Conjugation and the Class Equation

A group acting on itself by conjugation, $g\cdot a=ga{g}^{-1}$, behaves nothing like the left-multiplication action: it fixes the identity, respects the group operation, and splits $G$ into classes of unequal size. The sizes obey an arithmetic identity, the class equation, which constrains a group from its order alone.¹

The conjugation action

Define $G$ acting on itself by $g\cdot a=ga{g}^{-1}$. The axioms hold:

Two features distinguish this from left multiplication. First, unless $|G|=1$ the action is never transitive: $\{1\}$ is always its own class, since $g1g^{-1}=1$. Second, the singleton classes are precisely the central elements.

So the center is the union of the size-one orbits. An abelian group is all center: every class is a singleton, and conjugation is trivial.

The stabilizer of $a$ under conjugation is $\{g:ga{g}^{-1}=a\}=C_G(a)$, the centralizer. Orbit-stabilizer immediately gives the size of each class.

The subset statement uses $N_G(S)=\{g:gS{g}^{-1}=S\}$ as the stabilizer; for a single element, $N_G(\{s\})=C_G(s)$.² Every class size is therefore a divisor of $|G|$, a constraint used constantly below.

The class equation

Summing the sizes of all classes recovers $|G|$. Separating the singletons (the center) from the rest gives the central identity of finite group theory.

Each central element is its own class, contributing $|Z(G)|$; each non-central class $g_i$ contributes $[G:C_G(g_i)]>1$. The classes partition $G$, so the sizes sum to $|G|$.³ Every summand on the right divides $|G|$, and every non-central summand exceeds $1$; together these two constraints are strongly restrictive.

For the two nonabelian groups of order $8$, a shortcut speeds the computation: $⟨g⟩\le C_G(g)$ always, so a non-central element of $D_8$ or $Q_8$ has centralizer of order exactly $4$ and hence a class of size $2$.

• $Q_8$. The classes are $\{1\},\{-1\},\{\pm i\},\{\pm j\},\{\pm k\}$, with $Z(Q_8)=\{1,-1\}$. The class equation is $8=2+2+2+2$.
• $D_8$. The classes are $\{1\},\{r^2\},\{r,r^3\},\{s,s{r}^2\},\{sr,s{r}^3\}$, with $Z(D_8)=\{1,r^2\}$. The class equation is again $8=2+2+2+2$.

Both groups share the same class-size profile even though they are not isomorphic: the class equation constrains a group without determining it.

Prime-power order forces a center

The first consequence underlies the theory of $p$-groups.

Read the class equation modulo $p$. Each non-central summand $[P:C_P(g_i)]$ is a divisor of $p^a$ greater than $1$, hence divisible by $p$. Since $p\mid |P|$ as well, the equation forces $p\mid |Z(P)|$. In particular $|Z(P)|\ge p$, so the center is nontrivial.⁴ A group with $Z(P)=1$ cannot have prime-power order.

Since $Z(P)\ne 1$ and $|Z(P)|$ divides $p^2$, if $Z(P)\ne P$ then $P/Z(P)$ has order $p$, hence is cyclic — but a group with cyclic central quotient is abelian, a contradiction. So $P=Z(P)$ is abelian, and the classification of finite abelian groups splits it into the two listed types.⁵

Conjugacy in the symmetric group

Conjugation in $S_n$ has an explicit description: it relabels the entries of a cycle decomposition.⁶

The proof is one line: if $\sigma (i)=j$ then $\tau \sigma {\tau }^{-1}(\tau (i))=\tau (j)$, so the ordered pair $\tau (i),\tau (j)$ appears in the conjugate exactly where $i,j$ appeared in $\sigma$. The cycle type — the multiset of cycle lengths (including fixed points as $1$-cycles) — is a complete invariant.

Same cycle type gives a matching between the two decompositions; the permutation $\tau$ carrying one list to the other conjugates one into the other. A cycle type is precisely a partition of $n$, so the classes are indexed by partitions.⁷

The class sizes come from the centralizer formula. For an $m$-cycle $\sigma$ in $S_n$, the number of $m$-cycles is $\frac{n(n-1)\cdots (n-m+1)}{m}$, so $|C_{S_n}(\sigma )|=m\cdot (n-m)!$: the $m$-cycle commutes with its own powers and with any permutation disjoint from it. Working this out for $S_4$:

The sizes sum to $1+6+3+8+6=24=|S_4|$, and $Z(S_4)=1$ (the only singleton class is the identity). The even classes — sizes $1,3,8$ — sum to $12=|A_4|$, which is why normal subgroups, being unions of classes, are so constrained; this counting drives the simplicity arguments for the alternating groups.

One subtlety carries into the alternating group: a class of $S_n$ contained in $A_n$ need not remain a single class under conjugation by $A_n$ alone, since the permutation $\tau$ relabeling one representative to another may be odd. The class splits in two exactly when its cycle type consists of distinct odd integers. For $S_5$ the classes have sizes $1,10,15,20,20,30,24$ (partitions $1^5,21^3,2^{2}1,31^2,32,41,5$), and inside $A_5$ the even classes are $1,15,20$ and the $24$ five-cycles, which split into two classes of $12$. Those sizes $1,15,20,12,12$ are what make $A_5$ simple: no sub-collection containing the $1$ sums to a proper divisor of $60$.

Burnside's lemma: counting orbits

Orbit-stabilizer counts one orbit; Burnside's lemma counts all of them at once, by averaging fixed points. For an action of $G$ on $X$, let $X^g=\{x\in X:g\cdot x=x\}$ be the set fixed by $g$.

Count the pairs $(g,x)$ with $g\cdot x=x$ two ways. Summing over $g$ gives ${\sum }_g|X^g|$; summing over $x$ gives ${\sum }_x|G_x|$. For each orbit $\mathcal{O}$, orbit-stabilizer makes ${\sum }_{x\in \mathcal{O}}|G_x|=|\mathcal{O}|\cdot |G_x|=|G|$, so summing over the $k$ orbits gives $k\cdot |G|$.⁸ Equating the two counts and dividing by $|G|$ finishes it.

The application is counting configurations up to symmetry. Consider a necklace of four beads, each coloured black or white, where two colourings are the same if one rotates to the other. The group is the cyclic rotation group $C_4=\{e,\rho ,{\rho }^2,{\rho }^3\}$ acting on the $2^4=16$ colourings, and a colouring is fixed by ${\rho }^d$ exactly when it is constant on the orbits of ${\rho }^d$ on the beads.

The per-element fixed-colouring counts follow the cycle structure of each rotation acting on the four bead positions:

Averaging, $k=\frac{1}{4}(16+2+4+2)=6$: there are exactly six distinct two-colour necklaces. A fixed colouring under a rotation whose cycle decomposition has $c$ cycles must be constant on each cycle, giving $|X^g|=2^c$ colours — which is why the fixed counts are powers of two indexed by the number of cycles. The same average recurs whenever configurations must be counted up to symmetry.