So far a group has mostly been something we study on its own: its subgroups, its quotients, its homomorphisms. A group action puts a group to work moving the points of some set around. That does two things for us. It lets us put a group structure on things that don't look like groups at all: the moves of a Rubik's cube form a group, and that group acts on the set of positions of the cube. It also lets us see a group through the set it moves. For example, $SL_2(\mathbb{R})$ (the $2 \times 2$ real matrices of determinant $1$) acts on the upper half-plane $\{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$ by fractional linear transformations, $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot z = \frac{az + b}{cz + d}.$$
The Rubik's cube group has $43{,}252{,}003{,}274{,}489{,}856{,}000$ elements. I have solved it zero times.
On this page we'll set up the language (orbits, stabilizers, the permutation representation), prove the one counting fact everything else rests on, and then use actions to prove three theorems about groups that never mention a set at all.
An action is a rule for letting each element of $G$ move each point of $A$, in a way that respects the group's multiplication.
Suppose $G$ is a group and $A$ is a set. A group action of $G$ on $A$ is a map from $G \times A$ to $A$ (denoted $(g, a) \mapsto g \cdot a$ or $ga$) satisfying:
Notation: $G \curvearrowright A$ means $G$ acts on $A$. Read axiom (i) carefully: acting by $g_2$ and then by $g_1$ is the same as acting by the product $g_1 g_2$. The element on the right acts first, exactly like composing functions.
Label the vertices of a square $1, 2, 3, 4$ counterclockwise. Let $r$ be the quarter turn counterclockwise and $s$ the reflection across the diagonal through vertices $1$ and $3$. Show that $D_8 = \{e, r, r^2, r^3, s, sr, sr^2, sr^3\}$ acts on $A = \{1, 2, 3, 4\}$ by $g \cdot i = g(i)$, the vertex that $g$ moves $i$ to.
As permutations of the vertices, $$r = (1\ 2\ 3\ 4), \quad r^2 = (1\ 3)(2\ 4), \quad r^3 = (1\ 4\ 3\ 2), \quad s = (2\ 4),$$ $$sr = (1\ 4)(2\ 3), \quad sr^2 = (1\ 3), \quad sr^3 = (1\ 2)(3\ 4).$$ Products of motions are composed right to left, so $(g_1 g_2)(i) = g_1(g_2(i))$, which is axiom (i). The identity motion moves nothing, which is axiom (ii). $\checkmark$
Here's the first big observation. Freeze one $g$ and watch what it does to all of $A$ at once: you get a shuffle of $A$.
Suppose $G \curvearrowright A$. Then:
Proof. (i) $\forall\, g \in G$, $a \in A$, $$\begin{align*} (\pi_{g^{-1}} \circ \pi_g)(a) &= g^{-1} \cdot (g \cdot a) && \text{(def. of } \pi_{g^{-1}}, \pi_g) \\ &= (g^{-1} g) \cdot a && \text{(assoc. of } G \curvearrowright A) \\ &= e_G \cdot a \\ &= a && \text{(identity prop. of } G \curvearrowright A). \end{align*}$$ The same computation with $g$ and $g^{-1}$ swapped gives $(\pi_g \circ \pi_{g^{-1}})(a) = a$. So $\pi_g$ has a well-defined inverse function, $\pi_{g^{-1}}$, and is therefore a bijection: $\pi_g \in S_A$.
(ii) $\forall\, g_1, g_2 \in G$, $a \in A$, $$\begin{align*} \varphi(g_1 g_2)(a) = \pi_{g_1 g_2}(a) &= (g_1 g_2) \cdot a \\ &= g_1 \cdot (g_2 \cdot a) && \text{(assoc. of } G \curvearrowright A) \\ &= \pi_{g_1}(\pi_{g_2}(a)) = (\varphi(g_1) \circ \varphi(g_2))(a). \end{align*}$$ So $\varphi(g_1 g_2) = \varphi(g_1) \circ \varphi(g_2)$, and $\varphi$ is a homomorphism. $\blacksquare$
That homomorphism gets a name, and it's the main tool for turning facts about actions into facts about groups.
The homomorphism $\varphi : G \rightarrow S_A$, $\varphi(g) = \pi_g$, is called the permutation representation associated to the action.
Its kernel measures how much of $G$ the set $A$ can't see. Two extremes get names.
Let $\varphi$ be the permutation representation associated to $G \curvearrowright A$. If $\ker \varphi = G$, the action is the trivial action.
Trivial action: every group element shows up, looks around, and does nothing. Relatable.
Let $\varphi$ be the permutation representation associated to $G \curvearrowright A$. If $\ker \varphi = \{e\}$, then the group acts faithfully.
In words: the action is trivial when $g \cdot a = a$ for every $g$ and every $a$, and faithful when the only element that fixes every point is $e$. A faithful action means $A$ sees all of $G$: different elements always shuffle $A$ differently.
$D_8$ also acts on the two diagonals of the square, $d_1 = \{1, 3\}$ and $d_2 = \{2, 4\}$, by moving them: $g \cdot \{i, j\} = \{g(i), g(j)\}$. Is this action faithful? Is the action on the vertices faithful?
On the diagonals: $r^2 = (1\ 3)(2\ 4)$ sends $\{1, 3\}$ to $\{3, 1\}$ and $\{2, 4\}$ to $\{4, 2\}$, so it fixes both diagonals. So do $s = (2\ 4)$ and $sr^2 = (1\ 3)$. The others swap them. So $$\ker \varphi = \{e, r^2, s, sr^2\} \neq \{e\},$$ and the action is not faithful. The diagonals can't tell $e$ from $r^2$.
On the vertices: every element other than $e$ moves some vertex (see the list of permutations above), so $\ker \varphi = \{e\}$ and the action is faithful.
Now fix a point $a$ instead of an element $g$. Two natural questions: which elements of $G$ leave $a$ where it is, and where can $a$ end up?
$\forall\, a \in A$, the stabilizer of $a$ is $$G_a = \operatorname{stab}_G(a) = \{g \in G : g \cdot a = a\}.$$
$\forall\, a \in A$, the orbit of $a$ is $$\operatorname{orb}_G(a) = \{g \cdot a : g \in G\}.$$
The action is transitive if $\forall\, a, b \in A$, $\exists\, g \in G$ s.t. $g \cdot a = b$.
Equivalently, a transitive action has only one orbit: $\operatorname{orb}_G(a) = A$ for every $a \in A$. The action of $D_8$ on the vertices is transitive (rotations alone get any vertex to any other), and the stabilizer of vertex $1$ is $\{e, s\}$. Three facts make orbits and stabilizers useful. The first: a stabilizer isn't just a subset.
$\forall\, a \in A$, $G_a \le G$.
Proof. $e \in G_a$, so $G_a \neq \emptyset$. Suppose $g, h \in G_a$. Then $$g^{-1} \cdot a = g^{-1} \cdot (g \cdot a) = (g^{-1} g) \cdot a = e \cdot a = a \implies g^{-1} \in G_a,$$ using $a = g \cdot a$ (since $g \in G_a$) in the first step, and $$(gh) \cdot a = g \cdot (h \cdot a) = g \cdot a = a \implies gh \in G_a. \qquad \blacksquare$$
The second fact: orbits never overlap. Two orbits are either the same orbit or completely disjoint.
The orbits of $G \curvearrowright A$ form a partition of $A$. Equivalently: the relation $$a \sim_{orb} b \iff \exists\, g \in G \text{ s.t. } g \cdot a = b$$ is an equivalence relation on $A$.
On Logic & Equivalence Classes we saw that the equivalence classes of an equivalence relation always form a partition. So all we have to check is reflexive, symmetric, and transitive. The three group axioms we need are the identity, inverses, and closure.
Proof. We check the three properties.
So $\sim_{orb}$ is an equivalence relation, and the equivalence class of $a$ is $\{b \in A : \exists\, g \in G,\ g \cdot a = b\} = \operatorname{orb}_G(a)$. Equivalence classes partition $A$, so the orbits do. $\blacksquare$
Below, $A$ is the set of all $16$ ways to paint some of the vertices of the square, and a subgroup of $D_8$ moves the paint around with the square. Each colored band is one orbit.
Whatever group you pick, the bands never overlap and always use all $16$ paintings. Going from a subgroup to a bigger group containing it can only glue orbits together, never split one. Notice also that every time, the orbit size times the stabilizer size is the order of the group. That's the third fact.
$\forall\, a \in A$, $|\operatorname{orb}_G(a)| = |G : G_a|$.
The idea: the elements that send $a$ to the same place are exactly one left coset of $G_a$. So points of the orbit and cosets of the stabilizer match up one to one.
Proof. Fix $a \in A$. Define $f : \operatorname{orb}_G(a) \rightarrow G/G_a$ by $g \cdot a \mapsto gG_a$.
So $f$ is a bijection, and $|\operatorname{orb}_G(a)| = |G/G_a| = |G : G_a|$. $\blacksquare$
When $G$ is finite, Lagrange's theorem turns this into $|\operatorname{orb}_G(a)| \cdot |G_a| = |G|$. In particular, the size of every orbit divides $|G|$.
Check orbit–stabilizer for $D_8$ acting on the vertices and on the diagonals of the square, with $a = 1$ and $a = d_1 = \{1, 3\}$.
Vertices: $\operatorname{orb}_{D_8}(1) = \{1, 2, 3, 4\}$ (the rotations get there), and $G_1 = \{e, s\}$, since $s = (2\ 4)$ is the only non-identity element that fixes $1$. So $|G : G_1| = 8/2 = 4 = |\operatorname{orb}_{D_8}(1)|$. $\checkmark$
Diagonals: $\operatorname{orb}_{D_8}(d_1) = \{d_1, d_2\}$, and $G_{d_1} = \{e, r^2, s, sr^2\}$ (the kernel we found earlier: every element fixing $d_1$ also fixes $d_2$). So $|G : G_{d_1}| = 8/4 = 2$. $\checkmark$
Watch the bijection $f$ happen below: each coset of the stabilizer is a group of elements that all agree on where to send $a$.
McKay's proof of Cauchy's theorem on Finite Abelian Groups was secretly a group action. Turning a necklace is $C_p$ acting on the set $S$ of $p$-tuples, and the equivalence classes were its orbits. By orbit–stabilizer, every orbit has size dividing $|C_p| = p$, so size $1$ or $p$. That's the two kinds of class the proof found by hand.
The divisibility is a strong restriction. An action of a group of order $8$ can only have orbits of size $1$, $2$, $4$, or $8$, no matter what set it acts on.
Let $\mathbb{Z}$ (under addition) act on $\mathbb{R}$ by $n \cdot x = x + n$. Show that this is a group action. Is it faithful? Is it transitive? Describe the orbit and the stabilizer of each $x \in \mathbb{R}$, and check that orbit–stabilizer holds.
Action: the group operation is $+$ and the identity is $0$. For $m, n \in \mathbb{Z}$ and $x \in \mathbb{R}$, $m \cdot (n \cdot x) = (x + n) + m = x + (m + n) = (m + n) \cdot x$, and $0 \cdot x = x + 0 = x$. $\checkmark$
Faithful: yes. If $n \cdot x = x$ for every $x$ then $x + n = x$, so $n = 0$. The kernel is $\{0\}$.
Transitive: no. There is no $n \in \mathbb{Z}$ with $n \cdot 0 = \tfrac{1}{2}$.
Orbits and stabilizers: $\operatorname{orb}_{\mathbb{Z}}(x) = \{x + n : n \in \mathbb{Z}\} = x + \mathbb{Z}$, the points that differ from $x$ by an integer. The stabilizer is $\{n : x + n = x\} = \{0\}$. So $|\mathbb{Z} : \{0\}| = |\mathbb{Z}|$, and the orbit $x + \mathbb{Z}$ is in bijection with $\mathbb{Z}$ via $n \mapsto x + n$, as orbit–stabilizer says. There are infinitely many orbits, one for each $x \in [0, 1)$.
The most important examples come from letting a group act on itself (or on things built from itself). The simplest: multiply on the left.
Any group $G$ acts on itself by left multiplication: $\forall\, g \in G$, $a \in G$, $g \cdot a = ga$. This is called the left regular action on $G$.
The axioms are just associativity and the identity of $G$: $g \cdot (h \cdot a) = g(ha) = (gh)a = (gh) \cdot a$, and $e \cdot a = a$. Two things about this action:
What about multiplying on the right? It's tempting to define $g \cdot a = ag$. But right multiplication is not a group action, in general:
$$(gh) \cdot a = agh, \qquad \text{but} \qquad g \cdot (h \cdot a) = g \cdot (ah) = ahg.$$Cancel the $a$: these agree exactly when $gh = hg$. So in a non-Abelian group, axiom (i) fails.
On Symmetric Groups we composed permutations right to left: in $\sigma\tau$, $\tau$ goes first. A left action matches that order, since in $(gh) \cdot a = g \cdot (h \cdot a)$ the $h$ goes first. Right multiplication does $g$ first, so it reverses every product.
Compare the three candidate rules in $S_3$. Pick $g$, $h$, and $a$, and the table computes both sides of axiom (i) for each rule.
Right multiplication isn't an action? Fine. I'll multiply by $g^{-1}$ on the right. Petty, but legal.
The bottom row of the table is the fix: putting the inverse in, $g \cdot a = ag^{-1}$, reverses the order a second time, and the two reversals cancel.
Show that $g \cdot a = ag^{-1}$ ($g, a \in G$) is a group action of $G$ on itself. Is it faithful?
Associativity: for $g_1, g_2, a \in G$, $$(g_1 g_2) \cdot a = a(g_1 g_2)^{-1} = a g_2^{-1} g_1^{-1}, \qquad g_1 \cdot (g_2 \cdot a) = g_1 \cdot (a g_2^{-1}) = (a g_2^{-1}) g_1^{-1} = a g_2^{-1} g_1^{-1}.$$ These agree. $\checkmark$
Identity: $e \cdot a = ae^{-1} = a$. $\checkmark$
Faithful: yes. If $g \cdot a = a$ for even one $a$, then $ag^{-1} = a$, so $g^{-1} = e$ and $g = e$.
Now for the payoff. Feed the left regular action into the permutation representation.
Any group $G$ is isomorphic to a subgroup of $S_G$.
We proved Cayley's theorem on Homomorphisms by building the maps $\tau_g(h) = gh$ and checking everything by hand. Those $\tau_g$ are exactly the $\pi_g$ of the left regular action. Now the general theorem about actions does the checking for us.
Proof. Consider the left regular action of $G$ on itself. The action is faithful, so the permutation representation $\varphi : G \rightarrow S_G$ has $\ker \varphi = \{e\}$, i.e. $\varphi$ is injective. By the 1st Isomorphism Theorem, $$G \cong G / \ker \varphi \cong \varphi(G) \le S_G. \qquad \blacksquare$$
Same theorem, but the proof is now three lines, because the work moved into the general facts about actions. That's the pattern for the rest of the page: pick a clever set for $G$ to act on, then read off a fact about $G$.
Index $2$ subgroups are always normal. Here's the much stronger version, where the clever set is the set of cosets $G/H$.
Assume $|G| < \infty$. If $p$ is the smallest prime dividing $|G|$, then any subgroup of $G$ of index $p$ is normal.
Proof. Suppose $H \le G$, $|G : H| = p$. Let $G$ act on the quotient space $G/H$ by left multiplication: $g \cdot (kH) = (gk)H$. (This is well-defined: if $kH = k'H$ then $(gk)^{-1}(gk') = k^{-1}k' \in H$, so $(gk)H = (gk')H$. The axioms hold because $g \cdot (g' \cdot kH) = (gg'k)H = (gg') \cdot kH$ and $e \cdot kH = kH$.)
$|G/H| = |G : H| = p$, so after labeling the cosets $1, \dots, p$, the associated permutation representation is a homomorphism $\varphi : G \rightarrow S_p$. Let $K = \ker \varphi$.
Suppose $g \in K$. Then $g$ fixes every coset, in particular $g \cdot H = H$. That says $gH = H$, so $g \in H$. Therefore $K \subseteq H$. Note:
Since $pm \mid p!$, we get $m \mid (p - 1)!$, so every prime dividing $m$ is less than $p$. But $m = |H : K|$ divides $|H|$, which divides $|G|$, so every prime dividing $m$ divides $|G|$ and is at least $p$. So $m$ has no prime factors: $m = 1$, and $K = H$. Since $K = H$ is the kernel of a homomorphism, $K \unlhd G$, so $H \unlhd G$. $\blacksquare$
The proof squeezed $K$ between two facts: it sits inside $H$, and its index divides $p!$. Below, each element of $G$ shuffles the cosets of $H$, and the elements that shuffle nothing make up $K$.
Missions 2 and 4 show that the theorem needs both hypotheses. In both, the kernel collapses to $\{e\}$, so all the proof can say is that $K = \{e\}$ is normal, which says nothing about $H$.
(a) Use the theorem to show that every subgroup of index $2$ is normal. (b) Show that $H = \langle (1\ 2) \rangle \le S_3$ has prime index but is not normal. Which hypothesis of the theorem fails?
(a) If $|G : H| = 2$ and $G$ is finite, then $2$ divides $|G| = 2|H|$. No prime is smaller than $2$, so $2$ is the smallest prime dividing $|G|$, and the theorem says $H \unlhd G$.
(b) $|S_3 : H| = 6/2 = 3$, which is prime. But $$(1\ 3)(1\ 2)(1\ 3)^{-1} = (1\ 3)(1\ 2)(1\ 3) = (2\ 3) \notin H,$$ so $(1\ 3) H (1\ 3)^{-1} = \{e, (2\ 3)\} \neq H$, and $H$ is not normal. The theorem doesn't apply because $3$ is not the smallest prime dividing $|S_3| = 6$; that's $2$.
If a finite group moves the points of a set around transitively, is there always one element that moves every point? Yes, as long as there's more than one point. The proof needs a counting lemma about conjugate subgroups.
If $G$ is a finite group and $H$ is a proper subgroup of $G$, then $$\bigcup_{g \in G} gHg^{-1} \neq G.$$
Proof. Let $n = |G : H|$; $n \ge 2$ since $H \neq G$. The conjugate $gHg^{-1}$ only depends on the coset $gH$: for $h \in H$, $$(gh)H(gh)^{-1} = g(hHh^{-1})g^{-1} = gHg^{-1}.$$ So there are at most $n$ different conjugates. Each has $|H|$ elements, one of which is $e$. Counting $e$ once, $$\Big|\bigcup_{g \in G} gHg^{-1}\Big| \le 1 + n(|H| - 1) = 1 + |G| - n < |G|,$$ since $n|H| = |G|$ and $n \ge 2$. $\blacksquare$
For example, $H = \langle (1\ 2) \rangle$ in $S_3$ has three conjugates, $\{e, (1\ 2)\}$, $\{e, (1\ 3)\}$, $\{e, (2\ 3)\}$. Together they cover only $4$ of the $6$ elements of $S_3$.
Suppose $G$ is a finite group acting transitively on a finite set $A$ with $|A| > 1$. Then $\exists\, g \in G$ s.t. $\forall\, a \in A$, $g \cdot a \neq a$.
Elements like that get a name.
If $G \curvearrowright A$, an element $g \in G$ with $g \cdot a \neq a$ for all $a \in A$ is a fixed-point-free element.
The proof has two steps. First, every stabilizer is a conjugate of one fixed stabilizer $G_b$. Then the Lemma says those conjugates can't cover $G$, and any element left over fixes nothing.
Proof. First fix $b \in A$. Then $\forall\, c \in A$, $\exists\, g \in G$ s.t. $g \cdot b = c$ ($G$ acts transitively). For such a $g$, $$\begin{align*} G_c = \{h \in G : h \cdot c = c\} &= \{h \in G : h \cdot (g \cdot b) = g \cdot b\} \\ &= \{h \in G : (hg) \cdot b = g \cdot b\} && \text{(assoc. of } G \curvearrowright A) \\ &= \{h \in G : (g^{-1} h g) \cdot b = b\} && \text{(act by } g^{-1} \text{; assoc. + iden. of } G \curvearrowright A) \\ &= \{h \in G : g^{-1} h g \in G_b\} = g G_b g^{-1}. \end{align*}$$ This holds for every $g \in G$ (with $c = g \cdot b$), and by transitivity every $c \in A$ is some $g \cdot b$. So $$\bigcup_{c \in A} G_c = \bigcup_{g \in G} g G_b g^{-1}.$$ Note that $G_b \neq G$: since $|A| > 1$ there is some $c \neq b$, and since $G$ acts transitively some $g$ has $g \cdot b = c \neq b$, so $g \notin G_b$. Therefore $\bigcup_{g \in G} g G_b g^{-1} \neq G$, by the Lemma.
This implies that $\exists\, g \in G$ with $g \notin \bigcup_{c \in A} G_c$. Then $\forall\, c \in A$, $g \notin G_c$, i.e. $g \cdot c \neq c$. $\blacksquare$
Below, each row is an element of $G$ and each column a point of $A$, with a dot wherever $g \cdot c = c$. A fixed-point-free element is a row with no dots. Each column's dots are a stabilizer, so the proof says: the columns' dots can't reach every row.
In $S_n$, fixed-point-free permutations are called derangements. Someone named that and everyone just went along with it.
Mission 4 is why "transitive" is in the theorem: if some point is fixed by everybody, no element can be fixed-point-free. The kernel is also hiding in this grid: an element fixes every point exactly when its row is full of dots.
Let $\varphi$ be the permutation representation of $G \curvearrowright A$. Show that $$\ker \varphi = \bigcap_{a \in A} G_a.$$ Deduce that if the action is transitive and $b \in A$, then $\ker \varphi = \bigcap_{g \in G} g G_b g^{-1}$. What is $\ker \varphi$ in the proof of the smallest-prime-index theorem?
$g \in \ker \varphi \iff \pi_g = \operatorname{id}_A \iff g \cdot a = a\ \ \forall\, a \in A \iff g \in G_a\ \ \forall\, a \in A \iff g \in \bigcap_{a \in A} G_a.$
If the action is transitive, the proof of the theorem above showed that the stabilizers are exactly the conjugates: $\{G_c : c \in A\} = \{gG_bg^{-1} : g \in G\}$. So the two intersections are the same.
In the smallest-prime-index proof, $G$ acts transitively on $G/H$ (the element $k$ sends $H$ to $kH$), and the stabilizer of the coset $H$ is $\{g : gH = H\} = H$. So $$K = \ker \varphi = \bigcap_{g \in G} gHg^{-1},$$ the largest normal subgroup of $G$ contained in $H$ (any $N \unlhd G$ with $N \subseteq H$ has $N = gNg^{-1} \subseteq gHg^{-1}$ for every $g$). The theorem's proof showed that this intersection is all of $H$.
There's another way for a group to act on itself: by conjugation, $g \cdot a = gag^{-1}$. Its orbits are the conjugacy classes, and counting $G$ orbit by orbit gives the class equation. On The Class Equation & Burnside's Lemma we'll use it to show that groups of order $p^2$ are Abelian, and to count colorings of a square up to symmetry.
That's the toolkit: orbits partition the set, orbit sizes are indices of stabilizers, and the permutation representation turns an action into a homomorphism whose kernel we can study. The next page pushes all three much further.
The smallest-prime-index proof never really needed $p$ to be prime. Here's the general version.
Let $G$ be a finite group and $H \le G$ with $|G : H| = n > 1$. Prove that if $|G|$ does not divide $n!$, then $G$ is not simple. Use this to show that a group of order $24$ with a subgroup of order $8$ is not simple.
Hint: let $G$ act on $G/H$, as in the proof of the smallest-prime-index theorem.
Proof. Let $G$ act on $G/H$ by left multiplication, and let $\varphi : G \rightarrow S_n$ be the permutation representation (labeling the $n$ cosets $1, \dots, n$). Let $K = \ker \varphi$, which is a normal subgroup of $G$.
So $K$ is a normal subgroup of $G$ with $\{e\} \neq K \neq G$, and $G$ is not simple. $\blacksquare$
Order $24$: if $|G| = 24$ and $|H| = 8$, then $n = |G : H| = 3$ and $n! = 6$. Since $24 \nmid 6$, $G$ is not simple. In fact $|G : K|$ divides $6$ and is a multiple of $|G : H| = 3$ (because $K \subseteq H$), so $K$ is a normal subgroup of order $8$ or $4$ inside $H$.