Group Actions

A group action of $G$ on a set $X$ is a map $G \times X \to X$Written $(g, x) \mapsto g \cdot x$ Satisfying:

1. $e \cdot x = x$ for all $x \in X$.
2. $g \cdot (h \cdot x) = (gh) \cdot x$ for all $g, h \in G$ and $x \in X$.

The orbit of $x \in X$ is $\mathrm{Orb}(x) = \{g \cdot x : g \in G\}$.

The stabilizer of $x \in X$ is $\mathrm{Stab}(x) = \{g \in G : g \cdot x = x\}$.

Proposition 6.1. $\mathrm{Stab}(x)$ is a subgroup of $G$.

Theorem 6.2 (Orbit-Stabilizer Theorem). For any $x \in X$

$|\mathrm{Orb}(x)| = [G : \mathrm{Stab}(x)] = \frac{|G|}{|\mathrm{Stab}(x)|}$

Proof. Define $\phi : G \to \mathrm{Orb}(x)$ by $\phi(g) = g \cdot x$. Then $g$ and $h$ have the same Image iff $g \cdot x = h \cdot x$ iff $h^{-1}g \cdot x = x$ iff $h^{-1}g \in \mathrm{Stab}(x)$ Iff $g \in h\,\mathrm{Stab}(x)$. So the fibers of $\phi$ are precisely the cosets of $\mathrm{Stab}(x)$ And there are $[G : \mathrm{Stab}(x)]$ of them, each mapping to a distinct element of $\mathrm{Orb}(x)$. $\blacksquare$

Theorem 6.3 (Burnside’s Lemma). If a finite group $G$ acts on a finite set $X$Then the number Of orbits is

$\frac{1}{|G|} \sum_{g \in G} |\mathrm{Fix}(g)|$

Where $\mathrm{Fix}(g) = \{x \in X : g \cdot x = x\}$.

Proof. Count the set $S = \{(g, x) \in G \times X : g \cdot x = x\}$ in two ways. Grouping by $g$: $|S| = \sum_{g \in G} |\mathrm{Fix}(g)|$. Grouping by $x$: $|S| = \sum_{x \in X} |\mathrm{Stab}(x)|$. For $x$ in orbit $O$, $|\mathrm{Stab}(x)| = |G|/|O|$. So $\sum_{x \in O} |\mathrm{Stab}(x)| = |O| \cdot |G|/|O| = |G|$. Summing over all orbits: $|S| = |G| \cdot (\mathrm{number\ of\ orbits})$. $\blacksquare$

$G$ acts on itself by conjugation: $g \cdot x = gxg^{-1}$.

The orbits are called conjugacy classes. The stabilizer of $x$ is the centralizer $C_G(x) = \{g \in G : gx = xg\}$.

Theorem 6.4 (Class Equation). For a finite group $G$

$|G| = |Z(G)| + \sum_{i} [G : C_G(x_i)]$

Where the sum is over representatives $x_i$ of the non-central conjugacy classes.

Proof. The conjugacy classes partition $G$. Central elements form singleton classes. For a non-central element $x$, $|\mathrm{Orb}(x)| = [G : C_G(x)]$ by the orbit-stabilizer theorem. Summing gives the result. $\blacksquare$

Problem. The rotational symmetry group of a cube has $24$ elements. Use the orbit-stabilizer theorem To verify the sizes of the orbits of vertices, edges, and faces under this action.

Theorem 6.5. If $G$ is a non-trivial finite $p$-group (i.e., $|G| = p^n$ for some prime $p$ and $n \geq 1$), then $Z(G)$ is non-trivial: $|Z(G)| \geq p$.

Proof. By the class equation:

$|G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(x_i)]$

Where $x_1, \ldots, x_r$ are representatives of the non-central conjugacy classes. For each $i$ $x_i$ is non-central, so $C_G(x_i) \neq G$. Thus $[G : C_G(x_i)]$ is a divisor of $|G| = p^n$ That is strictly greater than $1$Hence $p$ divides $[G : C_G(x_i)]$. Since $p$ also divides $|G|$ We have:

$|Z(G)| = |G| - \sum_{i=1}^{r} [G : C_G(x_i)] \equiv 0 - 0 \equiv 0 \pmod{p}$

Since $e \in Z(G)$We have $|Z(G)| \geq 1$. Therefore $|Z(G)| \geq p$. $\blacksquare$

Corollary 6.6. Every group of order $p^2$ (where $p$ is prime) is abelian.

Proof. By Theorem 6.5, $|Z(G)| \geq p$. Since $Z(G) \leq G$, $|Z(G)|$ divides $p^2$So $|Z(G)| = p$ or $|Z(G)| = p^2$. If $|Z(G)| = p^2$Then $G = Z(G)$ is abelian. If $|Z(G)| = p$ Then $G/Z(G)$ has order $p$ and is therefore cyclic, say $G/Z(G) = \langle gZ(G) \rangle$. Then every element of $G$ has the form $g^k z$ for some $k \in \mathbb{Z}$ and $z \in Z(G)$. For any two such elements $(g^{k_1}z_1)(g^{k_2}z_2) = g^{k_1+k_2}z_1z_2 = g^{k_2+k_1}z_2z_1 = (g^{k_2}z_2)(g^{k_1}z_1)$ So $G$ is abelian, contradicting $|Z(G)| = p$. Thus $|Z(G)| = p^2$ and $G$ is abelian. $\blacksquare$

• Forgetting that the stabilizer is always a subgroup (it inherits identity and closure from the group axioms).
• Confusing the orbit of $x$ with the orbit of $G$: the orbit is a subset of $X$, not of $G$.
• Assuming that the number of orbits equals $|X|/|G|$; this is only true when the action is free.
• Misapplying Burnside’s lemma by using the wrong group action (e.g., using conjugation when the problem specifies a different action).
• Assuming that two elements in the same conjugacy class have the same centralizer; they have conjugate centralizers, but the sizes are equal.
• Confusing $[G : C_G(x)]$ (the index of the centralizer) with $|G|/|Z(G)|$ (the size of the quotient by the centre).