | $-1$ | $-1$ | | $\chi_3$ | $3$ | $-1$ | $0$ | $-1$ | $1$ | | $\chi_4$ | $3$ | $-1$ | $0$ | $1$ | $-1$ | We use the symmetric and exterior square to find the fifth. Compute $\chi_3(g^2)$: the squares of the conjugacy class representatives are $1$, $1$, $(1\,3\,2)$, $(1\,3)$, $1$, which have $\chi_3$-values $3, 3, 0, -1, 3$. Then: | | $1$ | $(12)(34)$ | $(123)$ | $(1234)$ | $(12)$ | |---|---|---|---|---|---| | $\chi_3^2$ | $9$ | $1$ | $0$ | $1$ | $1$ | | $\chi_3(g^2)$ | $3$ | $3$ | $0$ | $-1$ | $3$ | | $S^2\chi_3$ | $6$ | $2$ | $0$ | $0$ | $2$ | | $\Lambda^2\chi_3$ | $3$ | $-1$ | $0$ | $1$ | $-1$ | We recognise $\Lambda^2\chi_3 = \chi_4$. For $S^2\chi_3$: computing its inner product with itself gives $\langle S^2\chi_3, S^2\chi_3 \rangle = 3 > 1$, so it is not irreducible. Taking inner products with all four known irreducible characters gives $S^2\chi_3 = \chi_1 + \chi_3 + \chi_5$, where | | $1$ | $(12)(34)$ | $(123)$ | $(1234)$ | $(12)$ | |---|---|---|---|---|---| | $\chi_5$ | $2$ | $2$ | $-1$ | $0$ | $0$ | This $\chi_5$ is irreducible (inner product with itself is $1$ and $\chi_5(1) = 2 > 0$), completing the character table of $S_4$. [/example] ## Characters of $G \times H$ Given the irreducible representations of $G$ and those of $H$ separately, can we recover all irreducible representations of $G \times H$? This is not automatic: a direct product has more conjugacy classes than either factor, and naive attempts to combine characters need not produce irreducibles. The tensor product construction gives a definitive answer. [quotetheorem:2443] [citeproof:2443] This is one of the cleanest results in the subject: the representation theory of a direct product is completely determined by the representation theories of its two factors, with no additional data needed. The proof relies on the factored sum decomposition $G \times H = G \times H$ in a very concrete way — each conjugacy class of $G \times H$ is a product of a class of $G$ and a class of $H$, and the inner product factorises accordingly. This argument fails entirely for semidirect products $G \ltimes H$: there the conjugacy classes do not factorise, and the irreducibles of $G \ltimes H$ are not in general products of irreducibles of $G$ and $H$. Determining the representation theory of a semidirect product requires the Mackey machine (or Clifford theory), which is a genuinely harder task. The direct product theorem also connects forward to the Künneth perspective in homological algebra, where the same product formula appears for cohomology. ## Symmetric and Exterior Powers What symmetry types live inside $V^{\otimes n}$ for general $n$? For $n = 2$, the swap operator $\tau$ cut out two pieces. For $n \geq 3$, the entire symmetric group $S_n$ acts on $V^{\otimes n}$ by permuting tensor factors, and the eigenspaces of this action — the symmetric and antisymmetric tensors — are only two of many $S_n$-isotypical components. For a vector space $V$ over $\mathbb{F}$ with $\dim V = d$, define the $n$-fold tensor power \begin{align*} T^n V = V^{\otimes n} = \underbrace{V \otimes \cdots \otimes V}_{n \text{ times}}, \end{align*} with basis $\{v_{i_1} \otimes \cdots \otimes v_{i_n} : i_1, \ldots, i_n \in \{1, \ldots, d\}\}$. There is a natural action of the symmetric group $S_n$ on $T^n V$ permuting the tensor factors: for $\sigma \in S_n$, \begin{align*} \sigma(v_1 \otimes \cdots \otimes v_n) = v_{\sigma^{-1}(1)} \otimes \cdots \otimes v_{\sigma^{-1}(n)}. \end{align*} (The inverse appears to ensure we get a left action under the standard left-to-right composition of permutations.) If $V$ is also a $G$-space via $\rho: G \to \operatorname{GL}(V)$, then $G$ acts on $T^n V$ diagonally: \begin{align*} \rho^{\otimes n}(g)(v_1 \otimes \cdots \otimes v_n) = \rho(g)v_1 \otimes \cdots \otimes \rho(g)v_n. \end{align*} A key observation is that the $G$-action and the $S_n$-action on $T^n V$ **commute** with each other. This is because the $G$-action applies $\rho(g)$ to each factor independently, while $S_n$ merely reorders the factors. This commutativity is the starting point for Schur–Weyl duality, a deep theorem with wide-ranging consequences — though its full development lies beyond this course. [definition: Symmetric and Exterior Power] For a $G$-space $V$, the **$n$th symmetric power** and **$n$th exterior power** are \begin{align*} S^n V &= \{x \in V^{\otimes n} : \sigma(x) = x \text{ for all } \sigma \in S_n\}, \\ \Lambda^n V &= \{x \in V^{\otimes n} : \sigma(x) = (\operatorname{sgn}\sigma)\,x \text{ for all } \sigma \in S_n\}. \end{align*} Both are $G$-subspaces of $V^{\otimes n}$. [/definition] Since the $G$-action and $S_n$-action commute, $G$ preserves both $S^n V$ and $\Lambda^n V$, so these are indeed subrepresentations. [example: Bases and Dimensions of Symmetric and Exterior Powers] Fix a basis $e_1, \ldots, e_d$ of $V$. The symmetrised elements \begin{align*} \frac{1}{n!}\sum_{\sigma \in S_n} e_{i_{\sigma(1)}} \otimes \cdots \otimes e_{i_{\sigma(n)}}, \qquad 1 \leq i_1 \leq i_2 \leq \cdots \leq i_n \leq d, \end{align*} form a basis for $S^n V$, so \begin{align*} \dim S^n V = \binom{d + n - 1}{n}. \end{align*} The antisymmetrised elements \begin{align*} \frac{1}{n!}\sum_{\sigma \in S_n} (\operatorname{sgn}\sigma)\, e_{i_{\sigma(1)}} \otimes \cdots \otimes e_{i_{\sigma(n)}}, \qquad 1 \leq i_1 < i_2 < \cdots < i_n \leq d, \end{align*} form a basis for $\Lambda^n V$, so \begin{align*} \dim \Lambda^n V = \begin{cases} \binom{d}{n} & n \leq d, \\ 0 & n > d. \end{cases} \end{align*} The vanishing for $n > d$ reflects the fact that any antisymmetric tensor over a $d$-dimensional space must vanish when more than $d$ indices are chosen from $\{1, \ldots, d\}$. For $n = 2$, this recovers $\dim S^2 V = \frac{1}{2}d(d+1)$ and $\dim \Lambda^2 V = \frac{1}{2}d(d-1)$, consistent with the earlier decomposition $V^{\otimes 2} = S^2 V \oplus \Lambda^2 V$. For $n > 2$, however, $S^n V \oplus \Lambda^n V \subsetneq V^{\otimes n}$: the tensor power contains many other $S_n$-isotypical components corresponding to other irreducible representations of $S_n$ (classified by Young tableaux). Each such component is $G$-invariant, and understanding them all is a rich part of the subject. [/example] The symmetric and exterior powers each give a single $G$-representation. To package all powers together into one algebraic object, we assemble them into the tensor algebra. ## Tensor Algebra Can we encode all tensor powers of $V$ in a single algebraic object that makes the graded structure transparent? The answer is the tensor algebra — a non-commutative ring whose degree-$n$ piece is exactly $T^n V$, and whose quotients by appropriate ideals recover the symmetric and exterior constructions. [definition: Tensor Algebra] Let $V$ be a vector space over $\mathbb{F}$ (of characteristic $0$), and set $T^0 V = \mathbb{F}$ by convention. The **tensor algebra** of $V$ is the direct sum \begin{align*} T(V) = \bigoplus_{n \geq 0} T^n V, \end{align*} with ring multiplication given by the tensor product: for $x \in T^n V$ and $y \in T^m V$, set $x \cdot y = x \otimes y \in T^{n+m} V$. [/definition] This makes $T(V)$ into a graded, associative, non-commutative ring. Non-commutativity arises because $v \otimes w \neq w \otimes v$ in general. The grading means the degree-$n$ and degree-$m$ components multiply into the degree-$(n+m)$ component. By quotienting the tensor algebra by appropriate ideals, one obtains the symmetric and exterior algebras: [definition: Symmetric and Exterior Algebra] The **symmetric algebra** of $V$ is \begin{align*} S(V) = T(V) \big/ (\text{ideal generated by all } u \otimes v - v \otimes u), \end{align*} and the **exterior algebra** of $V$ is \begin{align*} \Lambda(V) = T(V) \big/ (\text{ideal generated by all } v \otimes v). \end{align*} Here $u, v$ range over arbitrary elements of $T^n V$ for any $n$, not just elements of $V$ itself. [/definition] The symmetric algebra $S(V)$ is commutative, and its degree-$n$ piece is exactly $S^n V$. The exterior algebra $\Lambda(V)$ is **graded-commutative**: for $x \in \Lambda^n(V)$ and $y \in \Lambda^m(V)$, one has $x \cdot y = (-1)^{nm} y \cdot x$. In particular, $v \cdot v = 0$ for every $v \in V \subset \Lambda(V)$. These algebras appear throughout geometry and algebra; in this course, they provide the algebraic framework for the symmetric and exterior powers of representations. ## Character Ring We have seen that sums and products of characters of $G$ are again characters. However, the difference of two characters need not be — it may take negative values at the identity, whereas a genuine character of a representation satisfies $\chi(1) = \dim V \geq 0$. To handle this, we introduce the character ring, which closes the characters under subtraction. To see concretely that differences of characters are not characters: if $\chi_1$ and $\chi_2$ are distinct irreducible characters of $G$, then $\psi = \chi_1 - \chi_2$ satisfies $\psi(1) = \chi_1(1) - \chi_2(1)$, which could be negative, zero, or positive. Even if $\psi(1) > 0$, the function $\psi$ need not be a character: for $G = S_3$ with irreducibles $\chi_1$ (the principal character, degree $1$), $\chi_2$ (the sign character, degree $1$), and $\chi_3$ (standard, degree $2$), the difference $\chi_3 - \chi_1$ has values $1, -1, -1$ on the three conjugacy classes $\{e\}, \{(12)\}, \{(123)\}$. This is not a character of any representation (no representation has degree $1$ with these values), but it is a perfectly well-defined integer linear combination of irreducibles. The character ring is built to contain exactly such objects. Recall that $\mathcal{C}(G)$ denotes the ring of all class functions on $G$. [definition: Character Ring] The **character ring** $R(G)$ is the $\mathbb{Z}$-submodule of $\mathcal{C}(G)$ spanned (over $\mathbb{Z}$) by the irreducible characters of $G$. [/definition] Since products of characters are characters, $R(G)$ is closed under multiplication as well as integer-linear combination, making it a ring. The irreducible characters $\{\chi_i\}$ form a $\mathbb{Z}$-basis for $R(G)$ as a free $\mathbb{Z}$-module: they generate $R(G)$ by definition, and they are $\mathbb{Z}$-linearly independent because they are orthonormal for the inner product on $\mathcal{C}(G)$. [definition: Generalised Character] A **generalised** or **virtual character** is an element of $R(G)$, i.e.\ a class function of the form \begin{align*} \psi = \sum_\chi n_\chi \chi, \end{align*} where the sum is over all irreducible characters $\chi$ and $n_\chi \in \mathbb{Z}$. [/definition] Every generalised character is a difference of two genuine characters: given $\psi = \sum n_\chi \chi$, set $\alpha = \sum_{n_\chi \geq 0} n_\chi \chi$ and $\beta = \sum_{n_\chi < 0} (-n_\chi)\chi$. Then $\alpha$ and $\beta$ are characters (non-negative integer linear combinations of irreducibles), and $\psi = \alpha - \beta$. The following lemma gives a useful criterion for detecting irreducible characters among generalised characters. [quotetheorem:2444] [citeproof:2444] The hypothesis $\alpha(1) > 0$ is necessary: the generalised character $-\chi_i$ also satisfies $\langle -\chi_i, -\chi_i \rangle = 1$, but $(-\chi_i)(1) = -\dim V_i < 0$. Without checking the sign at the identity, the norm-$1$ condition alone only guarantees $\alpha = \pm\chi_i$ for some irreducible $\chi_i$; the positivity condition selects the $+$ sign and thereby ensures $\alpha$ is a genuine character rather than the negative of one. This criterion is used throughout the subject: when one constructs a class function by some process (e.g.\ decomposing a tensor product), one can verify irreducibility by computing the inner product with itself and checking it equals $1$, without needing to compare against all other irreducible characters explicitly. The character ring thus provides both a clean algebraic structure for the set of all representations and a practical tool for character computations. Tensor products and duals are operations on a fixed group; now we need tools to move representations between a group and its subgroups. Chapter 10 proves Frobenius reciprocity and the centralizer formula, giving us precise formulas for how induction and restriction of representations interact. # 10. Induction and restriction This chapter develops the two fundamental operations that relate representations of a group $G$ to representations of a subgroup $H \leq G$: restriction and induction. Restriction is the easy direction — any representation of $G$ yields one of $H$ simply by forgetting about elements outside $H$. Induction is the reverse process, manufacturing a representation of the larger group $G$ from data defined only on $H$. The key tool connecting these two operations is Frobenius reciprocity, which expresses an adjunction between induction and restriction at the level of inner products of characters. Throughout this chapter we work over $\mathbb{F} = \mathbb{C}$. ## Restriction When we restrict attention from $G$ to a subgroup $H$, what happens to an irreducible representation? Does it remain irreducible, or does it break into smaller pieces? This is the foundational question of restriction theory: understanding how $G$-representations decompose when we forget the action of elements outside $H$. Given a subgroup $H \leq G$, every representation of $G$ restricts to a representation of $H$ in a completely natural way. [definition: Restriction] Let $\rho: G \to \operatorname{GL}(V)$ be a representation affording character $\chi$. The **restriction** of $\rho$ to $H$ is the representation \begin{align*} \operatorname{Res}^G_H \rho = \rho_H = \rho\downarrow_H : H \to \operatorname{GL}(V), \end{align*} obtained by viewing $V$ as an $H$-space via $\rho|_H$. It affords the character $\operatorname{Res}^G_H \chi = \chi\downarrow_H$, which is simply the restriction of the function $\chi: G \to \mathbb{C}$ to the subset $H$. [/definition] The character of the restricted representation is just $\chi$ evaluated at group elements of $H$ — there is nothing to compute. However, even when $\chi$ is irreducible as a character of $G$, its restriction $\chi\downarrow_H$ need not be irreducible as a character of $H$. For example, restricting any non-linear irreducible character to the trivial subgroup $\{1\}$ produces a reducible character of degree $> 1$, since the only irreducible character of the trivial group is the trivial character of degree $1$. The next two lemmas quantify how irreducible characters of $G$ relate to those of $H$ under restriction. [quotetheorem:2445] [citeproof:2445] This covering result guarantees that induction and restriction together account for the full representation theory of $H$: no irreducible of $H$ is invisible from $G$. It gives no multiplicity information, however — an irreducible $\psi$ of $H$ might appear in many restricted characters, or just one. For that finer control, Frobenius reciprocity (proved in the next section) is the right tool. [quotetheorem:2446] [citeproof:2446] [remark: Usefulness of the Coefficient Bound] This bound is most useful when the index $|G:H|$ is small, such as $2$ or $3$. If $|G:H| = 2$, then $\sum c_i^2 \leq 2$, which forces the restriction to have at most two irreducible constituents with multiplicity at most $\sqrt{2}$, hence at most two distinct constituents each with multiplicity $1$. [/remark] [example: Restriction from $S_3$ to $A_3$] Let $G = S_3$ and $H = A_3 \cong C_3$. The index is $|S_3 : A_3| = 2$, so the coefficient bound gives $\sum c_i^2 \leq 2$. The irreducible characters of $S_3$ are: the trivial character $1$, the sign character $\epsilon$, and the standard character $\chi$ of degree $2$. The irreducible characters of $A_3 \cong C_3$ are the three cube-root-of-unity characters $\psi_0, \psi_1, \psi_2$ where $\psi_k((123)) = \omega^k$ and $\omega = e^{2\pi i/3}$. For $1\downarrow_{A_3}$: every element of $A_3$ maps to $1$, so $1\downarrow_{A_3} = \psi_0$. Similarly $\epsilon\downarrow_{A_3} = \psi_0$ since the sign character is also $1$ on even permutations. Both the principal character and the sign character restrict to the same character of $A_3$ — a clear example where $G$-irreducibles merge under restriction. For $\chi\downarrow_{A_3}$: the degree-$2$ character $\chi$ takes values $\chi(1) = 2$, $\chi((123)) = -1$, $\chi((12)) = 0$. Restricting to $A_3 = \{1, (123), (132)\}$: the character values are $2, -1, -1$. Decomposing: $\langle \chi\downarrow_{A_3}, \psi_k\rangle_{A_3} = \frac{1}{3}(2\cdot 1 + (-1)\omega^{-k} + (-1)\omega^k) = \frac{1}{3}(2 - 2\mathrm{Re}(\omega^k))$. For $k=0$ this gives $\frac{1}{3}(2-2) = 0$. For $k=1$ and $k=2$: $\frac{1}{3}(2 - 2\cdot(-1/2)) = \frac{1}{3}(2 + 1) = 1$. So $\chi\downarrow_{A_3} = \psi_1 + \psi_2$, confirming $\sum c_i^2 = 1^2 + 1^2 = 2 = |G:H|$ — the bound is attained, and $\chi$ vanishes on $G \setminus A_3$ (transpositions) confirms the equality condition. [/example] ## Induced Class Functions Given a class function on $H$, we want to produce one on all of $G$. The naive attempt — extend by zero outside $H$ — immediately fails: the zero extension of a class function on $H$ is not in general a class function on $G$. To see why, take $\psi$ to be the trivial character of $H$ and form $\mathring{\psi}(y) = \mathbf{1}_H(y)$. For this to be a class function of $G$, it must be constant on $G$-conjugacy classes. But if $g \in H$ and $x \in G \setminus H$, the conjugate $x^{-1}gx$ might lie outside $H$, making $\mathring{\psi}(x^{-1}gx) = 0 \neq 1 = \mathring{\psi}(g)$. The zero extension is an $H$-class function extended to $G$, but $G$-class functions must respect $G$-conjugacy, not merely $H$-conjugacy. The fix is to symmetrise over all $G$-conjugates, averaging out this asymmetry. [definition: Induced Class Function] Let $\psi \in \mathcal{C}(H)$ be a class function of $H$. Define the **zero extension** $\mathring{\psi}: G \to \mathbb{C}$ by \begin{align*} \mathring{\psi}(y) = \begin{cases} \psi(y) & y \in H \\ 0 & y \notin H. \end{cases} \end{align*} The **induced class function** $\operatorname{Ind}^G_H \psi = \psi\uparrow^G = \psi^G$ is defined by \begin{align*} \operatorname{Ind}^G_H \psi(g) = \frac{1}{|H|} \sum_{x \in G} \mathring{\psi}(x^{-1}gx). \end{align*} [/definition] The sum ranges over all $x \in G$, but the term for $x$ is nonzero only when $x^{-1}gx \in H$. The averaging by $\frac{1}{|H|}$ compensates for the fact that elements within the same coset of $H$ give the same conjugate, as we will see. [quotetheorem:2447] [citeproof:2447] The degree formula $\operatorname{Ind}^G_H\psi(1) = |G:H|\psi(1)$ passes a basic sanity check: induction scales dimension by the index. If $\psi$ is the trivial character of $H$ (degree $1$), the induced representation should have dimension $|G:H|$ — equal to the number of cosets that $G$ permutes, which matches the permutation representation on $G/H$ that we will verify shortly. The formula involving the full sum over $G$ is not always easy to use. There is a cleaner formula using coset representatives. Let $n = |G:H|$ and let $t_1 = 1, t_2, \ldots, t_n$ be a **left transversal** of $H$ in $G$, meaning $t_1 H, t_2 H, \ldots, t_n H$ are precisely the $n$ left cosets of $H$ in $G$. [quotetheorem:2448] [citeproof:2448] The transversal formula makes clear that the induced class function picks up contributions only from those coset representatives $t_i$ for which $t_i^{-1}g t_i \in H$ — that is, for which $g$ is conjugate (in $G$) to something in $H$ via $t_i$. ## Frobenius Reciprocity Induction and restriction are inverse-direction operations, and one natural question is whether they are truly adjoint: if $\psi$ is a character of $H$ and $\varphi$ is a character of $G$, how does the multiplicity of $\psi$ in $\operatorname{Res}^G_H\varphi$ compare to the multiplicity of $\varphi$ in $\operatorname{Ind}^G_H\psi$? Frobenius reciprocity answers this completely — the two multiplicities are equal — and does so in the sharpest possible form. [quotetheorem:2449] [citeproof:2449] The content of Frobenius reciprocity is that computing constituents of induced characters reduces to computing constituents of restricted characters, and vice versa. In practice this means: if you want to know how many times an irreducible $\chi$ of $G$ appears in $\operatorname{Ind}^G_H\psi$, you compute $\langle \psi, \operatorname{Res}^G_H\chi\rangle_H$ — a calculation entirely within $H$, which is typically a much smaller group. This inversion of perspective is the principal computational power of the theorem. [remark: Categorical Meaning] Frobenius reciprocity says that restriction and induction are adjoint functors: restriction is a right adjoint to induction. Concretely, it equates inner products of a character of $G$ with the induction of $\psi$, and inner products of its restriction to $H$ with $\psi$ directly. [/remark] The most important consequence is that induction of a character is again a character. [quotetheorem:2450] [citeproof:2450] This result is not obvious from the definition: the induced class function is defined by a signed averaging formula, and it is a priori possible for multiplicities to be negative or fractional. Frobenius reciprocity is what rules this out. Concretely, the theorem provides a systematic way to build new irreducible characters of a large group $G$ by inducing characters from manageable subgroups — a strategy that underlies the construction of the character tables of symmetric groups via Young's theory. ## An Explicit Formula via Conjugacy Classes The transversal formula for $\operatorname{Ind}^G_H\psi$ requires explicit coset representatives and is cumbersome when working with a character table of $G$ whose conjugacy classes are the natural input. We need a formula expressed directly in terms of centralizer orders and intersections of conjugacy classes. When we intersect a $G$-conjugacy class $\mathcal{C}_G(g)$ with $H$, the result $\mathcal{C}_G(g) \cap H$ need not be a single $H$-conjugacy class. Two elements of $H$ that are conjugate in $G$ need not be conjugate in $H$, since the conjugating element might lie outside $H$. However, $\mathcal{C}_G(g) \cap H$ is always a union of $H$-conjugacy classes. [quotetheorem:2451] [citeproof:2451] The case $m = 0$ deserves emphasis: if the entire $G$-conjugacy class of $g$ is disjoint from $H$, the induced character vanishes at $g$ entirely. This is the mechanism behind many of the zero entries we computed in the $C_4$ to $S_4$ example below. The case $m \geq 1$ involves a sum over double cosets $C_G(g) \backslash G / H$: each such double coset contributes one $H$-conjugacy class inside $\mathcal{C}_G(g) \cap H$, and the centralizer orders record the relative sizes of these contributions. This double-coset structure is the bridge between the combinatorics of group conjugacy and the linear algebra of induction. [example: Inducing a Character from $C_4$ to $S_4$] Let $H = C_4 = \langle (1\;2\;3\;4) \rangle \leq G = S_4$. The index is $|S_4 : C_4| = 6$. Let $\alpha$ be the faithful one-dimensional character of $C_4$ defined by $\alpha((1\;2\;3\;4)) = i$. The character table of $\alpha$ on $C_4$: | $1$ | $(1\;2\;3\;4)$ | $(1\;3)(2\;4)$ | $(1\;4\;3\;2)$ | |-----|----------------|----------------|----------------| | $1$ | $i$ | $-1$ | $-i$ | We compute $\chi = \operatorname{Ind}^{S_4}_{C_4}(\alpha)$ using the centralizer formula. **Degree:** $\chi(1) = |G:H|\cdot\alpha(1) = 6$. **Transpositions $(1\;2)$:** We need to check whether any element of $S_4$ in the conjugacy class of $(1\;2)$ lies in $C_4$. The elements of $C_4$ are $\{1, (1\;2\;3\;4), (1\;3)(2\;4), (1\;4\;3\;2)\}$, which have cycle types $1^4$, $4$, $2^2$, and $4$ respectively. The transposition $(1\;2)$ has cycle type $2 + 1^2$, which does not appear among the cycle types in $C_4$. Since cycle type is preserved under conjugation in $S_4$, no element of $S_4$ conjugates $(1\;2)$ into $C_4$. Thus $m = 0$ and $\chi((1\;2)) = 0$. **Three-cycles $(1\;2\;3)$:** Similarly, three-cycles have cycle type $3 + 1$, which does not appear in $C_4$ (whose elements have cycle types $1^4$, $4$, $2^2$, $4$). So $m = 0$ and $\chi((1\;2\;3)) = 0$. **Class $(1\;2)(3\;4)$:** In $S_4$, there are $3$ elements in this conjugacy class, and exactly one lies in $C_4$, namely $(1\;3)(2\;4)$, which has cycle type $2^2$ matching that of $(1\;2)(3\;4)$. Since $C_4$ is abelian, $|C_{C_4}((1\;3)(2\;4))| = |C_4| = 4$. For the centralizer in $S_4$: $|C_{S_4}((1\;2)(3\;4))| = |S_4|/|\mathcal{C}_{S_4}((1\;2)(3\;4))| = 24/3 = 8$. Applying the formula: \begin{align*} \chi((1\;2)(3\;4)) = 8 \cdot \frac{\alpha((1\;3)(2\;4))}{4} = 8 \cdot \frac{-1}{4} = -2. \end{align*} **Class $(1\;2\;3\;4)$:** The conjugacy class of $4$-cycles in $S_4$ has $|S_4|/|C_{S_4}((1\;2\;3\;4))| = 24/4 = 6$ elements. Two of these lie in $C_4$: namely $(1\;2\;3\;4)$ with $\alpha$-value $i$ and $(1\;4\;3\;2)$ with $\alpha$-value $-i$. Both are self-conjugate in the abelian group $C_4$, so $|C_{C_4}((1\;2\;3\;4))| = |C_{C_4}((1\;4\;3\;2))| = 4$: \begin{align*} \chi((1\;2\;3\;4)) = 4\left(\frac{i}{4} + \frac{-i}{4}\right) = 0. \end{align*} The induced character table: | | $1$ | $(1\;2)$ | $(1\;2\;3)$ | $(1\;2)(3\;4)$ | $(1\;2\;3\;4)$ | |--|-----|----------|-------------|----------------|----------------| | class size | $1$ | $6$ | $8$ | $3$ | $6$ | | $\operatorname{Ind}^{S_4}_{C_4}(\alpha)$ | $6$ | $0$ | $0$ | $-2$ | $0$ | [/example] ## The Induced Representation So far we have constructed $\operatorname{Ind}^G_H\psi$ as a class function and shown it is a genuine character. Is induction connected to any concrete construction we already know — can we actually build a $G$-space, not just a function on $G$? Let $H \leq G$ have index $n$, and let $t_1 = 1, t_2, \ldots, t_n$ be a left transversal. Let $W$ be an $H$-space.  [definition: Induced Representation] The **induced representation** $\operatorname{Ind}^G_H W$ is the vector space \begin{align*} V = W \oplus (t_2 \otimes W) \oplus \cdots \oplus (t_n \otimes W), \end{align*} where $t_i \otimes W = \{t_i \otimes w : w \in W\}$ is a formal copy of $W$ labelled by the coset representative $t_i$. The $G$-action is defined as follows: for $g \in G$ and each coset representative $t_i$, there is a unique index $j$ such that $t_j^{-1} g t_i \in H$ (equivalently, $gt_i H = t_j H$), and we set \begin{align*} g(t_i \otimes w) = t_j \otimes \bigl((t_j^{-1}gt_i)w\bigr). \end{align*} [/definition] The action makes intuitive sense: $g$ first permutes the cosets of $H$ (moving $t_i H$ to $t_j H$), and then applies the $H$-action element $t_j^{-1}gt_i \in H$ within the copy of $W$ labelled by $t_j$. To verify this defines a genuine $G$-action, one checks that $g_1(g_2(t_i w)) = (g_1g_2)(t_i w)$. If $g_2 t_i H = t_j H$ and $g_1 t_j H = t_\ell H$, then $g_1 g_2 t_i H = t_\ell H$, and the computation \begin{align*} g_1(g_2(t_i w)) = g_1(t_j (t_j^{-1}g_2 t_i)w) = t_\ell (t_\ell^{-1}g_1 t_j)(t_j^{-1}g_2 t_i)w = t_\ell (t_\ell^{-1}(g_1g_2)t_i)w \end{align*} confirms the associativity requirement. This construction has the correct character. Suppose $W$ has character $\psi$. The action $g: t_i w \mapsto t_j(t_j^{-1}gt_i)w$ contributes to the trace only when $j = i$, i.e., when $t_i^{-1}gt_i \in H$. In that case the contribution is $\psi(t_i^{-1}gt_i)$. Summing over all $i$: \begin{align*} \chi_{\operatorname{Ind}^G_H W}(g) = \sum_{i=1}^n \mathring{\psi}(t_i^{-1}gt_i), \end{align*} which matches the transversal formula for $\operatorname{Ind}^G_H\psi$. So the construction is consistent. [remark: Module-Theoretic Description] The construction can be expressed more cleanly using the language of modules. If $W$ is a left $\mathbb{F}H$-module, then \begin{align*} \operatorname{Ind}^G_H W = \mathbb{F}G \otimes_{\mathbb{F}H} W, \end{align*} where $\mathbb{F}G$ is viewed as an $(\mathbb{F}G, \mathbb{F}H)$-bimodule. This is the **extension of scalars** from $\mathbb{F}H$ to $\mathbb{F}G$ — the same operation familiar from commutative algebra. The explicit coset-decomposition above is this module presented in a chosen basis. [/remark] ## The Trivial Character and Permutation Representations Is induction connected to anything we have already met in the course? Permutation representations — which arise from group actions on sets — were among the first examples of representations we studied. It turns out that inducing the trivial character from a subgroup $H$ recovers precisely the permutation character of $G$ acting on the coset space $G/H$. [quotetheorem:2452] [citeproof:2452] As an immediate check via Frobenius reciprocity: \begin{align*} \langle \pi_X, 1_G \rangle_G = \langle \operatorname{Ind}^G_H 1_H, 1_G \rangle_G = \langle 1_H, 1_H \rangle_H = 1, \end{align*} confirming that the trivial character $1_G$ appears with multiplicity $1$ in the permutation character — a result established earlier in the course by direct computation. With Frobenius reciprocity in hand, we can apply character theory to study groups with special structure. Chapter 11 applies characters to Frobenius groups — where a subgroup intersects each of its non-trivial conjugates only in the identity — to prove Frobenius's theorem on the existence of a normal Frobenius kernel. # 11. Frobenius groups This chapter applies the character theory developed throughout the course to prove one of the landmark theorems in finite group theory: Frobenius' theorem (1891). The theorem concerns transitive permutation groups whose non-identity elements fix at most one point, and it asserts that the derangements — together with the identity — form a normal subgroup. The striking feature of this result is that no purely group-theoretic proof is known; character theory appears to be an essential ingredient in every known argument. We then use the theorem to define Frobenius groups and Frobenius complements and kernels, and connect these to Thompson's later nilpotency result. ## Frobenius' Theorem and Its Proof via Characters The setup is a transitive permutation group acting on a finite set, subject to a strong constraint: no non-identity element fixes more than one point. Under this constraint, the group decomposes in a very controlled way — every non-identity element either fixes exactly one point (and lies in a stabilizer) or fixes no point at all (and is a derangement). Frobenius' theorem says the derangements, together with the identity, form a normal subgroup. [quotetheorem:2453]The theorem does not explicitly assume $G$ is finite, but the hypotheses force $G \leq S_n$, which implies finiteness. The surprising content of the theorem is that $K$ is a subgroup at all — normality, once subgroup-hood is established, turns out to be the easier part. The proof uses character theory to construct a character whose kernel is exactly $K$. [citeproof:2453] [remark: Unavoidability of Representation Theory] This theorem has a remarkable history: despite being a statement purely about group theory, every known proof invokes representation theory in some essential way. No elementary proof has been found in over a century. [/remark] The key idea in the proof deserves emphasis. The characters $\theta_i$ are engineered so that they equal $\psi_i(1)$ on all of $K$ and equal $\psi_i$ on $H$. The sum $\Theta = \sum_i \psi_i(1)\theta_i$ then uses column orthogonality of $H$'s character table to annihilate everything outside $K$. The observation that $K = \ker(\text{representation affording } \Theta)$ gives normality for free, since kernels of group homomorphisms are always normal. Note what the theorem does not say: it makes no claim that $K$ is abelian or nilpotent — those are deeper structural facts that require additional work. It also relies critically on transitivity: if $G$ acts intransitively on $X$ with non-identity elements fixing at most one point, the derangements need not form a subgroup. Finally, this proof is a vivid illustration of characters doing group-theoretic work that bare group theory seems unable to do unaided; the forward connection is Thompson's theorem on the structure of $K$, which we reach shortly. ## Frobenius Groups What abstract structure underlies Frobenius' theorem? The permutation-theoretic hypotheses impose a strong combinatorial constraint on the stabilisers — which groups can realise this constraint, and how do they arise? To see why the constraint matters, suppose we try to define the kernel set $K$ in a group $G$ with a subgroup $H$ that does not satisfy malnormality: that is, there exist $g \notin H$ with some $1 \neq x \in H \cap gHg^{-1}$. Then the product of two elements from different conjugates of $H$ might land in a third conjugate of $H$ rather than in $K$, so $K$ fails to be closed under multiplication. The disjointness condition is precisely what forces the complement of $\bigcup_{g \in G} gHg^{-1} \cup \{1\}$ to be closed. This motivates the following definition. [definition: Frobenius Group and Frobenius Complement] A finite group $G$ is a **Frobenius group** if it has a subgroup $H$ such that \begin{align*} H \cap gHg^{-1} = \{1\} \quad \text{for all } g \notin H. \end{align*} Such a subgroup $H$ is called a **Frobenius complement** of $G$. [/definition] The condition says that distinct conjugates of $H$ are as disjoint as possible: they share no non-identity elements. This is exactly the combinatorial property that made the proof of Frobenius' theorem work. [proposition: Frobenius Groups Satisfy the Theorem's Hypotheses] The left action of any finite Frobenius group $G$ on the cosets $G/H$ (where $H$ is a Frobenius complement) satisfies the hypotheses of Frobenius' theorem. In particular, each non-identity element of $G$ fixes at most one coset. [/proposition] [citeproof:2453] This proposition is the bridge between the abstract definition and the permutation-theoretic setting of Frobenius' theorem. Every abstract Frobenius group $(G, H)$ produces a concrete permutation representation — the left coset action on $G/H$ — in which the hypotheses of Frobenius' theorem are satisfied, and so the kernel $K$ is guaranteed to exist as a normal subgroup. In particular, one does not need to start with a group acting on a set: any group possessing a malnormal proper subgroup is already a Frobenius group, and the coset action makes the permutation picture available automatically. [definition: Frobenius Kernel] The **Frobenius kernel** of a Frobenius group $G$ with complement $H$ is the normal subgroup $K$ obtained by applying Frobenius' theorem to the action on $G/H$. [/definition] Explicitly, the Frobenius kernel is \begin{align*} K = \{1\} \cup \{g \in G : g \text{ fixes no coset of } H\} = G \setminus \bigcup_{g \in G} (gHg^{-1} \setminus \{1\}). \end{align*} By Frobenius' theorem, $|K| = [G : H]$, and $G = KH$ with $K \cap H = \{1\}$, so $G$ is a semidirect product $G = K \rtimes H$. [example: Dihedral Groups as Frobenius Groups] The dihedral group $D_{2p}$ of order $2p$ (with $p$ an odd prime) is a Frobenius group. Take $G = D_{2p}$ with generators $r$ (rotation by $2\pi/p$) and $s$ (a fixed reflection), so $|G| = 2p$. Let $H = \langle s \rangle \cong C_2$. The conjugates of $H$ are the subgroups $\langle r^k s \rangle$ for $0 \leq k \leq p-1$. Each such subgroup has order $2$, with unique non-identity element $r^k s$. For distinct $j \neq k$, the elements $r^j s$ and $r^k s$ are distinct reflections, so $\langle r^j s \rangle \cap \langle r^k s \rangle = \{1\}$. Therefore distinct conjugates of $H$ meet only in the identity, $H$ is a Frobenius complement, and the Frobenius kernel is $K = \langle r \rangle \cong C_p$, a cyclic group of prime order — which is indeed normal in $D_{2p}$. A canonical family of Frobenius groups is the affine group $\operatorname{AGL}(1, q)$ of transformations $x \mapsto ax + b$ on $\mathbb{F}_q$ (with $a \neq 0$). The stabiliser of $0$ is $H = \{x \mapsto ax : a \neq 0\} \cong C_{q-1}$, and two distinct stabilisers share only the identity, so $H$ is a Frobenius complement. The Frobenius kernel is the translation subgroup $K = \{x \mapsto x + b\} \cong (\mathbb{F}_q, +)$, and $\operatorname{AGL}(1, q) = K \rtimes H$. These groups make Frobenius groups concrete and abundant. [/example] ## Thompson's Theorem and Nilpotency of the Kernel A natural follow-up question is: what can we say about the structure of the Frobenius kernel $K$? The action of any element of $H$ on $K$ by conjugation is a fixed-point-free automorphism of $K$ (an automorphism with no non-identity fixed points), because if $h \in H$ and $h$ centralises some $1 \neq k \in K$, then $k \in H \cap kHk^{-1}$, contradicting the Frobenius complement condition. [quotetheorem:2454]The restriction to prime order is essential: for composite order, Higman showed that there exist non-nilpotent groups admitting fixed-point-free automorphisms, so the conclusion fails. The nilpotency of the Frobenius kernel, combined with the semidirect product structure $G = K \rtimes H$, gives a very precise picture of the internal organisation of any Frobenius group. The kernel is a "nice" (nilpotent) normal subgroup, and the complement acts on it without fixed points. This is a deep structural result that goes beyond what character theory alone can provide — Thompson's proof does not use characters at all, but instead works with commutators and Sylow theory directly. Frobenius reciprocity assumes we understand restriction in a single direction, but in more complex situations we need finer control. Chapter 12 generalizes the picture to Mackey theory, giving a formula for how representations restrict across a tower of subgroups, with applications to irreducibility criteria. # 12. Mackey theory Mackey theory addresses a natural question that arises once we understand induced representations: if we induce a representation $W$ from a subgroup $H$ up to $G$, and then restrict back down to another subgroup $K$, what do we get? The answer — Mackey's restriction formula — expresses $\operatorname{Res}_K^G \operatorname{Ind}_H^G W$ as a direct sum of induced representations indexed by the double cosets $K \backslash G / H$. This is one of the deepest results in the course, and its corollaries include a powerful irreducibility criterion for induced representations. The chapter works throughout over $\mathbb{C}$. ## Double Cosets and the Permutation Representation Case Before stating the general theorem, it is instructive to examine the special case where $W = \mathbf{1}_H$ is the trivial representation of $H$. In that case, $\operatorname{Ind}_H^G \mathbf{1}_H$ is the permutation representation $\mathbb{C}(G/H)$ — the vector space with basis the left cosets of $H$ in $G$, with $G$ acting by left multiplication. Now let $K \leq G$ also act on $G/H$ by restriction. The set $G/H$ breaks into $K$-orbits, and the $K$-orbit of $gH$ is \begin{align*} \{kgH : k \in K\} = KgH, \end{align*} the double coset of $g$ with respect to $K$ and $H$. The collection of all such double cosets, written $K \backslash G / H$, partitions $G$. What is the stabilizer of $gH$ under the $K$-action? A coset element $k \in K$ fixes $gH$ precisely when $k \in gHg^{-1}$, so \begin{align*} K_{gH} = gHg^{-1} \cap K. \end{align*} We denote this subgroup $H_g := gHg^{-1} \cap K$. By the orbit-stabilizer theorem, the $K$-action on the orbit $KgH/H$ is isomorphic to $K$ acting on the coset space $K/H_g$. Since the permutation representation of $K$ on $K/H_g$ is $\operatorname{Ind}_{H_g}^K \mathbf{1}_{H_g}$, we obtain the following special case. [quotetheorem:2455]This is not proved separately — it is a direct consequence of the general Mackey theorem proved below, applied to $W = \mathbf{1}_H$. In the trivial-representation case, the formula has a pleasant combinatorial reading: the number of direct summands equals the number of double cosets $|K \backslash G/H|$, and each summand is itself a permutation representation of $K$. When $K = H$, the number of double cosets also controls how many distinct irreducible constituents can appear in $\operatorname{Ind}_H^G \mathbf{1}_H$ by character orthogonality arguments — this connects Mackey theory with Burnside's lemma and orbit-counting. Looking ahead, the general theorem replaces the trivial character $\mathbf{1}$ with an arbitrary representation $W$, replacing each trivial summand with a conjugate of $W$ induced from the intersection $H_g$. ## Mackey's Restriction Formula The general theorem replaces the trivial representation $\mathbf{1}_H$ with an arbitrary representation $(\rho, W)$ of $H$. To state it, we need the notion of a conjugated representation. [definition: Conjugate Representation] Let $H, K \leq G$, $g \in G$, and $H_g = gHg^{-1} \cap K$. Given a representation $(\rho, W)$ of $H$, the **conjugate representation** $(\rho_g, W_g)$ is the representation of $H_g$ on the same underlying vector space $W$, with action \begin{align*} \rho_g(x) = \rho(g^{-1} x g), \quad x \in H_g. \end{align*} This is well-defined since $x \in H_g \leq gHg^{-1}$ implies $g^{-1}xg \in H$. [/definition] The element $g^{-1}xg$ lies in $H$ by construction, so $\rho(g^{-1}xg)$ makes sense. Note that $H_g \leq K$, so we can induce $W_g$ from $H_g$ up to $K$. [quotetheorem:2456] [citeproof:2456] [remark: Hardest Proof in the Course] The course notes describe this as possibly the hardest and most sophisticated proof in the course. The essential structure — partitioning the induced representation $V$ into $K$-stable pieces $V(g)$ indexed by double cosets, then identifying each piece with an induced representation — is a model of how double cosets serve as a bridge between two subgroups. [/remark] Mackey's restriction formula is a structural decomposition theorem: it tells us what $\operatorname{Res}_K^G \operatorname{Ind}_H^G W$ looks like as a $K$-representation, but it does not say this decomposition is into irreducibles, nor that it gives a canonical isomorphism independent of auxiliary choices. The hypothesis that $G$ is finite is used throughout — the transversal argument and the double coset partition both depend on finiteness. Crucially, the formula shows that naively expecting $\operatorname{Res}_K^G \operatorname{Ind}_H^G W \cong W$ is wrong in general: even if $K = H$, we get a sum over all double cosets in $H \backslash G / H$, and only the identity coset contributes a copy of something resembling $W$ itself. The remaining summands are genuinely new. The main payoff of this formula comes next, when we apply it with $K = H$ to derive a criterion for irreducibility. [example: A Non-Normal Failure] To see why the formula is necessary and non-trivial, consider $G = S_3$, $H = \langle (12) \rangle \cong \mathbb{Z}/2$, and $K = \langle (13) \rangle \cong \mathbb{Z}/2$. Let $W$ be the nontrivial (sign) character of $H$. The double cosets $K \backslash S_3 / H$ are two in number: one represented by the identity $e$ and one by the transposition $(23)$ (since $|S_3| = 6$ and $|K||H| = 4$, the orbit-counting argument gives two double cosets). Then $H_e = eHe^{-1} \cap K = H \cap K = \{e\}$ and $H_{(23)} = (23)H(23)^{-1} \cap K$. One computes $(23)(12)(23)^{-1} = (13)$, so $(23)H(23)^{-1} = \langle (13) \rangle = K$, giving $H_{(23)} = K$. Thus Mackey's formula gives \begin{align*} \operatorname{Res}_K^{S_3} \operatorname{Ind}_H^{S_3} W \cong \operatorname{Ind}_{\{e\}}^K \mathbf{1} \oplus \operatorname{Ind}_K^K W_{(23)}. \end{align*} The first summand is the regular representation of $K$ (combining the principal character and the sign character of $K$), and the second is $W_{(23)}$, the conjugate of $W$ by $(23)$. This is a direct sum of three one-dimensional $K$-representations — certainly not just $W$ itself. The formula accounts for all these contributions systematically, which would be invisible from first principles. [/example] ## The Character Version and Conjugate Characters Working at the level of representations gives structural information, but computing inner products and checking irreducibility is easier with characters. The character version of Mackey's theorem translates the representation-level decomposition into an equation of class functions, where the familiar inner product machinery of characters can be applied directly. [definition: Conjugate Character] Let $H, K \leq G$, $g \in G$, $H_g = gHg^{-1} \cap K$, and $\psi$ a character of a representation of $H$. The **conjugate character** $\psi_g$ is the class function on $H_g$ defined by \begin{align*} \psi_g(x) = \psi(g^{-1}xg), \quad x \in H_g. \end{align*} Since $g^{-1}xg \in H$, this is well-defined. Moreover, $\psi_g$ is a character of $H_g$ (it is the character of the conjugate representation $W_g$). [/definition] [quotetheorem:2457]This follows directly from Mackey's restriction formula: the character of a direct sum is the sum of characters, and the character of an induced representation is the induced character. The character version matters because it unlocks the inner product. If we want to know whether $\operatorname{Ind}_H^G W$ is irreducible, we compute $\langle \operatorname{Ind}_H^G \psi, \operatorname{Ind}_H^G \psi \rangle_G$. By Frobenius reciprocity, this becomes $\langle \psi, \operatorname{Res}_H^G \operatorname{Ind}_H^G \psi \rangle_H$, and Mackey's character formula expresses the restricted character as a sum of induced conjugate characters. Each inner product $\langle \psi, \operatorname{Ind}_{H_g}^H \psi_g \rangle_H$ then simplifies via Frobenius reciprocity to $\langle \operatorname{Res}_{H_g}^H \psi, \psi_g \rangle_{H_g}$. This chain of manipulations — Frobenius, then Mackey, then Frobenius again — is exactly what drives the irreducibility criterion proved in the next section. [remark: Dependence on Double Coset Representatives] The formula appears to depend on the choice of representatives $\mathcal{S}$. In fact, the isomorphism class of $W_g$ (for $g \in G$) depends only on the double coset $KgH$, not on the particular choice of representative $g$. So the direct sum $\bigoplus_{g \in \mathcal{S}} \operatorname{Ind}_{H_g}^K W_g$ is canonically determined by the double coset decomposition. [/remark] ## Mackey's Irreducibility Criterion Inducing an irreducible representation does not automatically produce an irreducible result. For instance, if $H = \{e\}$ and $W = \mathbf{1}$, then $\operatorname{Ind}_{\{e\}}^G \mathbf{1}$ is the regular representation, which decomposes into all irreducible representations of $G$ and is highly reducible whenever $|G| > 1$. More subtly, even when $H$ is a large subgroup and $W$ is irreducible, the induced representation can still be reducible — different double cosets can interact to produce repeated summands. What we need is a checkable condition that tells us precisely when $\operatorname{Ind}_H^G W$ is irreducible, and Mackey's theorem provides exactly that. Recall that for a character $\psi$ of $H$, the induced character $\operatorname{Ind}_H^G \psi$ is irreducible if and only if its inner product with itself is $1$. [quotetheorem:2458]Here $\mathcal{S}$ is any set of double coset representatives for $H \backslash G / H$. We may equivalently require condition (2) for all $g \in G \setminus H$ — checking more elements is redundant work but avoids having to choose $\mathcal{S}$ explicitly. [citeproof:2458] [remark: Why the Identity Coset Is Special] The only $g \in \mathcal{S}$ that lies in $H$ is $g = 1$: for any $h \in H$, the double coset $HhH = H \cdot 1 \cdot H$, so $h$ is in the same double coset as the identity. This is why the $g = 1$ term is separated from the rest — it always contributes the irreducibility of $W$ itself. [/remark] The criterion shows that irreducibility of $W$ is necessary but not sufficient for irreducibility of $\operatorname{Ind}_H^G W$. The additional conditions — that $W$ and each of its conjugates $W_g$ are disjoint when restricted to the intersections $H_g$ — can fail even for irreducible $W$. For example, if $G = \mathbb{Z}/4$ and $H = \mathbb{Z}/2$ (the unique subgroup of order $2$), with $\chi$ the nontrivial character of $H$, then $H \backslash G / H$ has two double cosets (the identity and the nontrivial element of order $4$). The conjugate $\chi_g = \chi$ for $g$ of order $4$ (since $H$ is central in $G$), so $\chi$ and $\chi_g$ are identical on $H_g = H$, and the criterion fails — indeed $\operatorname{Ind}_H^G \chi$ decomposes into two distinct one-dimensional representations of $G$. In practice, to apply the criterion one first enumerates the double cosets $H \backslash G / H$ (often manageable via the formula $|H \backslash G / H| = \sum_{g \in \mathcal{S}} 1$), then for each non-identity representative $g$ checks whether $\operatorname{Res}_{H_g}^H \psi$ and $\psi_g$ share a common irreducible constituent, which reduces to an inner product computation. ## The Normal Subgroup Case Applying the general Mackey criterion requires enumerating all double cosets $H \backslash G / H$ and checking disjointness at each intersection $H_g$ — a computation that can be involved for general subgroups. When $H$ is normal in $G$, the double coset structure collapses dramatically: every double coset is a single left coset, and every intersection $H_g$ equals $H$ itself. This produces a simple, explicit criterion in terms of conjugate characters alone, without any intersection computation. [quotetheorem:2459] [citeproof:2459] The normality hypothesis is genuinely essential here. Without it, different double cosets can have different intersections $H_g$, and the disjointness condition must be checked at each intersection separately — the condition $\psi \neq \psi_g$ on all of $H$ is not the right statement. The theorem also does not identify which element $g$ causes $\psi$ to collide with $\psi_g$ when the criterion fails; it only asserts irreducibility when no such collision occurs. The deeper theory of what happens when $\psi = \psi_g$ for some $g$ — how $\operatorname{Ind}_H^G \psi$ decomposes in that case — is developed in Clifford theory, which systematically analyzes representations of a group in terms of the orbit of an irreducible character of a normal subgroup under conjugation by $G$. [example: Splitting in the Alternating Group — the 5-cycle in $A_5$] Let $G = S_5$ and $H = A_5$. The subgroup $A_5$ has index $2$ in $S_5$ and is normal. We work with the $5$-cycle $\sigma = (12345) \in A_5$. **Step 1: Conjugacy in $S_5$.** In $S_5$, all $5$-cycles form a single conjugacy class of size $\frac{5!}{5} = 24$ (since a $5$-cycle has centralizer of order $5$, generated by the cycle itself). **Step 2: Splitting in $A_5$.** The centralizer of $\sigma$ in $S_5$ is $C_{S_5}(\sigma) = \langle \sigma \rangle \cong \mathbb{Z}/5$. Since $\langle \sigma \rangle \leq A_5$ (all powers of a $5$-cycle are even permutations), we have $C_{S_5}(\sigma) = C_{A_5}(\sigma)$. The $A_5$-conjugacy class of $\sigma$ has size $|A_5|/|C_{A_5}(\sigma)| = 60/5 = 12$. But the full $S_5$-class has $24$ elements. Since $24 = 12 + 12$, the $S_5$-class of $\sigma$ splits into exactly two $A_5$-conjugacy classes, each of size $12$: one containing $\sigma = (12345)$ and one containing $(12354)$ (any $5$-cycle not conjugate to $\sigma$ within $A_5$, e.g., $(13245)$, lies in the other class). **Step 3: A character that separates them.** Since $A_5 \cong$ the icosahedral group, its character table is known. It has $5$ irreducible characters of dimensions $1, 3, 3, 4, 5$. Let $\chi$ be one of the two $3$-dimensional irreducible characters of $A_5$. These two characters — call them $\chi$ and $\chi'$ — take the value $\frac{1+\sqrt{5}}{2}$ on one of the two $5$-cycle classes and $\frac{1-\sqrt{5}}{2}$ on the other (the golden ratio and its conjugate appear here from the icosahedral symmetry). So $\chi$ takes different values on the two $A_5$-conjugacy classes of $5$-cycles. **Step 4: Applying the theorem.** Let $g \in S_5 \setminus A_5$, e.g., $g = (12)$. The conjugate character $\chi_g$ satisfies $\chi_g(x) = \chi(g^{-1}xg) = \chi((12)x(12))$ for $x \in A_5$. Acting by $(12)$ on $(12345)$ gives $(12)(12345)(12)^{-1} = (21345) = (13452)$, which lies in the other $A_5$-conjugacy class of $5$-cycles. Therefore $\chi_g(\sigma) = \chi((12)\sigma(12)) = \frac{1-\sqrt{5}}{2} \neq \frac{1+\sqrt{5}}{2} = \chi(\sigma)$. Since $\chi$ and $\chi_g$ differ on an element of $A_5$, they are distinct irreducible characters of $A_5$, so $\chi_g \neq \chi$. **Conclusion.** Since the only non-identity double coset of $A_5 \backslash S_5 / A_5$ corresponds to elements of $S_5 \setminus A_5$ (there are only two double cosets since $[S_5 : A_5] = 2$), and $\chi_g \neq \chi$, the irreducibility criterion for normal subgroups guarantees that $\operatorname{Ind}_{A_5}^{S_5} \chi$ is an irreducible $3$-dimensional representation of $S_5$. The conjugate character $\chi_g = \chi'$ (the other $3$-dimensional irreducible of $A_5$), and $\operatorname{Ind}_{A_5}^{S_5} \chi' = \operatorname{Ind}_{A_5}^{S_5} \chi$ gives the same irreducible of $S_5$ — both $3$-dimensional irreducibles of $A_5$ induce to the same irreducible of $S_5$. [/example] [explanation: The Structure of Mackey Theory] Mackey theory can be summarized in a single conceptual picture. When we form $\operatorname{Ind}_H^G W$, we are spreading $W$ over all left cosets of $H$ in $G$. When we then restrict to $K$, we see how these cosets group into $K$-orbits. The $K$-orbits of $G/H$ are exactly the double cosets $KgH$, and the piece of the induced representation sitting over the double coset $KgH$ is isomorphic to the induced representation $\operatorname{Ind}_{H_g}^K W_g$. The conjugate representation $W_g$ appears because, to make the $K$-action on the double coset $KgH$ into an induction, we must conjugate $H$ into a subgroup of $K$ — specifically, the intersection $gHg^{-1} \cap K = H_g$. The action of $H_g$ on $W$ is twisted by conjugation by $g$ to account for this change of perspective. This explains why the irreducibility criterion involves checking that $W$ and its conjugates are "disjoint" over the various intersections $H_g$: the induced representation $\operatorname{Ind}_H^G W$ decomposes into these conjugate pieces upon restriction, and irreducibility requires no cancellation to occur. [/explanation] Having used characters extensively, we now ask about their arithmetic properties. Chapter 13 develops the theory of algebraic integers in the group algebra, showing that character values are algebraic integers and deriving the divisibility theorem that constrains representation dimensions. # 13. Integrality in the group algebra This chapter develops the arithmetic theory of characters, culminating in one of the deepest results in the subject: the degree of every irreducible character divides the order of the group. The proof is indirect and elegant — rather than working with the group directly, we pass through the center of the group algebra and use the theory of algebraic integers. The main tools are the class sums in $\mathbb{C}G$, Schur's lemma applied to the center, and a key integrality lemma. ## Algebraic Integers The proof of divisibility rests on a number-theoretic observation: character values are not merely complex numbers, but algebraic integers. This section lays the algebraic foundation. [definition: Algebraic Integer] A complex number $a \in \mathbb{C}$ is an **algebraic integer** if $a$ is a root of a monic polynomial with integer coefficients. Equivalently, $a$ is an algebraic integer if the subring \begin{align*} \mathbb{Z}[a] = \{f(a) : f \in \mathbb{Z}[X]\} \subset \mathbb{C} \end{align*} is finitely generated as a $\mathbb{Z}$-module. A third equivalent characterisation: $a$ is the eigenvalue of some matrix with all entries in $\mathbb{Z}$. [/definition] The three characterisations are each useful in different contexts. The monic polynomial definition is what we check in examples. The finitely generated module condition is what powers the key lemma below. We quote without proof the following standard properties of algebraic integers, whose proofs belong to algebraic number theory. [quotetheorem:2460]Property (2) is the key to the divisibility argument: if we can show $\frac{|G|}{\chi(1)}$ is an algebraic integer and it is also rational (which is immediate since both $|G|$ and $\chi(1)$ are positive integers), then it must be an ordinary integer. The first step is to establish that character values themselves are algebraic integers. [quotetheorem:2461] [citeproof:2461] Notice that irreducibility of $\chi$ plays no role here: the result holds for any character, reducible or not. What matters is that elements of finite groups have finite order, forcing all eigenvalues to be roots of unity. This theorem is the entry point for the entire arithmetic theory of characters. [example: Algebraic integers in practice] Consider $\sqrt{2}$, which is a root of $X^2 - 2$ and so is an algebraic integer. On the other hand, $\frac{1}{2}$ is rational but not an integer, and indeed $\frac{1}{2}$ satisfies no monic polynomial with integer coefficients (if $\frac{1}{2}$ satisfied $X^n + a_{n-1}X^{n-1} + \cdots + a_0 = 0$ with $a_i \in \mathbb{Z}$, multiplying through by $2^n$ gives $1 + 2a_{n-1} + \cdots + 2^n a_0 = 0$, so $1$ would be even — a contradiction). This illustrates property (2): rationals that are algebraic integers must already be integers. [/example] ## The Center of the Group Algebra The strategy for proving divisibility passes through the center $Z(\mathbb{C}G)$ of the group algebra. Recall that $\mathbb{C}G$ is the complex vector space with basis $G$: \begin{align*} \mathbb{C}G = \left\{ \sum_{g \in G} \alpha_g g : \alpha_g \in \mathbb{C} \right\}, \end{align*} with multiplication extending the multiplication of $G$ bilinearly. This makes $\mathbb{C}G$ an algebra. The center $Z(\mathbb{C}G)$ consists of those elements that commute with every element of $\mathbb{C}G$. The natural basis for this center is given by class sums. [definition: Class Sum] Let $G$ be a finite group with conjugacy classes $\mathcal{C}_1, \ldots, \mathcal{C}_k$ (where $\mathcal{C}_1 = \{1\}$). The **class sum** of $\mathcal{C}_j$ is \begin{align*} C_j = \sum_{g \in \mathcal{C}_j} g \in \mathbb{C}G. \end{align*} [/definition] Class sums centralise every element of $G$ because conjugation merely permutes the elements within each conjugacy class. Since central elements of $\mathbb{C}G$ are precisely those whose coefficients are constant on conjugacy classes, the class sums span $Z(\mathbb{C}G)$ and are also linearly independent — making them a basis. [quotetheorem:2462] [citeproof:2462] The integrality of $a_{ij\ell}$ is crucial: it is not a formal algebraic fact but a combinatorial one — the structure constants count pairs of group elements from specified conjugacy classes whose product lands in a third class. This counting interpretation will be reused when we prove that the central character values form a finitely generated $\mathbb{Z}$-module. [definition: Class Algebra Constants] The non-negative integers $a_{ij\ell}$ defined by $C_i C_j = \sum_\ell a_{ij\ell} C_\ell$ are called the **class algebra constants** or **structure constants** of $G$. [/definition] ## The Central Character The question is: how do $Z(\mathbb{C}G)$ and the irreducible characters of $G$ interact? Schur's lemma provides the answer — every irreducible representation forces the center to act by scalars, and those scalars assemble into an algebra homomorphism that encodes character-theoretic information. Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation affording the character $\chi$. Extending linearly, $\rho$ becomes an algebra homomorphism $\rho: \mathbb{C}G \to \operatorname{End}(V)$. Any $z \in Z(\mathbb{C}G)$ maps under $\rho$ to an operator $\rho(z)$ that commutes with all $\rho(g)$ for $g \in G$. By Schur's lemma, $\rho(z) = \lambda_z I$ for some scalar $\lambda_z \in \mathbb{C}$. This gives a well-defined map: [definition: Central Character] Let $\rho$ be an irreducible representation of $G$ with character $\chi$. The **central character** associated to $\chi$ is the algebra homomorphism \begin{align*} \omega_\chi: Z(\mathbb{C}G) &\to \mathbb{C}, \\ z &\mapsto \lambda_z, \end{align*} where $\rho(z) = \lambda_z I$. [/definition] Irreducibility is essential here. For a reducible representation $\rho = \rho_1 \oplus \rho_2$ with $\rho_1 \not\cong \rho_2$, a central element $z$ need not act by a scalar: $\rho(z)$ will in general be block-diagonal with different eigenvalues in each block, so there is no single $\lambda_z$ and the definition of $\omega_\chi$ breaks down. The central character can be computed explicitly from the character table. Applying $\rho(C_i) = \omega_\chi(C_i) I$ and taking traces of both sides: \begin{align*} \sum_{g \in \mathcal{C}_i} \chi(g) = \chi(1)\, \omega_\chi(C_i). \end{align*} Since $\chi$ is a class function, all terms in the sum equal $\chi(g_i)$ for any representative $g_i \in \mathcal{C}_i$, so \begin{align*} \omega_\chi(C_i) = \frac{|\mathcal{C}_i|\, \chi(g_i)}{\chi(1)}. \end{align*} This formula for $\omega_\chi(C_i)$ is the crux of everything: it is a ratio that we will show is an algebraic integer. ## The Key Integrality Lemma The fraction $\omega_\chi(C_i) = \frac{|\mathcal{C}_i|\,\chi(g_i)}{\chi(1)}$ is not obviously an algebraic integer — the numerator is an algebraic integer (a class size times a character value), but we are dividing by $\chi(1)$, a positive integer. Division by an integer can destroy integrality. What makes integrality survive here is that the $\omega_\chi(C_i)$ do not stand alone: they generate a subring of $\mathbb{C}$ that is finitely generated as a $\mathbb{Z}$-module, and the key is that the module-generation is witnessed by the integer structure constants. [quotetheorem:2463] [citeproof:2463] The proof is elegant precisely because the integrality of the structure constants $a_{ij\ell}$, which was a combinatorial observation in the previous section, now yields an algebraic conclusion via the module-theoretic criterion. The argument has a precise structure: the $\omega_\chi(C_i)$ satisfy a system of multiplication rules with integer coefficients, so they generate an integer lattice, and any element of an integer lattice in $\mathbb{C}$ is an algebraic integer. [remark: Structure Constants from the Character Table] The structure constants can be recovered from the character table via the formula \begin{align*} a_{ij\ell} = \frac{|G|}{|C_G(g_i)|\,|C_G(g_j)|} \sum_{s=1}^k \frac{\chi_s(g_i)\,\chi_s(g_j)\,\chi_s(g_\ell^{-1})}{\chi_s(1)}, \end{align*} where the sum is over all irreducible characters $\chi_s$. The key ingredient in the proof is column orthogonality. This formula is quoted here for completeness and will not be used further. [/remark] ## Divisibility of Group Order by Character Degrees Why should divisibility hold at all? A priori, $\chi(1)$ is constrained by the sum-of-squares identity $\sum_\chi \chi(1)^2 = |G|$, but this does not force each individual degree to divide $|G|$ — it only bounds the sum. The remarkable content of the theorem below is that the group structure imposes a much sharper arithmetic constraint: each degree must be an exact divisor of $|G|$. [quotetheorem:2464] [citeproof:2464] The argument is a beautiful illustration of the strategy: transform a divisibility question into an integrality question by passing through the center of the group algebra. Several points deserve emphasis. First, irreducibility is genuinely needed: the regular character of $G$ has degree $|G|$ and generally does not divide $|G|$, so the theorem fails without irreducibility. Second, the theorem tells us which divisors of $|G|$ are *excluded* from being character degrees (those not dividing $|G|$), but it says nothing about which divisors of $|G|$ actually arise — determining the precise set of character degrees is a much harder problem. Third, this divisibility theorem is one of the two key inputs to Burnside's celebrated theorem that every group of order $p^a q^b$ is soluble; the other input is a more refined application of character theory. ## Applications Divisibility constrains the structure of representations over a finite group in ways that are not obvious from the group's definition. The examples below show how to read off structural information — abelianness, $p$-power degrees — purely from the constraint $\chi(1) \mid |G|$. [example: Groups of order $p^2$ are abelian] Let $G$ be a group of order $p^2$ for a prime $p$. Since $|G| = p^2$, every irreducible character degree $\chi(1)$ must divide $p^2$, so $\chi(1) \in \{1, p, p^2\}$. The sum-of-squares identity gives \begin{align*} \sum_{\chi \text{ irred.}} \chi(1)^2 = |G| = p^2. \end{align*} If any irreducible character had degree $p$, its square alone contributes $p^2$, exhausting the entire sum and leaving no room for the trivial character (which must appear). If any irreducible character had degree $p^2$, its square contributes $p^4 > p^2$, which immediately violates the identity. So every irreducible character has degree $1$, contributing $1^2 = 1$ to the sum. Since the sum equals $p^2$, there are exactly $p^2$ irreducible characters, hence $p^2$ conjugacy classes. A finite group equals its own center if and only if every conjugacy class is a singleton, which happens if and only if the group is abelian. Having $p^2$ conjugacy classes in a group of order $p^2$ means every class has size $|G|/|Z(G)| \cdot \text{(class-size formula)} = 1$, which forces $G = Z(G)$, i.e. $G$ is abelian. [/example] The previous example illustrates a general pattern: when divisibility forces all character degrees to be $1$, the group is abelian. The next example is the $p$-group analogue for general $n$. [example: $p$-groups and character degrees] More generally, if $G$ is a $p$-group (i.e. $|G| = p^n$ for some $n$), then the degree of every irreducible character is a power of $p$. This follows from $\chi(1) \mid |G| = p^n$: every divisor of $p^n$ is of the form $p^k$ with $0 \leq k \leq n$. In particular, $\chi(1)$ can never be divisible by any prime other than $p$. This is a non-trivial constraint: knowing only the group order, we can rule out any character degree that is not a $p$-power. [/example] [remark: Simple Groups and Degree Two] No non-abelian simple group can have an irreducible character of degree $2$. The proof uses the divisibility theorem together with structural constraints on simple groups; the details are a worthwhile exercise in combining the tools developed in this chapter. [/remark] Burnside's theorem, a foundational result about group structure, states that groups of order $p^a q^b$ are soluble. Chapter 14 proves this theorem entirely using character theory and the divisibility constraints from Chapter 13, demonstrating the power of representation theory for group-theoretic questions. # 14. Burnside's theorem The final chapter of this course reaches one of the great applications of character theory to pure group theory. Burnside's $p^a q^b$ theorem asserts that any group whose order has at most two distinct prime factors cannot be simple (and in fact must be soluble). This is a deep structural result about finite groups, and what makes it remarkable is that the most natural proof — the one presented here — passes entirely through representation theory. The group-theoretic content is almost entirely encoded in the arithmetic of characters. ## The Statement and Its Optimality Before proving the theorem, it is worth understanding exactly what it says and why it is sharp. [quotetheorem:2465]The condition $a + b \geq 2$ rules out the trivial cases: if $a + b \leq 1$ then $|G| \in \{1, p, q\}$, and groups of prime order are cyclic hence simple by definition, while the trivial group is simple vacuously. These cases do not need the theorem. [remark: Solubility] In fact something stronger is true: every group of order $p^a q^b$ is soluble. This follows from the theorem by an easy induction. If $G$ is not simple, it has a proper normal subgroup $N$ with $1 < N < G$, and both $N$ and $G/N$ have orders dividing $p^a q^b$, so by induction they are soluble, hence $G$ is soluble. [/remark] [remark: Sharpness] The theorem is best possible with respect to the number of prime factors. The alternating group $A_5$ has order $60 = 2^2 \cdot 3 \cdot 5$, which involves three distinct primes, and $A_5$ is simple. So one cannot extend the theorem to allow three distinct prime divisors. In fact there are exactly eight simple groups whose order has precisely three distinct prime factors, and $A_5$ is the smallest. The boundary cases $a = 0$ or $b = 0$ (so $G$ is a $p$-group) are also easy: every non-trivial $p$-group has non-trivial centre, so it cannot be simple unless it has prime order. [/remark] [remark: Historical note] Burnside proved this theorem in 1904 using representation theory. In 1963, Feit and Thompson proved the far deeper result that every group of odd order is soluble; their proof ran to 255 pages. In 1972, Bender found the first purely group-theoretic proof of Burnside's theorem, but it is considerably more involved than the character-theoretic argument given here. [/remark] ## Two Key Lemmas How can one extract group-theoretic information — specifically simplicity — from purely numerical data about conjugacy class sizes? The answer runs through representation theory: character values at elements of a class carry arithmetic information about that class size, and divisibility constraints force characters to vanish or achieve their maximum. The proof of Burnside's theorem rests on two lemmas that make this precise. The first involves Galois theory and is non-examinable; the second is a purely algebraic consequence of the orthogonality relations. ### A Lemma from Galois Theory [quotetheorem:2466] [citeproof:2466] The algebraic-integer hypothesis on $\alpha$ is not optional — without it the lemma fails badly. The average $\frac{1}{m}\sum \lambda_j$ of roots of unity can be any complex number of absolute value at most $1$, and without the arithmetic constraint that forces $N(\alpha)$ to be an integer there is nothing to prevent it from taking an intermediate value. The lemma says that the algebraic-integer condition is so rigid that it collapses the continuum of possibilities to just two: zero or a root of unity. The theorem does not tell us which alternative holds — for that we need extra information about the context (the coprimality lemma below supplies it). In the proof of Burnside's theorem, the role of the algebraic-integer hypothesis is played by the fact that character values and class-sum eigenvalues are always algebraic integers, a general structural feature of group algebras. ### A Coprimality Lemma for Character Values The second lemma is the key bridge between divisibility of class sizes and vanishing of characters. [quotetheorem:2467] [citeproof:2467] The coprimality hypothesis is what makes this dichotomy sharp. If $\gcd(\chi(1), |\mathcal{C}|)$ were not $1$, the Bezout step would fail and $\alpha$ might no longer be an algebraic integer — one could not conclude either alternative. The result is striking because it says character values at a given element are not just constrained in absolute value by $\chi(1)$; coprimality forces the absolute value to be exactly $0$ or $\chi(1)$, with nothing in between. In the proof of Burnside's theorem, we will apply this to a class of prime-power size: the irreducible characters whose degrees are not divisible by $p$ (and hence are coprime to $p^r$) must either vanish or achieve their maximum absolute value at $g$, and then a further argument eliminates the maximum case. [example: $S_3$ and the coprimality lemma] Consider $G = S_3$, the symmetric group on three letters, which has order $6 = 2 \cdot 3$. Its character table has three irreducible characters: the trivial character $\mathbf{1}$ of degree $1$, the sign character $\varepsilon$ of degree $1$, and the standard $2$-dimensional character $\chi_{\mathrm{std}}$ of degree $2$. The conjugacy classes are: $\mathcal{C}_1 = \{e\}$ of size $1$, $\mathcal{C}_2 = \{(12),(13),(23)\}$ of size $3$ (transpositions), and $\mathcal{C}_3 = \{(123),(132)\}$ of size $2$ (3-cycles). Take $\chi = \chi_{\mathrm{std}}$ with $\chi(1) = 2$, and the class $\mathcal{C}_2$ of size $3$. Since $\gcd(2, 3) = 1$, the coprimality lemma applies: for any transposition $g = (12)$, either $\chi_{\mathrm{std}}(g) = 0$ or $|\chi_{\mathrm{std}}(g)| = 2$. Reading off the standard character table, $\chi_{\mathrm{std}}((12)) = 0$, confirming the first alternative. For the class $\mathcal{C}_3$ of size $2$, we have $\gcd(2, 2) = 2 \neq 1$, so the lemma gives no conclusion — and indeed $\chi_{\mathrm{std}}((123)) = -1$, which is neither $0$ nor $\pm 2$. This shows that coprimality is genuinely necessary. [/example] ## Non-Abelian Simplicity and Prime-Power Class Sizes Does simplicity constrain the sizes of conjugacy classes? Can a non-abelian simple group have a conjugacy class of prime-power size? The answer is no, and this is the proposition that does the real work in Burnside's theorem. [quotetheorem:2468] [citeproof:2468] The non-abelian hypothesis is essential: abelian simple groups do exist (the cyclic groups of prime order), and their conjugacy classes all have size $1 = p^0$. The proposition says nothing about classes of size $1$, and rightly so — an abelian group has every element in its own conjugacy class. The condition $r \geq 1$ (non-trivial prime power) is what excludes central elements. If $r = 0$ (class size $1$), the element is in $Z(G)$, and an abelian simple group is fine. For $r \geq 1$, Sylow theory will supply us with such an element in any $p^a q^b$ group: the centre of a Sylow $q$-subgroup provides a non-central element whose class has size a power of $p$. ## Proof of Burnside's Theorem How do we manufacture a conjugacy class of prime-power size from the hypothesis that $|G| = p^a q^b$? Sylow theory guarantees that a $q$-Sylow subgroup exists, its centre is non-trivial, and any non-identity element of that centre centralises the entire Sylow subgroup — which has order $q^b$. This forces the conjugacy class of that element to have size dividing $p^a$, giving the prime-power class we need. [citeproof:2468] [remark: End of finite group theory] This completes what the course has to say about finite groups via representation theory. The machinery developed — characters, orthogonality, the group algebra, induced representations — has culminated in a theorem that would be extremely difficult to prove by purely combinatorial means. The passage through Galois theory (in the first lemma) and algebraic integers (throughout) illustrates how analytic and arithmetic tools, once embedded in a representation-theoretic framework, become powerful instruments for understanding group structure. [/remark] Finally, we extend representation theory beyond finite groups to continuous groups. Chapter 15 studies representations of compact groups like $S^1$, $\mathrm{SU}(2)$, and $\mathrm{SO}(3)$, where Haar measure plays the role of averaging and characters again satisfy orthogonality relations. # 15. Representations of compact groups The preceding chapters developed a complete representation theory for finite groups, exploiting the fact that we can average over the group to produce invariant structures. The passage to infinite groups requires care: without some extra structure, a group like $S^1$ admits a wild, uncountable collection of irreducible representations that defies classification. The remedy is to equip the group with a topology, demand that representations be continuous, and — crucially — restrict to groups that are **compact**. Under these conditions every technique from the finite theory survives, with sums replaced by integrals via the Haar measure, and we obtain complete, explicit lists of irreducible representations for $S^1$ and $\operatorname{SU}(2)$. ## Topological Groups, Compactness, and Haar Measure [definition: Topological Group] A **topological group** is a group $G$ which is simultaneously a topological space such that the multiplication map $G \times G \to G$, $(h,g) \mapsto hg$, and the inversion map $G \to G$, $g \mapsto g^{-1}$, are both continuous. [/definition] The most important topological groups in this course are matrix groups. The general linear groups $\operatorname{GL}_n(\mathbb{R})$ and $\operatorname{GL}_n(\mathbb{C})$ are topological groups under the topology inherited from $\mathbb{R}^{n^2}$ and $\mathbb{C}^{n^2}$ respectively. Any finite group equipped with the discrete topology is also a topological group, so the finite theory is a special case. For representation theory to work well, we need the group to be "small" in a topological sense. [definition: Compact Group] A **compact group** is a topological group that is compact as a topological space. [/definition] The key examples are: [example: Circle Group] The **circle group** $S^1 = \{z \in \mathbb{C} : |z| = 1\}$ under complex multiplication is a compact group. More generally, any finite product $S^1 \times \cdots \times S^1$ (a torus) is compact. We have the isomorphism $S^1 \cong \mathbb{R}/\mathbb{Z}$ of abelian groups via $x \mapsto e^{2\pi i x}$. [/example] [example: Orthogonal and Special Orthogonal Groups] The orthogonal group $\operatorname{O}(n) \leq \operatorname{GL}_n(\mathbb{R})$ is compact: its entries are bounded in magnitude by $1$ (columns are orthonormal), so it sits as a closed bounded subset of $\mathbb{R}^{n^2}$. The special orthogonal group $\operatorname{SO}(n) = \{A \in \operatorname{O}(n) : \det A = 1\}$ is also compact. In particular, $\operatorname{SO}(2) \cong S^1$ via $\theta \mapsto e^{i\theta}$. [/example] [example: Unitary and Special Unitary Groups] The unitary group $\operatorname{U}(n) = \{A \in \operatorname{GL}_n(\mathbb{C}) : AA^\dagger = I\}$ is compact, and so is $\operatorname{SU}(n) = \{A \in \operatorname{U}(n) : \det A = 1\}$. We have $\operatorname{U}(1) \cong \operatorname{SO}(2) \cong S^1$ as topological groups. The group of primary interest is \begin{align*} \operatorname{SU}(2) = \left\{ \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} : a, b \in \mathbb{C},\ |a|^2 + |b|^2 = 1 \right\}, \end{align*} which is homeomorphic to the three-sphere $S^3 \subset \mathbb{R}^4$. (It is a non-trivial theorem from topology that the only spheres admitting group structures are $S^1$ and $S^3$.) [/example] We define what it means for a topological group to act on a vector space. [definition: Representation of a Topological Group] A **representation** of a topological group $G$ on a finite-dimensional complex vector space $V$ is a continuous group homomorphism $\rho: G \to \operatorname{GL}(V)$. Continuity of $\rho: G \to \operatorname{GL}_n(\mathbb{C})$ means that each component function $g \mapsto \rho(g)_{ij}$ is continuous. [/definition] The continuity requirement is not a technicality — it is essential. To see why, consider $S^1 \cong \mathbb{R}/\mathbb{Z}$ viewed as a bare abstract abelian group, ignoring its topology. One can use a Hamel basis $\mathcal{A}$ for $\mathbb{R}$ over $\mathbb{Q}$ (with $1 \in \mathcal{A}$) to decompose $\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_{\alpha \in \mathcal{A} \setminus \{1\}} \mathbb{Q}\alpha$. This yields uncountably many distinct one-dimensional representations \begin{align*} \rho_\lambda(e^{2\pi i \mu}) = \begin{cases} 1 & \mu \notin \mathbb{Q}\lambda, \\ e^{2\pi i \mu} & \mu \in \mathbb{Q}\lambda, \end{cases} \end{align*} one for each $\lambda \in \mathcal{A} \setminus \{1\}$, none of which are continuous. Imposing continuity tames this profusion completely. ### Representations of $S^1$ The full classification of continuous representations of $S^1$ rests on two lemmas from real analysis. [quotetheorem:2469] [citeproof:2469] [quotetheorem:2470] [citeproof:2470] [quotetheorem:2471] [citeproof:2471] [remark: Completeness via Fourier Theory] Since $S^1$ is abelian, every continuous irreducible representation is one-dimensional, so the $\rho_n$ for $n \in \mathbb{Z}$ give a complete list. The character of a representation $V$ of $S^1$ takes the form \begin{align*} \chi_V(z) = \sum_{n \in \mathbb{Z}} a_n z^n, \end{align*} with $a_n \in \mathbb{Z}_{\geq 0}$ and only finitely many nonzero. The coefficient $a_n = \langle \chi_n, \chi_V \rangle = \frac{1}{2\pi}\int_0^{2\pi} e^{-in\theta} \chi_V(e^{i\theta})\, d\theta$ is the multiplicity of $\rho_n$ in $V$. This is exactly a Fourier decomposition. The Peter–Weyl theorem, which is highly nontrivial, shows that the $\chi_n$ form a complete orthonormal set in $L^2(S^1)$. [/remark] ### Haar Measure and the Transfer of Finite-Group Techniques The success of the finite-group theory relied on the ability to average over $G$: many constructions involved $\frac{1}{|G|}\sum_{g \in G}$. For compact groups, the correct substitute is integration against a **Haar measure**. [definition: Haar Measure] Let $G$ be a compact topological group. Write $\mathcal{C}(G)$ for the space of continuous class functions $f: G \to \mathbb{C}$ (satisfying $f(gxg^{-1}) = f(x)$ for all $g, x \in G$). A **Haar measure** on $G$ is a nontrivial linear functional $\int_G: \mathcal{C}(G) \to \mathbb{C}$, written $f \mapsto \int_G f(g)\, dg$, satisfying: 1. **Normalisation:** $\int_G 1\, dg = 1$. 2. **Translation invariance:** $\int_G f(xg)\, dg = \int_G f(g)\, dg = \int_G f(gx)\, dg$ for all $x \in G$. [/definition] [example: Haar Measure on Finite Groups] For a finite group $G$, the Haar measure is \begin{align*} \int_G f(g)\, dg = \frac{1}{|G|} \sum_{g \in G} f(g). \end{align*} The factor $1/|G|$ enforces normalisation. [/example] [example: Haar Measure on $S^1$] For $G = S^1$, the Haar measure is \begin{align*} \int_{S^1} f(g)\, dg = \frac{1}{2\pi}\int_0^{2\pi} f(e^{i\theta})\, d\theta. \end{align*} The factor $1/(2\pi)$ accounts for the total length of the circle. [/example] [quotetheorem:1063]We will not prove this theorem. Henceforth "compact" means "compact and Hausdorff," so the theorem applies. For $\operatorname{SU}(2)$, the Haar measure will be constructed explicitly later (the Weyl integration formula). Once the Haar measure is in hand, the averaging arguments from the finite theory go through verbatim. [quotetheorem:2472] [citeproof:2472] [quotetheorem:2473] [citeproof:2473] Characters are well-defined: if $\rho: G \to \operatorname{GL}(V)$ is a continuous representation, then $\chi_\rho = \operatorname{tr}\circ\, \rho$ is a continuous class function. We can endow $\mathcal{C}(G)$ with the inner product \begin{align*} \langle f, f' \rangle = \int_G \overline{f(g)}\, f'(g)\, dg. \end{align*} Schur's lemma holds for compact groups by the same proof as in the finite case (it uses only the algebraic structure of $G$-homomorphisms), and this yields: [quotetheorem:2474]## Representations of $\operatorname{SU}(2)$ We now turn to the classification of irreducible representations of $\operatorname{SU}(2)$, the group at the heart of the compact theory. Recall: \begin{align*} \operatorname{SU}(2) = \left\{ \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} : a, b \in \mathbb{C},\ |a|^2 + |b|^2 = 1 \right\} \cong S^3. \end{align*} ### Quaternions and the Haar Measure It is illuminating to embed $\operatorname{SU}(2)$ inside **Hamilton's quaternion algebra**: \begin{align*} \mathbb{H} = \left\{ \begin{pmatrix} z & w \\ -\bar{w} & \bar{z} \end{pmatrix} : z, w \in \mathbb{C} \right\} \leq M_2(\mathbb{C}). \end{align*} This is a 4-dimensional real vector space (with real coordinates coming from $\mathrm{Re}(z)$, $\mathrm{Im}(z)$, $\mathrm{Re}(w)$, $\mathrm{Im}(w)$), carrying the norm $\|A\|^2 = \det A$. Under this norm, $\operatorname{SU}(2)$ is exactly the unit sphere $\{A \in \mathbb{H} : \|A\|^2 = 1\}$. Elements of $\operatorname{SU}(2)$ act as isometries on $\mathbb{H}$ by left multiplication, since $\|AX\| = \|A\|\|X\| = \|X\|$ for $A \in \operatorname{SU}(2)$. The Haar measure on $\operatorname{SU}(2)$ is given (after normalisation by $1/(2\pi^2)$) by the usual integration of functions on the sphere $S^3 \subset \mathbb{R}^4$. ### Conjugacy Classes The structure of conjugacy classes in $\operatorname{SU}(2)$ mirrors what we know about $S^1$, via the **maximal torus**: \begin{align*} T = \left\{ \begin{pmatrix} a & 0 \\ 0 & \bar{a} \end{pmatrix} : a \in \mathbb{C},\ |a|^2 = 1 \right\} \cong S^1. \end{align*} [quotetheorem:2475] [citeproof:2475] Write $\mathcal{C}_t = \{g \in G : \frac{1}{2}\operatorname{tr}\, g = t\}$ for $t \in [-1, 1]$. The extreme classes are $\mathcal{C}_1 = \{I\}$ and $\mathcal{C}_{-1} = \{-I\}$. For $t \in (-1,1)$, $\mathcal{C}_t \cong S^2$ as a topological space — the conjugacy class is a 2-sphere parameterised by the unit eigenvector axis of $g$. ### The Standard Representations $V_n$ For each $n \geq 0$, define $V_n$ to be the complex vector space of homogeneous polynomials of degree $n$ in two variables: \begin{align*} V_n = \{r_0 x^n + r_1 x^{n-1}y + \cdots + r_n y^n : r_j \in \mathbb{C}\}. \end{align*} This has dimension $n+1$, with basis $\{x^n, x^{n-1}y, \ldots, y^n\}$. We define the action of $\operatorname{GL}_2(\mathbb{C})$ on $V_n$ by \begin{align*} \rho_n\!\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) f(x,y) = f(ax + cy,\, bx + dy). \end{align*} In other words, the group element acts by the transpose action on the row vector $(x\; y)$. Restricting to $\operatorname{SU}(2) \leq \operatorname{GL}_2(\mathbb{C})$ gives a representation $\rho_n: \operatorname{SU}(2) \to \operatorname{GL}(V_n)$ of dimension $n+1$. [example: Low-Dimensional Cases] For $n = 0$: $V_0 \cong \mathbb{C}$ and $\rho_0$ is the principal (one-dimensional) representation. For $n = 1$: $V_1 = \mathbb{C}^2$ with basis $\{x, y\}$, and $\rho_1(g) = g$ is the standard two-dimensional representation. For $n = 2$: $V_2 = \mathbb{C}^3$ with basis $\{x^2, xy, y^2\}$. With $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we compute $\rho_2(g)(x^2) = (ax+cy)^2 = a^2 x^2 + 2ac\, xy + c^2 y^2$, and similarly for the other basis vectors, giving the matrix \begin{align*} \rho_2(g) = \begin{pmatrix} a^2 & ab & b^2 \\ 2ac & ad+bc & 2bd \\ c^2 & cd & d^2 \end{pmatrix}. \end{align*} [/example] ### Characters and Irreducibility Since every conjugacy class in $\operatorname{SU}(2)$ meets $T$, a continuous class function is completely determined by its restriction to $T$. The restriction is necessarily **even**: $f\!\left(\mathrm{diag}(\lambda, \lambda^{-1})\right) = f\!\left(\mathrm{diag}(\lambda^{-1}, \lambda)\right)$, since the two diagonal matrices are conjugate via $s$. For a representation $V$ of $\operatorname{SU}(2)$, the restriction $\operatorname{Res}_T^{\operatorname{SU}(2)} V$ is a representation of $T \cong S^1$. Its character is a Laurent polynomial in $\lambda$, i.e., a finite $\mathbb{Z}_{\geq 0}$-linear combination of functions $\mathrm{diag}(\lambda, \lambda^{-1}) \mapsto \lambda^n$. We write $\mathbb{N}[z, z^{-1}]_{\mathrm{ev}}$ for the set of even such Laurent polynomials. We now compute the character of $V_n$ on $T$. The diagonal matrix $\mathrm{diag}(z, z^{-1})$ acts on the basis monomial $x^{n-j}y^j$ by: \begin{align*} \rho_n\!\left(\begin{pmatrix} z & 0 \\ 0 & z^{-1} \end{pmatrix}\right)(x^{n-j}y^j) = (zx)^{n-j}(z^{-1}y)^j = z^{n-2j}\, x^{n-j}y^j. \end{align*} So each basis monomial is an eigenvector with eigenvalue $z^{n-2j}$, and the character is \begin{align*} \chi_n\!\left(\begin{pmatrix} z & 0 \\ 0 & z^{-1} \end{pmatrix}\right) = z^n + z^{n-2} + \cdots + z^{-n} = \frac{z^{n+1} - z^{-(n+1)}}{z - z^{-1}}, \end{align*} the last expression being valid for $z \neq \pm 1$. [quotetheorem:2476] [citeproof:2476] [remark: Alternative via Weyl Integration Formula] An alternative proof uses the **Weyl integration formula**: for a class function $f$ on $\operatorname{SU}(2)$, \begin{align*} \int_G f(g)\, dg = \frac{1}{2\pi^2}\int_0^{2\pi} \frac{1}{2}f\!\left(\begin{pmatrix}e^{i\theta} & 0 \\ 0 & e^{-i\theta}\end{pmatrix}\right) 4\pi \sin^2\!\theta\, d\theta. \end{align*} Using this, one computes $\langle \chi_n, \chi_n \rangle = 1$, which implies irreducibility. This formula arises from the identification of conjugacy classes $\mathcal{C}_{\cos\theta}$ with 2-spheres of radius $\sin\theta$. [/remark] [quotetheorem:2477] [citeproof:2477] This is a beautiful result: the representations $V_0, V_1, V_2, \ldots$ of dimensions $1, 2, 3, \ldots$ exhaust all irreducible representations of $\operatorname{SU}(2)$. ### Tensor Products and the Clebsch–Gordan Rule Since $\operatorname{Res}_T^{\operatorname{SU}(2)} V \cong \operatorname{Res}_T^{\operatorname{SU}(2)} W$ implies $V \cong W$, it suffices to work with $T \cong S^1$ when computing characters of tensor products. [quotetheorem:2478] [citeproof:2478] [example: Low-Dimensional Tensor Products] We compute: \begin{align*} \chi_{V_1 \otimes V_1}(z) &= (z + z^{-1})^2 = z^2 + 2 + z^{-2} = \chi_{V_2} + \chi_{V_0}, \end{align*} so $V_1 \otimes V_1 \cong V_2 \oplus V_0$. Similarly, \begin{align*} \chi_{V_1 \otimes V_2}(z) &= (z + z^{-1})(z^2 + 1 + z^{-2}) = z^3 + 2z + 2z^{-1} + z^{-3} = \chi_{V_3} + \chi_{V_1}, \end{align*} giving $V_1 \otimes V_2 \cong V_3 \oplus V_1$. [/example] The general pattern is the **Clebsch–Gordan rule**: [quotetheorem:2479] [citeproof:2479] ## Representations of $\operatorname{SO}(3)$, $\operatorname{SU}(2)$ and $\operatorname{U}(2)$ With the complete classification for $\operatorname{SU}(2)$ in hand, we can deduce the representation theory of several related groups using structural isomorphisms. The following facts are quoted from Lie theory; they follow from explicit constructions using quaternions and the observation that a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism. [quotetheorem:2480]### Representations of $\operatorname{SO}(3)$ Since $\operatorname{SO}(3) \cong \operatorname{SU}(2)/\{\pm I\}$, representations of $\operatorname{SO}(3)$ correspond exactly to representations of $\operatorname{SU}(2)$ on which $-I$ acts as the identity. We compute the action of $-I$ on $V_n$: since $\rho_n(-I)(x^{n-j}y^j) = ((-1)x)^{n-j}((-1)y)^j = (-1)^n x^{n-j}y^j$, we get $\rho_n(-I) = (-1)^n I$. So $-I$ acts as the identity on $V_n$ if and only if $n$ is even. [quotetheorem:2481]The odd-dimensional-index representations $V_1, V_3, V_5, \ldots$ of $\operatorname{SU}(2)$ do not descend to $\operatorname{SO}(3)$; they are the spinor representations, which see $\operatorname{SU}(2)$ as the non-trivial double cover of $\operatorname{SO}(3)$. ### The Isomorphism $\operatorname{SO}(3) \cong \operatorname{SU}(2)/\{\pm I\}$ Explicitly We sketch the construction of the isomorphism via quaternions. [quotetheorem:2482] [citeproof:2482] ### Representations of $\operatorname{SO}(4)$ and $\operatorname{U}(2)$ For products of compact groups, the complete list of irreducible representations is obtained by taking tensor products of irreducibles of each factor: if $G$ and $H$ are compact groups with irreducibles $\{V_i\}$ and $\{W_j\}$, then $\{V_i \otimes W_j\}$ are the irreducibles of $G \times H$. [quotetheorem:2483]The parity condition $m \equiv n \pmod 2$ ensures that the central element $-(I,I) \in \operatorname{SU}(2) \times \operatorname{SU}(2)$ acts as the identity (it acts as $(-1)^m \otimes (-1)^n = (-1)^{m+n}$, which is $+1$ iff $m+n$ is even). [quotetheorem:2484][remark: Summary of the Classification] The compact group $\operatorname{SU}(2)$ serves as the linchpin of this story. Its representation theory is entirely determined by the spaces $V_n$ of homogeneous polynomials, indexed by degree $n \geq 0$. The representations of $\operatorname{SO}(3)$ are the even-indexed subset; those of $\operatorname{SO}(4)$ and $\operatorname{U}(2)$ are built from $\operatorname{SU}(2)$ irreducibles by the structural quotient isomorphisms. The Clebsch–Gordan rule governs how tensor products decompose, playing the same role that the character ring played for finite groups. [/remark] ## References Based on lecture notes from Cambridge Part II Representation Theory (Lent 2016, S. Martin), with course material structured for the Androma wiki.

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