Suppose you hold a square card and are asked: *in how many ways can you put it back so that it occupies exactly the same region of the table?* You can rotate it by $0°$, $90°$, $180°$, or $270°$; you can also flip it about any of its four axes of symmetry. These eight operations can be composed (do one, then another) and undone (every flip and rotation has a reverse). This simple observation — that the symmetries of an object form a system closed under composition and inversion — is the starting point of **group theory**. The power of the subject lies in abstraction: the same algebraic structure that describes the symmetries of a square also describes clock arithmetic modulo $n$, the permutations of a deck of cards, the invertible matrices acting on a vector space, and the conformal automorphisms of the Riemann sphere. By isolating the common axioms — closure, associativity, identity, inverses — we can prove theorems once and apply them everywhere. These notes develop the theory of [groups](/page/Group) from first principles, assuming only familiarity with basic set theory, [functions](/page/Function), and elementary linear algebra. The eight chapters trace a path from the axioms to substantial structural results: basic definitions and examples; the dihedral and symmetric groups; cosets and Lagrange's theorem; normal subgroups, quotient groups, and the First Isomorphism Theorem; direct products and the classification of small groups; group actions, including Cayley's theorem, Cauchy's theorem, and the simplicity of $A_5$; matrix groups and the $\mathrm{SO}_3$ rotation theorem; and Möbius groups on the extended complex plane. Several ideas recur throughout and are worth flagging at the outset. The **homomorphism philosophy**: to understand a group $G$, study its homomorphisms — the kernel tells you what information is lost, the image tells you where $G$ can be "seen," and the [First Isomorphism Theorem](/theorems/791) makes this precise. **Lagrange's constraint**: the order of every subgroup (and every element) divides $|G|$, a simple divisibility condition that, combined with its partial converses (Cauchy, Sylow), drives much of finite group theory. **Actions reveal structure**: watching a group act on a set — its own elements, its cosets, or an external geometric object — exposes internal structure invisible from the axioms alone. # Basic Definitions and Examples Throughout mathematics, we encounter [sets](/page/Set) equipped with operations: integers under addition, matrices under multiplication, symmetries of a geometric figure under composition. A natural question arises: *what common structure do these systems share, and what can we deduce from that structure alone?* The answer is the concept of a **group** — an algebraic structure that captures the essence of "reversible combination." The axioms are minimal (closure, associativity, identity, inverses), yet the consequences are remarkably deep: from the solvability of polynomial equations to the classification of crystal symmetries. This chapter builds the foundational language. We begin by isolating the axioms, then immediately confront two questions that the axioms raise but do not obviously answer: *Is the identity unique? Are inverses unique?* After resolving these, we introduce the key structural tools — subgroups, homomorphisms, and cyclic groups — each motivated by a concrete problem that the preceding theory cannot yet solve. ## Groups and Their Axioms Before we can study groups, we need a precise notion of "combining two elements." The familiar operations $+$ on $\mathbb{Z}$ and $\times$ on $\mathbb{R}$ share an important feature: given any two inputs from the set, they produce a single, unambiguous output in the same set. We isolate this as a definition. [definition:Binary Operation] A **binary operation** $*$ on a set $X$ is a function $* : X \times X \to X$, $(x, y) \mapsto x * y$. [/definition] The requirement that $*$ maps into $X$ (not into some larger set) is the **closure** condition: the operation never "escapes" the set. Subtraction on $\mathbb{N} = \{1, 2, 3, \ldots\}$ fails this condition, since $1 - 2 = -1 \notin \mathbb{N}$. [definition:Group] A **group** is a pair $(G, *)$ where $G$ is a set and $*$ is a binary operation on $G$ satisfying: 1. **Closure:** $x * y \in G$ for all $x, y \in G$. 2. **Identity:** There exists $e \in G$ with $x * e = x = e * x$ for all $x \in G$. 3. **Inverses:** For each $x \in G$, there exists $y \in G$ with $x * y = e = y * x$. 4. **Associativity:** $(a * b) * c = a * (b * c)$ for all $a, b, c \in G$. [/definition] The axioms deliberately say nothing about commutativity — the equation $x * y = y * x$ may or may not hold. When it does, the group is called **abelian**. [definition:Abelian Group] A group $G$ is **abelian** (or commutative) if $xy = yx$ for all $x, y \in G$. [/definition] [example:Groups And Non Groups] We verify the axioms in several settings, highlighting what goes wrong when they fail. 1. $(\mathbb{Z}, +)$ is a group: $e = 0$, and $x^{-1} = -x$. It is abelian since $a + b = b + a$. 2. $(\mathbb{Q}, +)$ is a group, as is $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$. 3. $(\mathbb{Z}, -)$ is **not** a group: associativity fails, since $(1 - 2) - 3 = -4$ but $1 - (2 - 3) = 2$. 4. $(\mathbb{Z}, \times)$ is **not** a group: the integer $2$ has no multiplicative inverse in $\mathbb{Z}$ (there is no $n \in \mathbb{Z}$ with $2n = 1$). The set $\mathbb{Q} \setminus \{0\}$ under multiplication *is* a group, but $\mathbb{Z}$ is too small. 5. $(\{+1, -1\}, \times)$ is a group: $e = 1$, and each element is its own inverse. The Cayley table is: | $\times$ | $+1$ | $-1$ | |---|---|---| | $+1$ | $+1$ | $-1$ | | $-1$ | $-1$ | $+1$ | This is the smallest non-trivial group, isomorphic to $C_2$. [/example] [example:Modular Arithmetic Group] The set $\{0, 1, 2\}$ under addition modulo $3$ is a group with $e = 0$. The Cayley table is: | $+_3$ | $0$ | $1$ | $2$ | |---|---|---|---| | $0$ | $0$ | $1$ | $2$ | | $1$ | $1$ | $2$ | $0$ | | $2$ | $2$ | $0$ | $1$ | Every row and column contains each element exactly once — this is a general feature of group Cayley tables (it follows from the cancellation law: if $ga = gb$ then $a = b$, so the map $x \mapsto gx$ is a bijection). This group is cyclic: $1 + 1 = 2$ and $1 + 1 + 1 = 0$, so $\langle 1 \rangle = \{0, 1, 2\}$. [/example] [example:Symmetries of the Equilateral Triangle] Label the vertices of an equilateral triangle as $1, 2, 3$. The symmetries are: $e$ (identity), $r$ (rotation by $120°$, $1 \to 2 \to 3 \to 1$), $r^2$ (rotation by $240°$, $1 \to 3 \to 2 \to 1$), $s_1$ (reflection fixing $1$, swapping $2 \leftrightarrow 3$), $s_2$ (reflection fixing $2$, swapping $1 \leftrightarrow 3$), $s_3$ (reflection fixing $3$, swapping $1 \leftrightarrow 2$). To compute $r \circ s_1$: first $s_1$ sends $1 \to 1, 2 \to 3, 3 \to 2$; then $r$ sends $1 \to 2, 3 \to 1, 2 \to 3$. Composing: $1 \to 2$, $2 \to 1$, $3 \to 3$. This fixes $3$ and swaps $1 \leftrightarrow 2$, so $r \circ s_1 = s_3$. The full Cayley table is: | $\circ$ | $e$ | $r$ | $r^2$ | $s_1$ | $s_2$ | $s_3$ | |---|---|---|---|---|---|---| | $e$ | $e$ | $r$ | $r^2$ | $s_1$ | $s_2$ | $s_3$ | | $r$ | $r$ | $r^2$ | $e$ | $s_3$ | $s_1$ | $s_2$ | | $r^2$ | $r^2$ | $e$ | $r$ | $s_2$ | $s_3$ | $s_1$ | | $s_1$ | $s_1$ | $s_2$ | $s_3$ | $e$ | $r$ | $r^2$ | | $s_2$ | $s_2$ | $s_3$ | $s_1$ | $r^2$ | $e$ | $r$ | | $s_3$ | $s_3$ | $s_1$ | $s_2$ | $r$ | $r^2$ | $e$ | This group is $D_6 \cong S_3$, the smallest non-abelian group (note $r \circ s_1 = s_3 \neq s_2 = s_1 \circ r$). [/example] The axioms demand that an identity exists and that each element has an inverse, but they do *not* assert uniqueness of either. Could a group have two different identity elements? If so, the notation $e$ and $g^{-1}$ would be ambiguous. [quotetheorem:764] The proof is short but structurally important: the uniqueness of the identity uses both the left- and right-identity properties simultaneously, while the uniqueness of inverses relies on associativity to "sandwich" one candidate against another. Without associativity, the proof breaks down — in non-associative structures (like quasigroups), inverses can fail to be unique. [citeproof:764] With uniqueness established, the notation $e$ and $g^{-1}$ is unambiguous. Useful identities follow: $(xy)^{-1} = y^{-1}x^{-1}$ (the "socks-and-shoes" rule), $(x^{-1})^{-1} = x$, and the cancellation law $xy = xz \implies y = z$. ### The Equivalent Group Definition The full axioms require two-sided identity and inverses. A natural question: *can we get away with less?* [quotetheorem:765] The key idea is that if $h^2 = h$ in a structure with right inverses, then $h$ must be the identity. This bootstraps right inverses into left inverses, which promotes the right identity to a left identity. Practically useful because it halves the verification work. [citeproof:765] ## Subgroups Given a group $G$, we often want to study "smaller groups living inside $G$." Not every subset works: $\{0, 1\} \subset \mathbb{Z}$ fails closure ($1 + 1 = 2 \notin \{0, 1\}$). [definition:Subgroup] Let $(G, *)$ be a group and $H \subseteq G$. Then $H$ is a **subgroup** of $G$, written $H \leq G$, if: (1) $e_G \in H$; (2) $h, k \in H \implies hk \in H$; (3) $h \in H \implies h^{-1} \in H$. [/definition] [example:Subgroup Verification in S3] Is $H = \{e, s_1\} \leq S_3$? Identity: $e \in H$ ✓. Closure: $s_1 s_1 = e \in H$ ✓. Inverses: $s_1^{-1} = s_1 \in H$ ✓. So $H \leq S_3$. But $K = \{e, s_1, s_2\}$ fails: $s_1 s_2 = r \notin K$. [/example] *Can we build new subgroups from old ones?* Unions fail, but intersections always work. [quotetheorem:767] The proof verifies the three subgroup conditions using both $A$ and $B$. The intersection of *any* family of subgroups is a subgroup. This underlies constructions like the centre $Z(G) = \bigcap_{h \in G} C_G(h)$. [citeproof:767] ### Subgroups of $\mathbb{Z}$ The integers under addition are the most familiar infinite group, yet even here the subgroup structure is not immediately obvious. We know $2\mathbb{Z}$ (even integers), $3\mathbb{Z}$ (multiples of $3$), and so on are subgroups — but are there others? Could some exotic subset of $\mathbb{Z}$, closed under addition and negation but not of the form $n\mathbb{Z}$, slip through? [quotetheorem:766] The answer is no: the subgroup lattice of $\mathbb{Z}$ is completely rigid. The proof hinges on the Division Algorithm — the same tool that underpins the Euclidean algorithm and unique factorisation in $\mathbb{Z}$. If a subgroup $H$ contains any nonzero element, then it contains a smallest positive element $n$ (by well-ordering), and division by $n$ shows every element of $H$ is a multiple of $n$. This result is the prototype for a recurring theme: in sufficiently "well-ordered" algebraic structures, every subobject is generated by a single element. The classification also explains why $\mathbb{Z}/n\mathbb{Z}$ is the only way to form a quotient of $\mathbb{Z}$ — every normal subgroup (and in $\mathbb{Z}$, every subgroup is normal since $\mathbb{Z}$ is abelian) is some $n\mathbb{Z}$. [citeproof:766] ## Homomorphisms and Isomorphisms *How do we compare two groups?* We need structure-preserving maps. [definition:Homomorphism] A function $\vartheta : G \to H$ between groups is a **homomorphism** if $\vartheta(xy) = \vartheta(x)\vartheta(y)$ for all $x, y \in G$. [/definition] The definition of a homomorphism asks only that products are preserved. But a group has more structure than just a product — it has an identity and inverses. Do we need to add extra conditions to ensure these are preserved, or does product-preservation alone suffice? If we had to check separately that $\vartheta(e_G) = e_H$ and $\vartheta(g^{-1}) = \vartheta(g)^{-1}$, the definition would be more cumbersome and harder to verify. Fortunately, neither extra condition is needed. [quotetheorem:768] The proof exploits a simple but powerful idea: $e_G \cdot e_G = e_G$, so applying $\vartheta$ produces an element that squares to itself. In a group, the only idempotent is the identity (by cancellation), so $\vartheta(e_G)$ is forced to be $e_H$. The same argument, applied to $g \cdot g^{-1} = e_G$, shows $\vartheta(g^{-1}) = \vartheta(g)^{-1}$. This is a recurring principle in algebra: operations that are uniquely determined by the group axioms (identity, inverses) are automatically preserved by any map that respects the product. It also provides a quick non-example test: if $f(e_G) \neq e_H$, then $f$ cannot be a homomorphism. [citeproof:768] The **image** $\operatorname{Im}(\vartheta) = \{\vartheta(g) : g \in G\}$ and the **kernel** $\ker(\vartheta) = \{g : \vartheta(g) = e_H\}$ are the two fundamental subgroups associated to any homomorphism. But why should the image be a subgroup at all? An arbitrary subset of $H$ need not be closed under products or inverses — the image inherits these properties precisely because $\vartheta$ translates closure in $G$ into closure in $H$. [quotetheorem:769] Each step of the proof traces a subgroup axiom through the homomorphism condition: the identity lands in the image by [Homomorphisms Preserve Identity](/theorems/768); products stay in the image because $\vartheta(c)\vartheta(d) = \vartheta(cd)$; and inverses stay because $\vartheta(b)^{-1} = \vartheta(b^{-1})$. The kernel has an even stronger property — it is not just a subgroup but a *normal* subgroup of $G$, since $\vartheta(g^{-1}kg) = \vartheta(g)^{-1} e_H \vartheta(g) = e_H$ for any $k \in \ker(\vartheta)$. This normality is what makes quotient constructions possible, and it is the entry point for the First Isomorphism Theorem in the next chapter. [citeproof:769] [example:Concrete Homomorphism Computation] $\vartheta : \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$, $n \mapsto n \bmod 4$. Homomorphism: $\vartheta(a+b) = \overline{a+b} = \bar{a} + \bar{b}$. Image: $\mathbb{Z}/4\mathbb{Z}$. Kernel: $4\mathbb{Z}$. Non-example: $f(n) = n + 1$ is not a homomorphism since $f(0) = 1 \neq 0$, violating [Homomorphisms Preserve Identity](/theorems/768). [/example] [definition:Isomorphism] A bijective homomorphism is an **isomorphism**. Write $G \cong H$ if one exists. [/definition] We want $\cong$ to behave like an equivalence relation — reflexivity is trivial (the identity map is an isomorphism), but symmetry and transitivity require proof. Symmetry asks: if $\vartheta : G \to H$ is an isomorphism, is the set-theoretic inverse $\vartheta^{-1} : H \to G$ also a homomorphism? There is no a priori reason why the inverse of a structure-preserving bijection should also preserve structure — this is a special feature of groups (and fails, for instance, for [continuous](/page/Continuity) bijections between [topological](/page/Topology) spaces). [quotetheorem:770] Part (i) verifies that composing two homomorphisms yields a homomorphism, and that bijectivity is inherited by the composition — establishing transitivity of $\cong$. Part (ii) proves the inverse of an isomorphism is an isomorphism by using the homomorphism property of $\vartheta$ "in reverse": if $\vartheta(cd) = \vartheta(c)\vartheta(d)$, then applying $\vartheta^{-1}$ to both sides gives $\vartheta^{-1}(ab) = \vartheta^{-1}(a)\vartheta^{-1}(b)$. Together, these two facts confirm that $\cong$ is an [equivalence relation](/page/Equivalence%20Relation) on groups, justifying the practice of treating isomorphic groups as "the same." [citeproof:770] [definition:Element Order] The **order** of $g \in G$, written $o(g)$, is the smallest positive $n$ with $g^n = e$ (if it exists). [/definition] Computing orders directly — testing $g, g^2, g^3, \ldots$ until the identity appears — is tedious. A more useful question is: given that $o(g) = m$, *which* powers $g^n$ equal $e$? If $m \mid n$ the answer is immediate: $g^n = (g^m)^{n/m} = e^{n/m} = e$. But the converse — that $g^n = e$ forces $m \mid n$ — is less obvious and requires the Division Algorithm. [quotetheorem:771] This lemma is the workhorse of finite group theory. Its forward direction uses minimality of the order: writing $n = qm + r$ with $0 \leq r < m$, if $g^n = e$ then $g^r = e$, and since $r < m = o(g)$, the only possibility is $r = 0$. The backward direction is immediate. The result appears wherever element orders interact with divisibility — in [Lagrange's Theorem](/theorems/780), in the classification of cyclic groups ($C_n$ has exactly one subgroup of each order dividing $n$), and in the theory of finite abelian groups. It also gives a quick way to compute orders: $o(g)$ is the smallest positive divisor $d$ of $|G|$ with $g^d = e$. [citeproof:771] If two groups are isomorphic, they should be "algebraically identical." But how do we *prove* two groups are *not* isomorphic? We need invariants — quantities preserved by any isomorphism. The order of the group is the crudest invariant ($|G| \neq |H|$ immediately rules out $G \cong H$). Element orders provide a much finer invariant. [quotetheorem:772] The proof uses the [Order Division Lemma](/theorems/771) twice: once through $\vartheta$ to show $o(\vartheta(a)) \mid o(a)$, and once through $\vartheta^{-1}$ (which is an isomorphism by [Functional Properties of Isomorphisms](/theorems/770)) to show $o(a) \mid o(\vartheta(a))$. Mutual divisibility of positive integers forces equality. This result is the primary tool for showing two groups of the same order are *not* isomorphic: if one group contains an element of order $k$ and the other does not, no isomorphism can exist. It also shows that properties defined in terms of element orders — such as the exponent of a group (the lcm of all element orders) — are isomorphism invariants. [citeproof:772] [example:Distinguishing C4 from C2 x C2] $C_4$ has elements of order $4$; $C_2 \times C_2$ has maximum order $2$ (by [Order in Direct Products](/theorems/793)). So $C_4 \not\cong C_2 \times C_2$ by [Isomorphisms Preserve Element Order](/theorems/772). [/example] ## Cyclic Groups Many of the groups we have encountered so far — $(\mathbb{Z}/n\mathbb{Z}, +_n)$, the $n$th roots of unity under multiplication, any group of prime order — share a remarkable property: every element is a power of a single generator. Such groups are called cyclic, and they are the simplest class of groups to classify completely. [definition:Cyclic Group] $H$ is **cyclic** if $H = \langle h \rangle = \{h^m : m \in \mathbb{Z}\}$ for some generator $h$. [/definition] Every cyclic group is abelian, since $h^a h^b = h^{a+b} = h^{b+a} = h^b h^a$. The abstract cyclic group of order $n$ is denoted $C_n$, and any cyclic group of order $n$ is isomorphic to $C_n$. To see this, define $\vartheta : \mathbb{Z} \to \langle g \rangle$ by $\vartheta(k) = g^k$; this is a surjective homomorphism with kernel $n\mathbb{Z}$ (by the [Order Division Lemma](/theorems/771), $g^k = e \iff n \mid k$), so the [First Isomorphism Theorem](/theorems/791) gives $\mathbb{Z}/n\mathbb{Z} \cong \langle g \rangle$. In particular, $C_n$ has exactly one subgroup of each order dividing $n$, and no others — this follows from the [Subgroups of the Integers](/theorems/766) classification, since subgroups of $\mathbb{Z}/n\mathbb{Z}$ correspond to subgroups of $\mathbb{Z}$ containing $n\mathbb{Z}$. [example:Cyclic Group Generators] $C_6 = \langle g \rangle$ has generators $g$ and $g^5$ (the elements coprime to $6$). Its subgroup lattice is: $\langle e \rangle \leq \langle g^3 \rangle \leq C_6$ and $\langle e \rangle \leq \langle g^2 \rangle \leq C_6$, corresponding to divisors $1 \mid 3 \mid 6$ and $1 \mid 2 \mid 6$. There is no subgroup of order $4$ or $5$, consistent with the fact that $4 \nmid 6$ and $5 \nmid 6$. [/example] When a group is not cyclic, we can still describe it using generators and relations. [definition:Presentation] A **presentation** $G = \langle g_1, \ldots, g_m \mid \text{relations} \rangle$ describes $G$ via generators and relations. For example, $D_{2n} = \langle r, t \mid r^n = t^2 = e, trt = r^{-1} \rangle$. [/definition] # The Dihedral and Symmetric Groups We now turn to two concrete families: the **dihedral groups** $D_{2n}$ (symmetries of polygons) and the **symmetric groups** $S_n$ (all permutations of a finite set). The symmetric groups are universal: [Cayley's Theorem](/theorems/795) shows every group embeds into some $S_n$. The central technical achievement is **cycle notation**, from which we read off the order, sign, and conjugacy class of any permutation. ## Dihedral Groups *What are all symmetries of a regular $n$-gon?* We can see rotations easily: $n$ of them, each by a multiple of $360°/n$. Reflections are also visible: $n$ axes of symmetry. But could there be *other* isometries we've missed — some exotic distance-preserving map of the polygon to itself? [citetheorem:773] The answer is no: exactly $2n$ symmetries exist. The proof works by placing the polygon at roots of unity in $\mathbb{C}$ and using a rigidity argument: any isometry is determined by where it sends two adjacent vertices, so there are at most $2n$ possibilities, and all $2n$ are realized by rotations and reflections. This "determined by two adjacent points" principle is the geometric heart of the result — it is the same rigidity that forces isometries of $\mathbb{R}^n$ to be affine maps. The relations $r^n = t^2 = e$ and $trt = r^{-1}$ encode the full multiplication table of $D_{2n}$; the last relation says that conjugating a rotation by a reflection reverses the angle. [citeproof:773] The group has presentation $D_{2n} = \langle r, t \mid r^n = t^2 = e, trt = r^{-1} \rangle$. The rotation subgroup $\langle r \rangle \cong C_n$ has index $2$, hence is normal by [Index Two Implies Normal](/theorems/789). ## The Symmetric Group [definition:Symmetric Group] $S_n = \operatorname{Sym}(\{1, \ldots, n\})$, the group of all bijections under composition. $|S_n| = n!$. [/definition] ### Cycle Notation [definition:Cycle] The **$k$-cycle** $(a_1\; \ldots\; a_k)$ sends $a_i \mapsto a_{i+1}$ (indices mod $k$) and fixes all other elements. A $2$-cycle is a **transposition**. [/definition] [example:Cycle Multiplication] Non-disjoint cycles don't commute. $(1\;2\;3)(1\;3)$: track $1 \to 3 \to 1$, $2 \to 2 \to 3$, $3 \to 1 \to 2$, giving $(2\;3)$. But $(1\;3)(1\;2\;3)$: track $1 \to 2 \to 2$, $2 \to 3 \to 1$, $3 \to 1 \to 3$, giving $(1\;2)$. Different results since supports overlap. [/example] Disjoint cycles commute — since they act on non-overlapping subsets of $\{1, \ldots, n\}$, their effects cannot interfere. This is not a trivial observation: non-disjoint cycles generally do *not* commute, as the preceding example shows. [citetheorem:774] The proof splits $\{1, \ldots, n\}$ into three sets — the support of $\sigma$, the support of $\tau$, and the fixed points of both — and verifies that $\sigma\tau$ and $\tau\sigma$ agree on each. The disjointness hypothesis is used at every step: without it, $\sigma$ could rearrange elements that $\tau$ is about to move, producing a different outcome. This commutativity is what makes the disjoint cycle decomposition canonical and is the foundation for the lcm formula for orders. [citeproof:774] Every permutation has a canonical "normal form" — a product of disjoint cycles. The existence of this decomposition is what makes cycle notation so powerful: it converts an arbitrary bijection into a collection of independent circular motions. [citetheorem:775] The proof is algorithmic: trace the orbit of any element under $\sigma$ until it returns to the start (which it must, by finiteness), record the resulting cycle, then repeat on untouched elements. Injectivity of $\sigma$ guarantees the cycles are disjoint — if two orbits shared an element, the orbit-tracing procedure would have merged them. Uniqueness (up to ordering) follows because the cycles are exactly the orbits of $\sigma$, which are determined by $\sigma$ alone. This decomposition underpins everything that follows: cycle type determines conjugacy class, and the number of cycles determines the sign. [citeproof:775] Given a permutation in disjoint cycle form, what is its order? Testing powers $\sigma, \sigma^2, \sigma^3, \ldots$ is inefficient. The cycle decomposition gives an immediate answer. [citetheorem:776] The proof uses two ingredients: commutativity of disjoint cycles (from [Disjoint Cycles Commute](/theorems/774), which lets us compute $(\sigma\tau)^n = \sigma^n\tau^n$) and the [Order Division Lemma](/theorems/771) to show the lcm is both an upper bound and the exact order. The key insight in the minimality argument is that $(\sigma\tau)^m = e$ forces $\sigma^m = \tau^{-m}$, but since $\sigma$ and $\tau$ have disjoint supports, both sides must equal the identity — so $o(\sigma) \mid m$ and $o(\tau) \mid m$ simultaneously, forcing $\operatorname{lcm}(o(\sigma), o(\tau)) \mid m$. [citeproof:776] ### Transpositions and the Sign The disjoint cycle decomposition is canonical, but cycles themselves are "composite" objects — a $k$-cycle permutes $k$ elements in a single loop. Is there a smaller building block? The simplest non-trivial permutation is a transposition (a $2$-cycle), and it turns out these generate the entire symmetric group. [citetheorem:777] The proof reduces the problem to cycles via the [Disjoint Cycle Decomposition](/theorems/775), then decomposes each $k$-cycle as $(a_1\; a_2)(a_2\; a_3) \cdots (a_{k-1}\; a_k)$, a product of $k - 1$ transpositions. The decomposition is *not* unique — for instance, $(1\;2\;3) = (1\;2)(2\;3) = (1\;3)(1\;2)$ — but the *parity* of the number of transpositions turns out to be an invariant. This invariance is not at all obvious and requires a separate proof. [citeproof:777] The non-uniqueness of transposition decompositions raises a fundamental question: is the parity (even or odd number of transpositions) an invariant of the permutation, or could the same permutation be written as both an even and an odd product? The sign homomorphism resolves this. [citetheorem:778] The proof introduces a clever invariant: $c(\sigma)$, the number of cycles (including $1$-cycles) in the disjoint cycle decomposition. The key claim is that multiplying by *any* transposition changes the cycle count by exactly $\pm 1$ — either merging two cycles or splitting one. Since the identity has $c(\mathrm{id}) = n$, a product of $k$ transpositions has cycle count $\equiv n + k \pmod{2}$. Two different decompositions of the same $\sigma$ must produce the same cycle count $c(\sigma)$, forcing the transposition counts to have the same parity. The homomorphism property is then immediate: concatenating transposition decompositions adds their lengths. [citeproof:778] [example:Computing the Sign] $\sigma = (1\;3\;5\;2)(4\;6) \in S_6$: transposition count is $(3 + 1) = 4$ (even), so $\operatorname{sgn}(\sigma) = +1$. Equivalently, $(-1)^{4-1}(-1)^{2-1} = (-1)^4 = +1$. [/example] ### The Alternating Group The sign homomorphism $\operatorname{sgn} : S_n \to \{+1, -1\}$ splits $S_n$ into two halves: the even permutations ($\operatorname{sgn} = +1$) and the odd permutations ($\operatorname{sgn} = -1$). The even permutations form the kernel of a homomorphism — so by general theory, they form a normal subgroup. This subgroup turns out to be one of the most important objects in all of group theory. [citetheorem:779] The proof identifies $A_n = \ker(\operatorname{sgn})$, which is automatically a normal subgroup (kernels always are). Since $\operatorname{sgn}$ is surjective, the [First Isomorphism Theorem](/theorems/791) gives $S_n/A_n \cong \{+1, -1\} \cong C_2$, so $|A_n| = |S_n|/2 = n!/2$. The equal-count argument (there are as many even as odd permutations) is proved by exhibiting a bijection: left-multiplying by any fixed transposition sends even permutations to odd ones and vice versa. For $n \geq 5$, $A_n$ is simple — a deep result proved via conjugacy class analysis in the [group actions chapter](#group-actions). The simplicity of $A_5$ is historically significant: it is the obstruction to solving the general quintic by radicals, as Galois discovered. [citeproof:779] # Cosets and Lagrange's Theorem *Given a finite group $G$ and a subgroup $H$, what is the relationship between $|H|$ and $|G|$?* The answer — $|H|$ divides $|G|$ — is **Lagrange's Theorem**. The proof: left cosets of $H$ partition $G$ into equal-sized pieces. ## Cosets Given a subgroup $H \leq G$, we want to understand how $H$ "sits inside" $G$. The natural approach is to look at the translates of $H$ — the sets obtained by multiplying every element of $H$ by a fixed group element. [definition:Left Coset] The **left coset** of $H \leq G$ by $g$ is $gH = \{gh : h \in H\}$. The **index** $|G : H|$ is the number of distinct left cosets. [/definition] If we want to count how many cosets there are and use them to relate $|H|$ to $|G|$, we need two facts: all cosets have the same size, and they don't overlap. The first is straightforward. [citetheorem:780] The bijection $h \mapsto gh$ is simply left multiplication — an operation that is always injective in a group (by the cancellation law $gh_1 = gh_2 \implies h_1 = h_2$). Since it maps $H$ onto $gH$ and is injective, it is a bijection. This means every coset has exactly $|H|$ elements, regardless of the representative $g$. The proof is short, but the consequence is powerful: it converts the problem of counting group elements into the problem of counting cosets. [citeproof:780] The second ingredient is that cosets cannot partially overlap — any two cosets are either identical or completely disjoint. This is the key structural fact that turns cosets into a partition. [citetheorem:781] The proof uses a "shared element" argument: if $c \in aH \cap bH$, then $aH \subseteq cH$ (since $a$ can be expressed in terms of $c$ and an element of $H$) and $cH \subseteq aH$ by the reverse argument, giving $aH = cH$. Applying the same reasoning with $b$ gives $bH = cH$, so $aH = bH$. The logic is elegant: a single shared element forces full equality, because every element of a coset determines the entire coset. Together with Step 1 (every $g \in G$ lies in $gH$, since $e \in H$), this proves the cosets partition $G$. [citeproof:781] [example:Coset Partition of S3] $G = S_3$, $H = \{e, r, r^2\}$. Two cosets: $H = \{e, r, r^2\}$ and $s_1 H = \{s_1, s_2, s_3\}$. Partition: rotations $\dot\cup$ reflections, each of size $3$. With $K = \{e, s_1\}$: three cosets $\{e, s_1\}$, $\{r, s_3\}$, $\{r^2, s_2\}$. Notice $rK = \{r, s_3\} \neq Kr = \{r, s_2\}$ — the left and right cosets differ because $K$ is *not* normal. [/example] In practice, one often needs to test whether two cosets are equal without listing all their elements. The following criterion reduces this to a single membership test. [citetheorem:786] The criterion $aH = bH \iff a^{-1}b \in H$ says that coset equality is governed by the "ratio" $a^{-1}b$: two elements represent the same coset precisely when they differ by an element of $H$. The forward direction is immediate: if $aH = bH$ then $b = ae$ lies in $aH$, so $b = ah$ for some $h \in H$, giving $a^{-1}b = h \in H$. The backward direction follows from the partition property: if $a^{-1}b \in H$ then $b \in aH \cap bH$, forcing $aH = bH$. This criterion is used constantly — in proving that coset multiplication is well-defined for quotient groups, and in the [Orbit-Stabiliser Theorem](/theorems/796). [citeproof:786] ## Lagrange's Theorem With the coset machinery in place — equal-sized cosets that partition $G$ — the counting argument writes itself. The result is one of the most powerful constraints in finite group theory: the order of any subgroup must divide the order of the group. [citetheorem:782] The proof is a one-line combination of the two coset facts: $G$ is the disjoint union of $|G : H|$ cosets, each of size $|H|$, so $|G| = |G : H| \cdot |H|$. Despite its simplicity, the theorem has remarkably far-reaching consequences. It immediately constrains which subgroups can exist: a group of order $12$ can only have subgroups of order $1, 2, 3, 4, 6$, or $12$. The converse is false in general — $A_4$ has order $12$ but no subgroup of order $6$ — and understanding exactly which divisors arise requires deeper tools: [Cauchy's Theorem](/theorems/797) for prime divisors and Sylow's theorems for prime-power divisors. [citeproof:782] A first application: every element's order divides the group order. This connects the "local" datum $o(g)$ (a property of a single element) to the "global" datum $|G|$ (a property of the whole group). [citetheorem:783] The proof applies [Lagrange's Theorem](/theorems/782) to the cyclic subgroup $\langle g \rangle$, which has order $o(g)$. Since $\langle g \rangle \leq G$, we get $o(g) \mid |G|$. Writing $|G| = k \cdot o(g)$, we obtain the exponent identity $g^{|G|} = (g^{o(g)})^k = e^k = e$. This universal exponent — every element raised to the $|G|$-th power gives the identity — is the group-theoretic backbone of the Fermat–Euler theorem in number theory. [citeproof:783] The constraint $o(g) \mid |G|$ is especially powerful when $|G|$ is prime, because then the only divisors are $1$ and $|G|$ itself. [citetheorem:784] Any non-identity element $g$ satisfies $o(g) \mid p$, so $o(g) = p$ (since $o(g) \neq 1$). Therefore $\langle g \rangle = G$, and $G$ is cyclic. This is the complete classification of groups of prime order: there is exactly one, up to isomorphism, namely $C_p$. The result also shows that groups of prime order are automatically abelian and simple — they have no nontrivial proper subgroups at all. [citeproof:784] The most celebrated number-theoretic application of Lagrange's theorem is the Fermat–Euler theorem, which translates the abstract group-theoretic identity $g^{|G|} = e$ into a concrete congruence. [citetheorem:785] The proof works by identifying the correct group: the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^\times$ of units modulo $n$, which has order $\varphi(n)$ by definition. The coprimality condition $\gcd(a, n) = 1$ is precisely the statement that $a$ is a unit modulo $n$ — i.e., $a \in (\mathbb{Z}/n\mathbb{Z})^\times$. Applying [Element Order Divides Group Order](/theorems/783) gives $a^{\varphi(n)} \equiv 1 \pmod{n}$. The special case $n = p$ prime gives Fermat's little theorem $a^{p-1} \equiv 1 \pmod{p}$, since $\varphi(p) = p - 1$. This is a striking example of abstract algebra solving a concrete number-theoretic problem: the group axioms do the work that would otherwise require delicate modular arithmetic. [citeproof:785] # Normal Subgroups, Quotient Groups, and Homomorphisms Lagrange's theorem tells us that a subgroup $H$ partitions $G$ into cosets of equal size. A natural next question is: *can we make the set of cosets into a group?* The answer is not always yes. If we try to define $(aH)(bH) = abH$, different representatives of the same coset may yield different products — the operation is not well-defined for arbitrary subgroups. The concept of a **normal subgroup** resolves this obstacle, and the resulting **quotient group** $G/K$ is one of the central constructions in all of algebra. The **First Isomorphism Theorem** then connects quotient groups to homomorphisms, establishing that every quotient is the image of some homomorphism and every image is a quotient. [example:Coset Multiplication Failure] In $S_3$ with $H = \{e, (1\;2)\}$: $(1\;3)(2\;3) = (1\;2\;3)$ gives coset $(1\;2\;3)H$, but $(1\;2\;3)(1\;3\;2) = e$ gives coset $H$. The operation is not well-defined because $H$ is not normal. [/example] ## Normal Subgroups [definition:Normal Subgroup] $K \unlhd G$ if $gK = Kg$ for all $g \in G$. [/definition] Not every subgroup is normal — the example above shows that $\{e, (1\;2)\} \leq S_3$ fails $gK = Kg$. But normality can be checked in several equivalent ways, and it is useful to have all of them available. [citetheorem:787] The three characterisations serve different purposes. Condition (i) — $gK = Kg$ — is the one needed to make quotient group multiplication well-defined. Condition (iii) — $gkg^{-1} \in K$ for all $k \in K$, $g \in G$ — is easiest to verify in practice, since it only requires checking individual elements rather than entire cosets. Condition (ii) — $gKg^{-1} = K$ — is useful for connecting normality to conjugation: $K$ is normal precisely when it is invariant under every inner automorphism $\gamma_g : x \mapsto gxg^{-1}$. The proof that these are equivalent is a clean cyclic implication: (i) $\Rightarrow$ (ii) by multiplying on the right by $g^{-1}$, (ii) $\Rightarrow$ (iii) by element-chasing, and (iii) $\Rightarrow$ (i) by showing both inclusions $gK \subseteq Kg$ and $Kg \subseteq gK$. [citeproof:787] Where do normal subgroups come from in practice? The most natural source is kernels of homomorphisms. In fact, the relationship is a perfect correspondence: every kernel is normal, and every normal subgroup is the kernel of some homomorphism. [citetheorem:788] The proof that $\ker(\vartheta) \unlhd G$ is a direct computation: $\vartheta(gkg^{-1}) = \vartheta(g) e_H \vartheta(g)^{-1} = e_H$, so $gkg^{-1} \in \ker(\vartheta)$. The converse — that every normal subgroup $K$ is the kernel of some homomorphism — is supplied by the quotient map $\pi : G \to G/K$, $g \mapsto gK$, which has $\ker(\pi) = K$. This establishes a tight link between the algebraic notion of normality and the structural notion of "what gets collapsed by a homomorphism." In the next section, we will see that the First Isomorphism Theorem makes this link even tighter. [citeproof:788] A simple but widely used sufficient condition for normality: any subgroup of index $2$ is automatically normal. [citetheorem:789] The proof is a counting argument with no computation at all. With only two left cosets ($K$ and $G \setminus K$) and two right cosets (also $K$ and $G \setminus K$), the non-$K$ cosets must agree: $gK = G \setminus K = Kg$ for any $g \notin K$. This applies immediately to several important subgroups: $A_n \leq S_n$ (index $2$), the rotation subgroup $\langle r \rangle \leq D_{2n}$ (index $2$), and $\mathrm{SL}_n \leq \mathrm{GL}_n$ (index $|\mathbb{R}^*| = \infty$, so this criterion does not apply there — but $\mathrm{SL}_n$ is still normal as the kernel of $\det$). [citeproof:789] ## Quotient Groups We have seen that cosets of a normal subgroup behave well — left and right cosets coincide, and the "ratio" criterion $aK = bK \iff a^{-1}b \in K$ gives a clean equivalence. The natural question is: can we make the set of cosets itself into a group? The answer is yes, provided $K$ is normal. [citetheorem:790] The crucial step is well-definedness: if we replace $a$ by another representative $a' \in aK$ and $b$ by $b' \in bK$, does $a'b'K = abK$? The answer depends on normality. Writing $a' = ak_1$ and $b' = bk_2$, we get $a'b' = ak_1bk_2$. For this to lie in $abK$, we need $b^{-1}k_1b \in K$ — which is exactly the normality condition. Without normality, the operation $(aK)(bK) = abK$ is ambiguous, as the $S_3$ example at the start of this chapter demonstrates. Once well-definedness is established, the group axioms are inherited from $G$ in a routine verification. The quotient group $G/K$ is a "coarsened" version of $G$ in which elements of $K$ have been identified with the identity. [citeproof:790] [example:Quotient Group Computation] $S_3 / \langle r \rangle$: two cosets $\{e,r,r^2\}$ and $\{s_1,s_2,s_3\}$, multiplication table of $C_2$. The quotient "forgets" which rotation/reflection, remembering only orientation. $\mathbb{Z}/3\mathbb{Z}$: three cosets, giving $C_3$ — modular arithmetic. [/example] ## The First Isomorphism Theorem We now have all the ingredients — homomorphisms, kernels, quotient groups — and the First Isomorphism Theorem ties them together into a single powerful statement. The problem it solves: given a homomorphism $\vartheta : G \to H$, we want to understand $\operatorname{Im}(\vartheta)$, which lives inside the potentially complicated group $H$. The theorem says we can instead study $G/\ker(\vartheta)$, which lives inside $G$ where we have more control. [citetheorem:791] The proof constructs the induced map $\bar\vartheta : G/\ker(\vartheta) \to \operatorname{Im}(\vartheta)$ by $gK \mapsto \vartheta(g)$. Well-definedness is the key step: two elements in the same coset have the same image (since they differ by a kernel element, which maps to $e_H$). Injectivity is then automatic: if $\bar\vartheta(aK) = \bar\vartheta(bK)$, then $\vartheta(a) = \vartheta(b)$, so $a^{-1}b \in \ker(\vartheta) = K$, giving $aK = bK$. Surjectivity onto $\operatorname{Im}(\vartheta)$ is immediate. The theorem is used constantly to identify quotient groups: to show $G/K \cong T$, one constructs a surjective homomorphism $G \to T$ with kernel $K$. This converts potentially difficult "quotient group" questions into "find the right homomorphism" questions. [citeproof:791] A closely related result characterises injectivity in terms of the kernel, providing the most efficient test for whether a homomorphism is injective. [citetheorem:792] The forward direction is trivial: if $\vartheta$ is injective and $\vartheta(g) = e_H = \vartheta(e_G)$, then $g = e_G$. The backward direction uses the homomorphism property: $\vartheta(a) = \vartheta(b)$ gives $\vartheta(a^{-1}b) = e_H$, so $a^{-1}b \in \ker(\vartheta) = \{e\}$, forcing $a = b$. This criterion is the mechanism behind [Cayley's Theorem](/theorems/795): the left-multiplication action $G \to \operatorname{Sym}(G)$ has trivial kernel (if $gx = x$ for all $x$, then $g = e$), so it is injective, embedding $G$ into a symmetric group. [citeproof:792] [example:First Isomorphism Theorem Applied to Sign] $\operatorname{sgn} : S_n \to \{+1,-1\}$ is surjective with kernel $A_n$. By the FIT: $S_n/A_n \cong C_2$, confirming $|A_n| = n!/2$. [/example] # Direct Products and Small Groups With the structural toolkit now in place — subgroups, quotient groups, homomorphisms, the First Isomorphism Theorem — we can attack a concrete classification problem: *what are all groups of small order, up to isomorphism?* The main construction is the **direct product** $H \times K$, which builds a new group from two old ones by operating componentwise. The **Internal Direct Product Theorem** gives conditions under which a group decomposes as such a product, and applying it systematically classifies all groups up to order $10$ (and beyond, with additional tools from group actions). ## The Direct Product Given two groups $H$ and $K$, the simplest way to build a new group is to take all ordered pairs $(h, k)$ and combine them componentwise. This is the external direct product. [definition:External Direct Product] The **(external) direct product** $H \times K$ is the set $\{(h, k) : h \in H, k \in K\}$ with operation $(h_1, k_1)(h_2, k_2) = (h_1 h_2, k_1 k_2)$. Identity: $(e_H, e_K)$. Inverse: $(h,k)^{-1} = (h^{-1}, k^{-1})$. [/definition] To use direct products for classification, we need to know the element orders. When does $(h, k)^n = (e_H, e_K)$? This requires $h^n = e_H$ *and* $k^n = e_K$ simultaneously — so $n$ must be divisible by both $o(h)$ and $o(k)$. [citetheorem:793] The proof uses the [Order Division Lemma](/theorems/771) three times: once to show $\operatorname{lcm}(o(h), o(k))$ is an exponent that kills $(h,k)$, twice more to show it is minimal. The result immediately distinguishes $C_4$ from $C_2 \times C_2$: in $C_4$, the generator has order $4$, but in $C_2 \times C_2$, every element $(a,b)$ satisfies $o((a,b)) = \operatorname{lcm}(o(a), o(b)) \leq \operatorname{lcm}(2, 2) = 2$. So $C_2 \times C_2$ has no element of order $4$ and cannot be isomorphic to $C_4$ — this is the **Klein four-group** $V_4$. More generally, the formula shows $C_m \times C_n \cong C_{mn}$ if and only if $\gcd(m,n) = 1$ (the [Chinese Remainder Theorem](/theorems/734) for groups). [citeproof:793] ### The Internal Direct Product The direct product constructs a new group from two separate groups. But often we encounter a group $G$ that we suspect is a direct product of two of its own subgroups. When can we conclude $G \cong H \times K$ for subgroups $H, K \leq G$? The answer requires three conditions. [citetheorem:794] The most interesting step in the proof is showing that elements of $H$ and $K$ commute: the commutator $hkh^{-1}k^{-1}$ lies in both $H$ and $K$ (using normality of each), so it lies in $H \cap K = \{e\}$, forcing $hk = kh$. Once commutativity is established, the map $(h, k) \mapsto hk$ is shown to be a homomorphism (commutativity is needed to verify $(h_1k_1)(h_2k_2) = (h_1h_2)(k_1k_2)$), injective (if $hk = e$ then $h = k^{-1} \in H \cap K = \{e\}$), and surjective (by the hypothesis $HK = G$). This theorem is the primary tool for classifying small groups: one identifies normal subgroups satisfying the three conditions and reads off the direct product decomposition. [citeproof:794] ## Classifying Small Groups **Primes ($2, 3, 5, 7$):** $|G| = p \implies G \cong C_p$ by [Prime Order Implies Cyclic](/theorems/784). **Order $4$:** Either $G \cong C_4$ (some element of order $4$) or $G \cong C_2 \times C_2$ (all non-identity elements order $2$, apply [Internal Direct Product](/theorems/794)). **Order $6$:** Either $G \cong C_6$ (element of order $6$) or $G \cong D_6 \cong S_3$. In the latter case, an element $r$ of order $3$ gives $\langle r \rangle \unlhd G$ (index $2$), and an element $s$ of order $2$ satisfies $s^{-1}rs \in \{e, r, r^2\}$. The cases $s^{-1}rs = e$ and $s^{-1}rs = r$ both lead to contradictions, forcing $s^{-1}rs = r^{-1}$ — the dihedral relation. [example:Why Order 6 Has No Third Group] Normality of $\langle r \rangle$ restricts $s^{-1}rs$ to the three-element set $\{e, r, r^2\}$. Two options are eliminated, and the third is forced. This illustrates how Lagrange, Cauchy, and normality leave no room for a third group of order $6$. [/example] **Order $8$:** Five groups: $C_8$, $C_4 \times C_2$, $C_2^3$, $D_8$, and the quaternion group $Q_8$. [definition:Quaternion Group] $Q_8 = \langle a, b : a^4 = e, b^2 = a^2, bab^{-1} = a^{-1} \rangle$, concretely $\{1, -1, i, -i, j, -j, k, -k\}$ with $i^2 = j^2 = k^2 = ijk = -1$. Distinguished from $D_8$ by having only $1$ element of order $2$ (versus $5$ for $D_8$). [/definition] For $|G| = p^2$: always abelian by the [Classification of Groups of Order p Squared](/theorems/818), giving $C_{p^2}$ or $C_p \times C_p$. # Group Actions Every group arises in practice as the symmetries of *something* — a polygon, a molecule, a set of solutions. So far, we have studied groups in isolation, extracting structure from the axioms alone. **Group actions** reverse this perspective: we let a group $G$ operate on a set $X$ and read off information about $G$ from the combinatorics of orbits and stabilisers. The main results are the Orbit-Stabiliser Theorem (the group-action analogue of Lagrange), Cauchy's Theorem (a partial converse to Lagrange for prime divisors), and the class equation (which unlocks the structure of $p$-groups). ## Group Actions [definition:Group Action] A **(left) group action** of $G$ on $X$ is a function $G \times X \to X$, written $g \cdot x$, satisfying $g \cdot (h \cdot x) = (gh) \cdot x$ and $e \cdot x = x$. The action is **faithful** if only $e$ fixes every point. [/definition] [definition:Orbit] The **orbit** of $x$ is $\operatorname{Orb}_G(x) = \{g \cdot x : g \in G\}$. [/definition] [definition:Stabiliser] The **stabiliser** of $x$ is $\operatorname{Stab}_G(x) = \{g \in G : g \cdot x = x\} \leq G$. [/definition] The distinct orbits partition $X$. Four fundamental actions: left multiplication ($g \cdot h = gh$, always faithful), conjugation ($g \cdot h = ghg^{-1}$, kernel is $Z(G)$), coset action, and geometric actions. [example:Orbits and Stabilisers in D8] $D_8$ acts on the vertices $\{1,2,3,4\}$ of a square. The orbit of vertex $1$: $e(1)=1$, $r(1)=2$, $r^2(1)=3$, $r^3(1)=4$, so $\operatorname{Orb}(1) = \{1,2,3,4\}$ — transitive. The stabiliser: checking all $8$ elements, only $e$ and the reflection $t$ (which fixes $1$ and $3$) stabilise vertex $1$, so $\operatorname{Stab}(1) = \{e, t\} \cong C_2$. Check: $|\operatorname{Orb}| \cdot |\operatorname{Stab}| = 4 \cdot 2 = 8 = |D_8|$. ✓ [/example] ### Cayley's Theorem Every group acts on itself by left multiplication: $g \cdot h = gh$. This action is always faithful — if $gh = h$ for all $h$, then $g = e$. The resulting embedding of $G$ into a symmetric group is Cayley's theorem, which answers a foundational question: *are abstract groups genuinely more general than permutation groups?* [citetheorem:795] The answer is no. The left regular action gives a homomorphism $\Phi : G \to \operatorname{Sym}(G)$ with $\ker(\Phi) = \{e\}$ (faithfulness). By the [First Isomorphism Theorem](/theorems/791), $G \cong \operatorname{Im}(\Phi) \leq \operatorname{Sym}(G)$. The theorem is conceptually important — it shows that every group is a permutation group — but the embedding is usually inefficient: $\operatorname{Sym}(G)$ has order $|G|!$, far larger than $|G|$. More efficient embeddings come from more carefully chosen actions, as we will see with the conjugation and coset actions below. [citeproof:795] ### The Orbit–Stabiliser Theorem The orbit of a point $x \in X$ records *where* $G$ can send $x$, and the stabiliser records *which elements of $G$ leave $x$ fixed*. The Orbit-Stabiliser Theorem quantifies the tradeoff between these two: a larger orbit means a smaller stabiliser, and vice versa, with the product always equal to $|G|$. [citetheorem:796] The proof constructs an explicit bijection $\operatorname{Orb}_G(x) \leftrightarrow G/\operatorname{Stab}_G(x)$: the map sends $g \cdot x$ to the coset $g\operatorname{Stab}_G(x)$. This is well-defined because $g \cdot x = h \cdot x$ precisely when $h^{-1}g$ fixes $x$, i.e., when $g$ and $h$ lie in the same coset of the stabiliser. Combining with [Lagrange's Theorem](/theorems/782) gives $|G| = |\operatorname{Orb}| \cdot |\operatorname{Stab}|$. This formula is the primary counting tool for the rest of the course: it determines the order of symmetry groups (via transitive actions), the sizes of conjugacy classes (via the conjugation action), and the number of Sylow subgroups (via conjugation on Sylow subgroups). [citeproof:796] [example:Symmetries of the Tetrahedron] The symmetry group $G$ of the regular tetrahedron acts transitively on $4$ vertices. $\operatorname{Stab}(1)$ consists of the $6$ symmetries of the opposite face. So $|G| = 4 \cdot 6 = 24 = |S_4|$. Since the action is faithful, $G \cong S_4$. [/example] ## Cauchy's Theorem Lagrange's theorem says $|H|$ divides $|G|$ for any subgroup $H$. The converse fails in general ($A_4$ has no subgroup of order $6$), but a partial converse holds for *prime* divisors: if $p \mid |G|$, then $G$ has a subgroup of order $p$ — equivalently, an element of order $p$. This is Cauchy's theorem, and its proof is one of the most elegant applications of group actions. [citetheorem:797] The proof sets up an action of $C_p$ on the set $X$ of $p$-tuples $(x_1, \ldots, x_p)$ with $x_1 \cdots x_p = e$, where $C_p$ acts by cyclic permutation. Since $|X| = |G|^{p-1}$ is divisible by $p$, and each orbit has size $1$ or $p$ (as $p$ is prime), the number of fixed points must be divisible by $p$. The trivial fixed point $(e, \ldots, e)$ contributes one, so there must be at least $p - 1$ others, each of the form $(g, g, \ldots, g)$ with $g^p = e$ and $g \neq e$. The elegance lies in reducing a structural question (existence of an element of order $p$) to a counting argument modulo $p$. Cauchy's theorem is the first step toward Sylow's theorems, which extend the result to prime-power divisors. [citeproof:797] ## The Conjugacy Action The orbits of conjugation $g \cdot h = ghg^{-1}$ are the **conjugacy classes** $\operatorname{ccl}_G(h)$. The stabiliser of $h$ is the **centraliser** $C_G(h)$, and elements with singleton conjugacy classes form the **centre** $Z(G)$. [definition:Conjugacy Class] $\operatorname{ccl}_G(h) = \{ghg^{-1} : g \in G\}$. [/definition] The **class equation** is $|G| = |Z(G)| + \sum_{\text{non-central}} |\operatorname{ccl}_G(x)|$. [example:Conjugacy Classes of S4] By the [Conjugacy and Cycle Type](/theorems/798) theorem, conjugacy classes in $S_n$ correspond to cycle types. In $S_4$: identity ($1$), transpositions ($6$), double transpositions ($3$), $3$-cycles ($8$), $4$-cycles ($6$). Check: $1+6+3+8+6 = 24$. Since only the identity has a singleton class, $Z(S_4) = \{e\}$. [/example] In symmetric groups, conjugacy has a particularly clean description: conjugation simply relabels the entries of each cycle. This means conjugacy classes in $S_n$ are determined entirely by cycle type — a combinatorial datum that is easy to read off. [citetheorem:798] The proof is a direct computation: if $\sigma = (a_1\; \ldots\; a_k)(b_1\; \ldots\; b_m) \cdots$, then $\tau\sigma\tau^{-1} = (\tau(a_1)\; \ldots\; \tau(a_k))(\tau(b_1)\; \ldots\; \tau(b_m)) \cdots$. Conjugation replaces each symbol by its image under $\tau$, preserving the cycle structure. The converse — that permutations with the same cycle type are conjugate — follows by choosing $\tau$ to map one set of cycle entries to the other. This characterisation is the basis for counting conjugacy classes in $S_n$: the number of classes equals the number of partitions of $n$, since each partition corresponds to a unique cycle type. [citeproof:798] ### $p$-Groups and the Centre A $p$-group is a group whose order is a power of a prime $p$. The class equation becomes especially powerful for these groups, because every non-central conjugacy class has size divisible by $p$. This forces the centre to be nontrivial — a constraint with far-reaching consequences. [citetheorem:799] The proof reads straight off the class equation: $|G| = |Z(G)| + \sum |G : C_G(x_i)|$, where the sum runs over non-central conjugacy class representatives. Since $|G| = p^n$, the [Orbit-Stabiliser Theorem](/theorems/796) gives $|G : C_G(x_i)| = p^{a_i}$ with $a_i \geq 1$ for non-central $x_i$. So $p$ divides every term in the sum and divides $|G|$, hence $p \mid |Z(G)|$. In particular $|Z(G)| \geq p > 1$. This immediately implies that no $p$-group of order $p^n$ with $n \geq 2$ is simple: $Z(G)$ is a nontrivial normal subgroup. More importantly, it enables an inductive strategy: quotient out by $Z(G)$ to get a strictly smaller $p$-group, and apply the theorem again. [citeproof:799] The inductive strategy just described yields a surprisingly strong result when applied to groups of order $p^2$. The key intermediate step is a lemma about cyclic quotients by the centre. [citetheorem:817] The proof is short but conceptually important. If $G/Z(G) = \langle yZ(G) \rangle$, then every $g \in G$ can be written as $g = y^i z$ for some $z \in Z(G)$. Since central elements commute with everything, any two elements $g = y^i z_1$ and $h = y^j z_2$ satisfy $gh = y^{i+j} z_1 z_2 = hg$. The conclusion is paradoxical at first glance: if $G/Z(G)$ is cyclic, then $G$ is abelian, which means $Z(G) = G$, which means $G/Z(G)$ is trivial — not just cyclic, but the trivial group. The lemma is used in its contrapositive: if $G$ is non-abelian, then $G/Z(G)$ is not cyclic. [citeproof:817] Combining the two preceding results immediately classifies all groups of order $p^2$. [citetheorem:818] By [Centre of a p-Group Is Nontrivial](/theorems/799), $|Z(G)| \in \{p, p^2\}$. If $|Z(G)| = p^2$, then $G$ is abelian. If $|Z(G)| = p$, then $|G/Z(G)| = p$, which is cyclic — but then [Cyclic Quotient by Centre Implies Abelian](/theorems/817) forces $G$ to be abelian, a contradiction. So $G$ is always abelian. For the classification: if $G$ has an element of order $p^2$, then $G \cong C_{p^2}$; otherwise every non-identity element has order $p$, and a direct product argument (using the [Internal Direct Product Theorem](/theorems/794)) gives $G \cong C_p \times C_p$. These are the only two groups of order $p^2$, for any prime $p$. [citeproof:818] ### Simplicity of $A_5$ The alternating group $A_5$ has order $60$. Is it simple? For the groups $A_2, A_3, A_4$, the answer is no ($A_4$ has the normal subgroup $V_4 = \{e, (1\;2)(3\;4), (1\;3)(2\;4), (1\;4)(2\;3)\}$). But $A_5$ is different, and proving its simplicity requires a detailed conjugacy class analysis. [citetheorem:819] The proof computes the five conjugacy class sizes in $A_5$: $1, 15, 20, 12, 12$ (the $5$-cycles split into two classes in $A_5$ because no odd permutation centralises a $5$-cycle). Any normal subgroup must be a union of conjugacy classes containing the identity, so its order has the form $1 + 15\lambda + 20\mu + 12\alpha + 12\beta$ with $\lambda, \mu, \alpha, \beta \in \{0, 1\}$. By [Lagrange's Theorem](/theorems/782), this must divide $60$. Checking all $16$ combinations, only $1$ and $60$ work — so the only normal subgroups are $\{e\}$ and $A_5$ itself. This makes $A_5$ the smallest non-abelian simple group, and its simplicity is the algebraic reason that the general quintic polynomial cannot be solved by radicals. [citeproof:819] # Matrix Groups Groups whose elements are matrices and whose operation is matrix multiplication bridge abstract algebra and geometry. The **determinant** is a group homomorphism whose kernel and image decompose matrix groups into geometrically meaningful pieces. ## The General Linear Group [definition:General Linear Group] $\mathrm{GL}_n(\mathbb{R}) = \{A \in M_n(\mathbb{R}) : \det(A) \neq 0\}$ under matrix multiplication. [/definition] The group axioms hold: closure ($\det(AB) = \det(A)\det(B) \neq 0$), identity ($I_n$), inverses ($A^{-1}$ exists when $\det A \neq 0$), associativity (matrix multiplication). Non-abelian for $n \geq 2$. The determinant $\det : \mathrm{GL}_n(\mathbb{R}) \to (\mathbb{R}^*, \times)$ is a surjective homomorphism. By the [First Isomorphism Theorem](/theorems/791): $\mathrm{GL}_n / \ker(\det) \cong \mathbb{R}^*$. [definition:Special Linear Group] $\mathrm{SL}_n(\mathbb{R}) = \ker(\det) = \{A \in \mathrm{GL}_n(\mathbb{R}) : \det(A) = 1\} \unlhd \mathrm{GL}_n(\mathbb{R})$. [/definition] ## The Orthogonal Group [definition:Orthogonal Group] $O_n(\mathbb{R}) = \{A : A^T A = I_n\}$. The **special orthogonal group** is $\mathrm{SO}_n(\mathbb{R}) = O_n \cap \mathrm{SL}_n$. [/definition] If $A^T A = I$ then $\det(A)^2 = 1$, so $\det(A) = \pm 1$. Thus $\mathrm{SO}_n$ (det $+1$, rotations) has index $2$ in $O_n$, hence is normal. Orthogonal matrices preserve inner products: $\langle Ax, Ay \rangle = x^T A^T A y = x^T y$, hence distances and angles. [example:A Concrete Orthogonal Matrix] $A = \frac{1}{3}\begin{pmatrix} 2 & -1 & 2 \\ 2 & 2 & -1 \\ -1 & 2 & 2 \end{pmatrix}$. Verify $A^T A = I_3$: e.g., column $1$ dot column $1$ is $(4+4+1)/9 = 1$, column $1$ dot column $2$ is $(-2+4-2)/9 = 0$. Determinant: $\det(A) = 1$, so $A \in \mathrm{SO}_3(\mathbb{R})$. Eigenvalue $1$: solve $(A-I)v = 0$ to get axis $(1,1,1)^T$. Angle: $\operatorname{tr}(A) = 2 = 1 + 2\cos\theta$, giving $\theta = 2\pi/3$. So $A$ is a $120°$ rotation about the main diagonal of the cube. [/example] ## Isometries in Two and Three Dimensions In $2$D: $\mathrm{SO}_2(\mathbb{R}) = \{R_\theta : \theta \in [0, 2\pi)\}$ where $R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$. This is abelian and isomorphic to the circle group. Elements of $O_2 \setminus \mathrm{SO}_2$ are reflections. The dihedral group $D_{2n}$ embeds as a finite subgroup of $O_2$. In $3$D, every element of $\mathrm{SO}_3(\mathbb{R})$ is a rotation about an axis. This is geometrically intuitive — a continuous rotation of $3$-space must fix a line — but the proof requires a purely algebraic argument. [citetheorem:820] The proof hinges on an elegant parity trick: $\det(A - I) = -\det(A - I)$ for any $3 \times 3$ orthogonal matrix of determinant $1$, forcing $\det(A - I) = 0$ and guaranteeing an eigenvalue of $1$. The corresponding eigenvector defines the rotation axis, and in the orthogonal complement (a $2$-dimensional plane), $A$ restricts to an element of $\mathrm{SO}_2(\mathbb{R})$, which is a rotation by some angle $\theta$. The "odd-dimensional trick" $\det(A - I) = -\det(A - I)$ fails in even dimensions — indeed, $-I \in \mathrm{SO}_2(\mathbb{R})$ has no eigenvalue $1$ — so this is genuinely a three-dimensional phenomenon. The result extends: every element of $O_3 \setminus \mathrm{SO}_3$ is a rotoreflection (a rotation composed with the central inversion $-I$). [citeproof:820] The subgroup lattice: $\mathrm{SL}_n \unlhd \mathrm{GL}_n$, $\mathrm{SO}_n \unlhd O_n \leq \mathrm{GL}_n$, $\mathrm{SO}_n = O_n \cap \mathrm{SL}_n$. All normality via the [Kernel Is Normal](/theorems/788) theorem. Finite groups embed as subgroups: $D_{2n} \hookrightarrow O_2$, $S_4 \hookrightarrow O_3$ (tetrahedron symmetries). # Möbius Groups The final chapter extends the group-theoretic framework from real matrices to complex matrices acting on the **extended complex plane** $\mathbb{C}_\infty = \mathbb{C} \cup \{\infty\}$. The **Möbius transformations** $z \mapsto \frac{az+b}{cz+d}$ are the conformal automorphisms of the Riemann sphere. Their algebraic structure — captured by the projective linear group $\mathrm{PGL}_2(\mathbb{C})$ — connects group theory to complex analysis and hyperbolic geometry. ## Möbius Transformations [definition:Mobius Transformation] $f(z) = \frac{az+b}{cz+d}$ with $a,b,c,d \in \mathbb{C}$, $ad - bc \neq 0$. [/definition] The condition $ad - bc \neq 0$ prevents the map from being constant. But what algebraic structure do these transformations carry? They are closed under composition — substituting one Möbius map into another produces another Möbius map — and the [composition corresponds to matrix multiplication](/theorems/383). This suggests a deep link between the Möbius group and the general linear group. [citetheorem:821] The map $\Phi : \mathrm{GL}_2(\mathbb{C}) \to \mathcal{M}$ sending $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (z \mapsto \frac{az+b}{cz+d})$ is a surjective homomorphism whose kernel consists of the scalar matrices $\lambda I$. By the [First Isomorphism Theorem](/theorems/791), $\mathrm{GL}_2(\mathbb{C})/Z \cong \mathcal{M}$. Alternatively, since every matrix can be rescaled to have determinant $1$ (at the cost of a sign ambiguity), we also get $\mathrm{SL}_2(\mathbb{C})/\{\pm I\} \cong \mathcal{M}$. The kernel being scalar matrices rather than just $\{I\}$ reflects the fact that $f(z) = \frac{az+b}{cz+d}$ and $f(z) = \frac{\lambda az + \lambda b}{\lambda cz + \lambda d}$ define the same transformation — the "projective" nature of the action. [citeproof:821] [example:Composing Mobius Maps] $f(z) = \frac{z+1}{z-1}$, $g(z) = \frac{2z}{z+1}$. Matrices: $A = \begin{pmatrix} 1&1\\1&-1 \end{pmatrix}$, $B = \begin{pmatrix} 2&0\\1&1 \end{pmatrix}$. Product $BA = \begin{pmatrix} 2&2\\2&0 \end{pmatrix}$, giving $g \circ f(z) = \frac{2z+2}{2z} = 1 + 1/z$. Direct substitution confirms. Fixed points of $f$: $z^2 - 2z - 1 = 0$, giving $z = 1 \pm \sqrt{2}$. Normalised trace: $\operatorname{tr}^2 = 0$, so $\tau = 0 \in [0,4)$ — the map is **elliptic** (rotation by $\pi$ about the fixed points). [/example] Every Möbius map decomposes into translations $z \mapsto z + b$, dilations $z \mapsto az$, and inversion $z \mapsto 1/z$. ## Sharp Triple Transitivity A group action is **transitive** if any point can be sent to any other; it is **sharply $k$-transitive** if any ordered $k$-tuple of distinct points can be sent to any other by a *unique* group element. The Möbius group achieves the strongest form of this for $k = 3$. [citetheorem:822] The proof of existence constructs the map explicitly via the cross-ratio: the auxiliary map $g(z) = \frac{(z - z_0)(z_1 - z_\infty)}{(z - z_\infty)(z_1 - z_0)}$ sends $z_0 \mapsto 0$, $z_1 \mapsto 1$, $z_\infty \mapsto \infty$, and composing with the inverse of the analogous map for $(w_0, w_1, w_\infty)$ gives the desired transformation. Uniqueness follows from the fact that the only Möbius map fixing $0, 1, \infty$ simultaneously is the identity — a short computation from the form $\frac{az+b}{cz+d}$. Sharp triple transitivity is a remarkably strong property: it means Möbius maps are completely determined by three data points, just as affine maps in $\mathbb{R}$ are determined by two. [citeproof:822] ### The Cross-Ratio [definition:Cross-Ratio] $[z; z_0, z_1, z_\infty] = \frac{(z - z_0)(z_1 - z_\infty)}{(z - z_\infty)(z_1 - z_0)}$. [/definition] The cross-ratio is invariant under Möbius transformations. Four points lie on a common circle in $\mathbb{C}_\infty$ if and only if their cross-ratio is real. ## Fixed Points and Conjugacy A non-identity Möbius map with normalised representative $A \in \mathrm{SL}_2(\mathbb{C})$ is classified by $\tau = \operatorname{tr}(A)^2$: [definition:Classification of Mobius Maps] **Parabolic:** $\tau = 4$ (one fixed point), conjugate to $z \mapsto z + 1$. **Elliptic:** $\tau \in [0,4)$ real (two fixed points, $|$multiplier$| = 1$), conjugate to $z \mapsto e^{i\theta}z$. **Loxodromic:** $\tau \notin [0,4]$ (two fixed points), conjugate to $z \mapsto \lambda z$ with $|\lambda| \neq 1$. [/definition] Two Möbius maps are conjugate iff they have the same $\tau$. Möbius maps preserve the family of circles in $\mathbb{C}_\infty$ (where "circle" includes lines through $\infty$), connecting the algebraic structure of $\mathrm{GL}_2(\mathbb{C})$ to the conformal geometry of the Riemann sphere.