Group actions are one of the central ideas in algebra, and their power is easy to understate. At first glance, an action is just a way of making a group "do something" to a set — move points around, permute coordinates, rotate a polygon. But underneath this simple picture lies a precise mechanism for converting abstract group theory into geometric or combinatorial information. Counting theorems that seemed combinatorially intractable become immediate. Subgroup structure that is opaque in isolation becomes visible through the orbits and stabilisers induced on a well-chosen set. The Sylow theorems, Burnside's lemma, the classification of transitive actions — all of these emerge from the single idea that a group can act. To see why this is more than a linguistic convenience, consider the symmetric group $S_3$ acting on the set $\{1, 2, 3\}$ by permuting its elements. The group "knows" something about the set: every element of $S_3$ is a bijection of $\{1, 2, 3\}$, and composing group elements corresponds to composing bijections. The group structure and the set structure are linked. Now ask: how many ways can we colour the vertices of an equilateral triangle with two colours, up to rotational symmetry? The answer is not $2^3 = 8$, because many of these colourings are equivalent under rotation. The correct answer — found immediately by Burnside's lemma — requires exactly this machinery. Without group actions, the question has no clean formulation. [example: Rotations of a Triangle] Let $G = \mathbb{Z}/3\mathbb{Z} = \{0, 1, 2\}$ (with addition mod $3$) and let $X = \{v_1, v_2, v_3\}$ be the vertices of an equilateral triangle. Define an action of $G$ on $X$ by \begin{align*} 0 \cdot v_j &= v_j \\ 1 \cdot v_j &= v_{j+1} \\ 2 \cdot v_j &= v_{j+2} \end{align*} where indices are taken mod $3$ (so $v_4 = v_1$, $v_5 = v_2$). We verify the two axioms directly. The identity element $0$ fixes every vertex, so $0 \cdot v_j = v_j$ for all $j$. For compatibility: $(1 + 1) \cdot v_j = 2 \cdot v_j = v_{j+2}$, while $1 \cdot (1 \cdot v_j) = 1 \cdot v_{j+1} = v_{j+2}$. These agree. Similarly $(2 + 1) \cdot v_j = 0 \cdot v_j = v_j$, and $2 \cdot (1 \cdot v_j) = 2 \cdot v_{j+1} = v_{j+3} = v_j$. All checks pass. The single orbit is $\{v_1, v_2, v_3\}$ — every vertex is reachable from every other. The stabiliser of $v_1$ is $G_{v_1} = \{0\}$, since only the identity fixes $v_1$. This reflects the fact that no nontrivial rotation fixes any vertex. [/example] ## Definition What makes the example above an "action" rather than an arbitrary assignment of group elements to permutations? The answer is the two compatibility axioms that link the group law to the function on $X$. Without them, there is no reason the group structure should tell us anything useful about $X$. The axioms are exactly what is needed to make the assignment a group homomorphism into the symmetric group of $X$. [definition: Group Action] Let $(G, \cdot)$ be a group and let $X$ be a set. A **left group action** of $G$ on $X$ is a function \begin{align*} G \times X &\to X \\ (g, x) &\mapsto g \cdot x \end{align*} satisfying the following two axioms: 1. **Identity:** $e \cdot x = x$ for all $x \in X$, where $e \in G$ is the identity element. 2. **Compatibility:** $(gh) \cdot x = g \cdot (h \cdot x)$ for all $g, h \in G$ and all $x \in X$. When these axioms hold, we say $G$ **acts on** $X$ (on the left), and write $G \curvearrowright X$. The set $X$ is called a **$G$-set**. [/definition] [remark: Right Actions] A **right action** is a map $X \times G \to X$, $(x, g) \mapsto x \cdot g$, satisfying $x \cdot e = x$ and $x \cdot (gh) = (x \cdot g) \cdot h$. Every left action gives a right action by setting $x \cdot g := g^{-1} \cdot x$, and vice versa. Unless stated otherwise, all actions in this chapter are left actions. [/remark] The compatibility axiom is the heart of the definition. It says that acting by $g$ and then by $h$ is the same as acting by the product $gh$, in that order. This is precisely the statement that the map $\rho: G \to \operatorname{Sym}(X)$ sending $g$ to the function $x \mapsto g \cdot x$ is a group homomorphism. This reformulation is worth making explicit, because it shows that actions are really the same thing as representations of $G$ by bijections. [definition: Permutation Representation] Let $G \curvearrowright X$ be a group action. The associated **permutation representation** is the group homomorphism \begin{align*} \rho: G &\to \operatorname{Sym}(X) \\ g &\mapsto \rho_g \end{align*} where $\rho_g: X \to X$ is the bijection $x \mapsto g \cdot x$. [/definition] Conversely, every group homomorphism $\rho: G \to \operatorname{Sym}(X)$ defines an action by $g \cdot x := \rho_g(x)$. This correspondence between actions and homomorphisms is more than a notational convenience — it reframes every question about the action as a question about the image and kernel of $\rho$ inside $\operatorname{Sym}(X)$. [explanation: Why Every Action is a Homomorphism] We verify that $\rho$ as defined is indeed a group homomorphism. For any $g, h \in G$ and any $x \in X$: \begin{align*} \rho_{gh}(x) = (gh) \cdot x = g \cdot (h \cdot x) = \rho_g(\rho_h(x)) = (\rho_g \circ \rho_h)(x). \end{align*} So $\rho_{gh} = \rho_g \circ \rho_h$, which is exactly the homomorphism property. Moreover, $\rho_e = \operatorname{id}_X$ by the identity axiom, and $\rho_{g^{-1}} = \rho_g^{-1}$, so each $\rho_g$ is indeed a bijection. The equivalence of actions with homomorphisms into $\operatorname{Sym}(X)$ makes the kernel of $\rho$ visible: it is $\ker \rho = \{g \in G : g \cdot x = x \text{ for all } x \in X\}$, the subgroup of elements that fix every point of $X$ — those whose permutation is the identity. [/explanation] With the definition in hand, we can immediately isolate the two extreme cases, which mark opposite ends of the spectrum for how informative an action can be. The first extreme: what if every group element fixes every point? This is the least informative action possible — the group "acts" but does nothing. [definition: Trivial Action] The **trivial action** of $G$ on $X$ is the action defined by $g \cdot x = x$ for all $g \in G$ and all $x \in X$. The associated permutation representation is the trivial homomorphism $\rho: G \to \operatorname{Sym}(X)$, $g \mapsto \operatorname{id}_X$. [/definition] The second extreme: what if only the identity element fixes every point? This is the most faithful action possible, in the sense that distinct group elements induce distinct permutations. [definition: Faithful Action] A group action $G \curvearrowright X$ is **faithful** if the only element of $G$ that fixes every point of $X$ is the identity. [/definition] Equivalently, the permutation representation $\rho: G \to \operatorname{Sym}(X)$ is injective — that is, $\ker \rho = \{e\}$. Faithful actions arise naturally: whenever $G$ acts on some structure that is rich enough to distinguish all group elements, the action is faithful. The most fundamental example — and the one that anchors the entire theory — is the action of $G$ on itself. [quotetheorem:795] ## Orbits and Stabilisers Given an action $G \curvearrowright X$, the group imposes an equivalence on $X$: two points $x$ and $y$ are equivalent if some group element carries one to the other. Without making this equivalence precise, we cannot even state what it means for two configurations to be "the same up to symmetry" — the very question Burnside's lemma is built to answer. The orbit of a point collects everything $G$ can reach from that point, and it is the right equivalence class for the action: not too fine (two points in the same orbit really are related by the group), and not too coarse (points in different orbits cannot be connected by any group element at all). The set of all images $g \cdot x$ as $g$ ranges over $G$ is the orbit of $x$, and the subgroup of elements that fix $x$ is the stabiliser. These two invariants are not independent — they are linked by a fundamental counting formula, the Orbit–Stabiliser theorem, that turns abstract orbit sizes into computable indices. [definition: Orbit] Let $G \curvearrowright X$. The **orbit** of a point $x \in X$ is the set \begin{align*} G \cdot x = \{g \cdot x : g \in G\} \subset X. \end{align*} [/definition] The orbit tells us how much of $X$ the group can reach from a given starting point. To understand which group elements are responsible for fixing a point in place, we look at the stabiliser. [definition: Stabiliser] Let $G \curvearrowright X$. The **stabiliser** (or **isotropy subgroup**) of $x \in X$ is \begin{align*} G_x = \{g \in G : g \cdot x = x\}. \end{align*} The stabiliser $G_x$ is a subgroup of $G$: it contains $e$ (since $e \cdot x = x$), is closed under multiplication (if $g \cdot x = x$ and $h \cdot x = x$ then $(gh) \cdot x = g \cdot (h \cdot x) = g \cdot x = x$), and is closed under inversion (if $g \cdot x = x$ then $g^{-1} \cdot x = g^{-1} \cdot (g \cdot x) = (g^{-1}g) \cdot x = x$). [/definition] The stabiliser answers a question about a fixed point $x$: which group elements leave it alone? But there is a complementary question with a different flavour — which points does the entire group leave alone simultaneously? Fixed points are exactly the points where the action is trivial, and they play a special role in counting arguments. In the class equation, for instance, the size of the centre $Z(G)$ counts the fixed points of the conjugation action; and in Burnside's lemma, the orbit formula reduces to counting fixed points of each individual group element. Isolating the fixed-point set gives a name and a notation to this stratum of $X$ that the group cannot move at all. [definition: Fixed Point Set] The **fixed point set** of $G$ in $X$ is \begin{align*} X^G = \{x \in X : g \cdot x = x \text{ for all } g \in G\}. \end{align*} A point in $X^G$ has stabiliser equal to all of $G$. Its orbit is the singleton $\{x\}$. [/definition] The orbits of a group action partition $X$ into disjoint equivalence classes. The relation $x \sim y$ if and only if $y = g \cdot x$ for some $g \in G$ is an equivalence relation: reflexivity follows from $e \cdot x = x$, symmetry from the existence of $g^{-1}$, and transitivity from the group law. This partition is fundamental to the counting arguments that follow. <!-- illustration-needed: orbit partition of a set X — show X as a collection of points grouped into disjoint blobs (orbits), with arrows showing elements of G mapping points within each orbit to each other --> The relationship between the size of an orbit and the size of a stabiliser is given by the Orbit–Stabiliser theorem. Its proof is clean: the elements of the orbit $G \cdot x$ are in bijection with the left cosets of $G_x$ in $G$, because $g \cdot x = h \cdot x$ if and only if $h^{-1}g \cdot x = x$, if and only if $h^{-1}g \in G_x$, if and only if $gG_x = hG_x$. [quotetheorem:796] The theorem has immediate consequences. Every orbit size divides $|G|$. The only orbits of size $1$ are fixed points. And if we know the stabiliser of one point in an orbit, we can read off the size of the orbit directly. [example: Stabilisers in the Dihedral Group] Let $G = D_6$ (the dihedral group of the regular triangle, with $|G| = 6$), acting on the set $X = \{v_1, v_2, v_3\}$ of vertices by the natural geometric action (rotations and reflections). Label the elements: $e, r, r^2$ are rotations by $0°, 120°, 240°$, and $s_1, s_2, s_3$ are reflections across the three axes of symmetry, where $s_i$ fixes $v_i$. The stabiliser of $v_1$ is $G_{v_1} = \{e, s_1\}$, since only the identity and the reflection across the axis through $v_1$ fix $v_1$. We have $|G_{v_1}| = 2$, so by the Orbit–Stabiliser theorem: \begin{align*} |G \cdot v_1| = \frac{|G|}{|G_{v_1}|} = \frac{6}{2} = 3. \end{align*} Since $|X| = 3$ and the orbit of $v_1$ has size $3$, the entire set $X$ is a single orbit: $D_6$ acts transitively on the vertices. This is geometrically obvious — any rotation or reflection can carry any vertex to any other — but the computation makes it quantitatively precise. [/example] ## Transitive Actions and Coset Spaces Some actions are especially structured: the group can carry any point to any other in a single step. These are the transitive actions, and they turn out to be completely classified by subgroups of $G$. [definition: Transitive Action] A group action $G \curvearrowright X$ is **transitive** if $X$ consists of a single orbit, i.e., for every $x, y \in X$ there exists $g \in G$ with $g \cdot x = y$. [/definition] The importance of transitivity is that it completely characterises the action up to isomorphism. If $G$ acts transitively on $X$ and we fix any basepoint $x_0 \in X$, the stabiliser $H = G_{x_0}$ determines the action entirely: $X$ is in bijection with the coset space $G/H$, and the action of $G$ on $X$ corresponds to the natural action of $G$ on $G/H$ by left multiplication. [definition: Coset Action] Let $G$ be a group and $H \le G$ a subgroup. The **coset action** of $G$ on the left coset space $G/H = \{gH : g \in G\}$ is defined by \begin{align*} g \cdot (kH) := (gk)H \end{align*} for $g \in G$ and $kH \in G/H$. This is well-defined: if $kH = k'H$ then $k' = kh$ for some $h \in H$, and $(gk')H = (gkh)H = (gk)H$ since $h \in H$. [/definition] [quotetheorem:3240] What fails when the action is not transitive? Nothing breaks — the action simply decomposes into a disjoint union of transitive pieces. If $X = \bigsqcup_i O_i$ is the orbit decomposition, then the action restricted to each orbit $O_i$ is transitive, and each $O_i$ is isomorphic (as a $G$-set) to a coset space $G/G_{x_i}$ for any representative $x_i \in O_i$. [example: Conjugation Action on Subgroups] Let $G$ be a finite group, and let $X$ be the set of all subgroups of $G$. Define an action of $G$ on $X$ by conjugation: \begin{align*} g \cdot H := gHg^{-1}. \end{align*} This is an action: $e \cdot H = eHe^{-1} = H$, and $(gh) \cdot H = (gh)H(gh)^{-1} = g(hHh^{-1})g^{-1} = g \cdot (h \cdot H)$. The orbit of a subgroup $H$ under this action is its **conjugacy class** of subgroups $\{gHg^{-1} : g \in G\}$. The stabiliser of $H$ is the set of $g \in G$ with $gHg^{-1} = H$, which is the **normaliser** $N_G(H) = \{g \in G : gHg^{-1} = H\}$. By the Orbit–Stabiliser theorem, the number of conjugates of $H$ is \begin{align*} |\{gHg^{-1} : g \in G\}| = [G : N_G(H)]. \end{align*} In particular: $H$ is normal in $G$ if and only if it has exactly one conjugate (itself), which happens exactly when $N_G(H) = G$. [/example] ## Burnside's Lemma and Counting One of the most elegant applications of group actions is to counting problems: how many distinct configurations exist, up to symmetry? The key insight is that "distinct up to symmetry" means "in different orbits under the symmetry group." The question becomes: how many orbits does a given action have? Burnside's lemma — sometimes called the Cauchy–Frobenius lemma — gives a formula in terms of fixed-point counts. Before stating it, let us see why naive counting goes wrong. If $G = \mathbb{Z}/4\mathbb{Z}$ acts on the $4$ corners of a square by rotation, and we colour corners with $k$ colours, there are $k^4$ total colourings. But many are equivalent under rotation, and the orbits have different sizes (size $1$ for uniform colourings, size $4$ for generic ones). Simply dividing $k^4$ by $4 = |G|$ would be wrong — we cannot assume orbits have equal size. [quotetheorem:3241] The proof rests on double-counting: count the set of pairs $\{(g, x) \in G \times X : g \cdot x = x\}$ in two ways. Summed over $g$, this is $\sum_{g \in G} |X^g|$. Summed over $x$, it is $\sum_{x \in X} |G_x|$. By the Orbit–Stabiliser theorem, for each $x$ in an orbit $O$ we have $|G_x| = |G|/|O|$, so summing over $x \in O$ gives $|O| \cdot |G|/|O| = |G|$. Summing over all orbits gives $|G| \cdot |X/G|$. Equating the two counts and dividing by $|G|$ yields the formula. [example: Colouring a Square's Corners] Let $G = \mathbb{Z}/4\mathbb{Z} = \{0, 1, 2, 3\}$ act on the $4$ corners $X = \{c_1, c_2, c_3, c_4\}$ of a square by rotation: element $k$ rotates by $k \times 90°$. We colour each corner red or blue ($2$ colours), so the full set of colourings is $X = \{R, B\}^4$, a set of size $2^4 = 16$. The group acts on this set of colourings by $k \cdot f(c_j) = f(k^{-1} \cdot c_j)$ (rotating the colouring pattern). We compute $|X^g|$ for each $g \in G$: - $g = 0$ (identity): fixes all $16$ colourings. $|X^0| = 16$. - $g = 1$ (rotation by $90°$): a colouring is fixed iff all four corners have the same colour. There are $2$ such colourings (all red, all blue). $|X^1| = 2$. - $g = 2$ (rotation by $180°$): a colouring is fixed iff $f(c_1) = f(c_3)$ and $f(c_2) = f(c_4)$. There are $2 \times 2 = 4$ such colourings. $|X^2| = 4$. - $g = 3$ (rotation by $270°$): same as rotation by $90°$ (since $g = 3$ generates the same orbit-equivalence condition). $|X^3| = 2$. Burnside's lemma gives: \begin{align*} |X/G| = \frac{1}{4}(16 + 2 + 4 + 2) = \frac{24}{4} = 6. \end{align*} There are exactly $6$ distinct $2$-colourings of the square's corners up to rotation. Listed explicitly, the six equivalence classes are: 1. All red: $RRRR$. 2. All blue: $BBBB$. 3. Three red, one blue: $RRRB$ (and its rotations form a single orbit). 4. Three blue, one red: $BBBR$ (and its rotations form a single orbit). 5. Two adjacent red, two adjacent blue: $RRBB$ (corners $c_1, c_2$ red; $c_3, c_4$ blue). 6. Two opposite red, two opposite blue: $RBRB$ (corners $c_1, c_3$ red; $c_2, c_4$ blue). The Burnside count confirms there are exactly $6$ classes, matching this list. [/example] ## The Class Equation When $G$ acts on itself by conjugation — the action $g \cdot h = ghg^{-1}$ — the orbit of $h$ is its **conjugacy class** $\operatorname{Cl}(h) = \{ghg^{-1} : g \in G\}$, and the stabiliser of $h$ is the **centraliser** $C_G(h) = \{g \in G : gh = hg\}$. The Orbit–Stabiliser theorem gives $|\operatorname{Cl}(h)| = [G : C_G(h)]$. The orbit decomposition of $G$ under conjugation leads directly to the class equation, one of the most powerful elementary tools in finite group theory. [quotetheorem:3242] The class equation arises by partitioning $G$ into conjugacy classes. Elements of $Z(G)$ form singleton conjugacy classes (since $ghg^{-1} = h$ for all $g$ iff $h \in Z(G)$). The non-central elements fall into orbits of size $[G : C_G(g_i)] \ge 2$, and summing over orbit representatives gives the formula. [explanation: Why the Class Equation is Powerful] The class equation is a divisibility statement dressed as an identity. Each term $[G : C_G(g_i)]$ divides $|G|$. If $|G| = p^n$ for a prime $p$, every term on the right-hand side is a power of $p$ (including $1$). Since $|G| = p^n$ is divisible by $p$, and each term $[G : C_G(g_i)] \ge 2$ is divisible by $p$, it follows that $|Z(G)|$ must also be divisible by $p$. In particular, $Z(G) \neq \{e\}$: every nontrivial $p$-group has a nontrivial centre. This one deduction — a two-line consequence of the class equation — implies that $p$-groups are solvable, and opens the door to the Sylow theory. [/explanation] [example: The Centre of a p-Group] Let $G$ be a group of order $p^2$ for a prime $p$. We show $G$ is abelian. The class equation gives \begin{align*} p^2 = |Z(G)| + \sum_{i=1}^r [G : C_G(g_i)]. \end{align*} Each term $[G : C_G(g_i)]$ divides $|G| = p^2$ and is greater than $1$, so each such term is either $p$ or $p^2$. Since $[G : C_G(g_i)] = p^2$ would mean $C_G(g_i) = \{e\}$, which is impossible (the identity is always central, so $e \in C_G(g_i)$ for all $g_i$, giving $|C_G(g_i)| \ge 1$; but more precisely $g_i \in C_G(g_i)$, so if $g_i \notin Z(G)$ then $|C_G(g_i)| \ge 2$ and $[G : C_G(g_i)] \le p^2/2 < p^2$). So each non-central term is exactly $p$. The equation $p^2 = |Z(G)| + kp$ for some non-negative integer $k$ forces $p \mid |Z(G)|$. So $|Z(G)| \in \{p, p^2\}$. If $|Z(G)| = p^2$, then $Z(G) = G$ and $G$ is abelian. If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, so $G/Z(G)$ is cyclic. But a group whose quotient by its centre is cyclic must itself be abelian: if $G/Z(G) = \langle gZ(G) \rangle$, then every element of $G$ has the form $g^k z$ for some $k \in \mathbb{Z}$ and $z \in Z(G)$, and two such elements commute because $z$ is central and $g^k g^m = g^m g^k$. This contradicts $|Z(G)| = p < p^2 = |G|$. Therefore $|Z(G)| = p^2$ and $G$ is abelian. [/example] ## Sylow Theory via Actions The Sylow theorems answer the question: given a prime $p$ dividing $|G|$, what can we say about subgroups of order $p^k$? The existence, conjugacy, and count of Sylow subgroups all emerge from carefully chosen group actions. [definition: Sylow Subgroup] Let $G$ be a finite group and $p$ a prime. Write $|G| = p^a m$ where $\gcd(p, m) = 1$. A **Sylow $p$-subgroup** of $G$ is a subgroup of order $p^a$ — a subgroup whose order is the full $p$-power dividing $|G|$. [/definition] The existence of Sylow subgroups is not obvious. Lagrange's theorem tells us that subgroup orders divide $|G|$, but it says nothing about which divisors actually occur as subgroup orders. Cauchy's theorem handles the case $|H| = p$; the Sylow theorem extends this to the maximal $p$-power. [quotetheorem:847] The action behind existence is $G$ acting on the set $\mathcal{S}$ of all subsets of $G$ of size $p^a$ by left multiplication: $g \cdot S = gS$. The key is that $p$ does not divide $\binom{p^a m}{p^a}$, so some orbit has size not divisible by $p$, and the stabiliser of any element in that orbit turns out to be a subgroup of order $p^a$. The conjugacy and count follow from the action of $G$ on its Sylow subgroups by conjugation. [example: Sylow Subgroups of S4] Let $G = S_4$, so $|G| = 24 = 2^3 \cdot 3$. The Sylow $3$-subgroups have order $3$, and the Sylow theorem says $n_3 \equiv 1 \pmod{3}$ and $n_3 \mid 8$. The divisors of $8$ that are $\equiv 1 \pmod{3}$ are $1$ and $4$. We check: does $S_4$ have a unique Sylow $3$-subgroup? A Sylow $3$-subgroup is generated by a $3$-cycle. The $3$-cycles in $S_4$ are $(123), (132), (124), (142), (134), (143), (234), (243)$, giving $8$ three-cycles in $4$ pairs $\{(abc), (acb)\}$, each pair generating a Sylow $3$-subgroup. So $n_3 = 4$. The Sylow $2$-subgroups have order $8$. We have $n_2 \equiv 1 \pmod{2}$ and $n_2 \mid 3$, so $n_2 \in \{1, 3\}$. The three distinct Sylow $2$-subgroups are the following dihedral groups of order $8$ inside $S_4$, each being the symmetry group of one of the three ways to partition $\{1,2,3,4\}$ into two pairs: \begin{align*} P_1 &= \langle (1234), (13) \rangle = \{e, (1234), (13)(24), (1432), (13), (24), (12)(34), (14)(23)\}, \\ P_2 &= \langle (1324), (12) \rangle = \{e, (1324), (12)(34), (1423), (12), (34), (13)(24), (14)(23)\}, \\ P_3 &= \langle (1243), (14) \rangle = \{e, (1243), (14)(23), (1342), (14), (23), (12)(34), (13)(24)\}. \end{align*} Each $P_i$ has order $8$, so $n_2 = 3$. [/example] ## References - David S. Dummit and Richard M. Foote, *Abstract Algebra* (2004). - Joseph J. Rotman, *An Introduction to the Theory of Groups* (1995). - Michael Artin, *Algebra* (2011). - John S. Rose, *A Course on Group Theory* (1978).