Exercise 4.3.19 (A conjugacy class ${\mathcal K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k = |G : HC_G(x)|$)

Proof.

(a)

Since $𝒦$ is a conjugacy class of $G$, $G$ acts transitively by conjugation on the finite set $𝒦$. Let ${𝒪}_1,{𝒪}_2,\dots ,{𝒪}_k$ be the distinct conjugation classes of $H$ on $𝒦$, i.e., if $H=\{{\sigma }_1,{\sigma }_2,\dots ,{\sigma }_r\}$, then the orbit of $a\in 𝒦$ for the action of $H$ on $A$ is $$𝒪(a)=\{{\sigma }_{1}a{\sigma }_1^{-1},{\sigma }_{2}a{\sigma }_2^{-1},\dots ,{\sigma }_{r}a{\sigma }_r^{-1}\}=\{\mathit{𝜎𝑎}{\sigma }^{-1}\mid \sigma \in H\}.$$

Then ${𝒪}_i,1\le i\le k$, are distinct conjugacy classes, and form a partition of $𝒦$:

$$\begin{array}{llll}\hfill & 𝒦=\underset{0\le i\le k}{\bigcup }{𝒪}_i,& \hfill & \\ \hfill & i\ne j\Rightarrow {𝒪}_i\cap {𝒪}_j=\varnothing .& \hfill & \end{array}$$

If $𝒪$ is a conjugacy class for $H$, and if $\sigma \in G$, let $\sigma 𝒪{\sigma }^{-1}$ denote

$$\sigma 𝒪{\sigma }^{-1}=\{\mathit{𝜎𝑎}{\sigma }^{-1}\mid a\in 𝒪\}.$$

Consider a complete system of representatives $\{a_1,a_2,\dots ,a_k\}$ of the conjugacy classes, so that $a_i\in {𝒪}_i$ and ${𝒪}_i=𝒪(a_i)$ and consider some particular conjugacy class ${𝒪}_i$:

$${𝒪}_i=\{\tau a_i{\tau }^{-1}\mid \tau \in H\}.$$

Since $H⊴G$, then for all $\sigma \in G$, the map

$$f_{\sigma }\{\begin{array}{ccc}\hfill H\hfill & \hfill \to \hfill & H\hfill \\ \hfill \tau \hfill & \hfill \mapsto \hfill & \mathit{𝜎𝜏}{\sigma }^{-1}\hfill \end{array}$$

is bijective, therefore

$$\begin{array}{llllll}\hfill \sigma {𝒪}_i{\sigma }^{-1}& =\{\sigma (\tau a_i{\tau }^{-1}){\sigma }^{-1}\mid \tau \in H\}& \hfill & & \hfill & \\ \hfill & =\{(\mathit{𝜎𝜏}{\sigma }^{-1})(\sigma a_i{\sigma }^{-1}){(\mathit{𝜎𝜏}{\sigma }^{-1})}^{-1}\mid \tau \in H\}& \hfill & & \hfill & \\ \hfill & =\{\lambda (\sigma a_i{\sigma }^{-1}){\lambda }^{-1}\mid \lambda \in H\}& \hfill (\lambda =\mathit{𝜎𝜏}{\sigma }^{-1})& & \hfill & & \hfill \\ \hfill & & \hfill & & \hfill \end{array}$$

So $\sigma {𝒪}_i{\sigma }^{-1}$ is the orbit of $\sigma a_i{\sigma }^{-1}$ under the action of $H$ on $𝒦$, thus

$$\sigma {𝒪}_i{\sigma }^{-1}={𝒪}_j\text{for some }j\in [[1,k]].$$

If $R=\{{𝒪}_1,{𝒪}_2,\dots ,{𝒪}_k\}$, the map $f:G\times R\to R$ defined by $(\sigma ,{𝒪}_i)\mapsto \sigma {𝒪}_i{\sigma }^{-1}$ is an action of $G$ on $R$, so $G$ permutes the sets ${𝒪}_1,{𝒪}_2,\dots ,{𝒪}_k$: the map ${\phi }_{\sigma }:R\to R$ defined by ${𝒪}_i\mapsto \sigma {𝒪}_i{\sigma }^{-1}$ is a permutation of $R$, i.e., ${\phi }_{\sigma }$ is a bijection for every $g\in G$.

We show that the action of $G$ on $R$ is transitive. Consider ${𝒪}_i,{𝒪}_j\in R$. Since these orbits are non empty, pick $a\in {𝒪}_i$ and $b\in {𝒪}_j$. Since $𝒦$ is a conjugacy class of $G$, there is some $\sigma \in G$ such that $\mathit{𝜎𝑎}{\sigma }^{-1}=b$. Then $b\in {𝒪}_j\cap \sigma {𝒪}_i{\sigma }^{-1}$, therefore these two orbits are non disjoint, hence they are equal: ${𝒪}_j=\sigma {𝒪}_i{\sigma }^{-1}$. This shows that the action of $G$ on $R$ is transitive.

Since the map $f:{𝒪}_i\to {𝒪}_j$ defined by $\sigma \mapsto \mathit{𝜎𝜏}{\sigma }^{-1}$ for all $\tau \in {𝒪}_i$ is bijective, $|{𝒪}_j|=|\sigma {𝒪}_i{\sigma }^{-1}|=|{𝒪}_i|$. This is true for every $i,j$, so all orbits of $H$ on $A$ have the same cardinality.

Pick some element $x\in 𝒦$, and let ${𝒪}_1$ be the orbit of $x$ (the conjugation class of $x$ for $H$). Note that for all $\sigma \in G$, and for all $x\in 𝒦$,

$$\begin{array}{llll}\hfill \sigma \in C_H(x)& ⟺\sigma \in H\text{ and }\mathit{𝜎𝑥}{\sigma }^{-1}=x& \hfill & \\ \hfill & ⟺\sigma \in H\cap C_G(x).& \hfill & \end{array}$$

So $C_H(x)=H\cap C_G(x)$.

Since $C_H(x)$ is the stabilizer of $x$ for the action of $H$ on $𝒦$, the orbit-stabilizer formula (Proposition 2) shows that

$$\begin{array}{lll}\hfill |{𝒪}_1|=|H:C_H(x)|=|H:H\cap C_G(x)|.& & \hfill \text{(1)}\end{array}$$

Since $G$ acts transitively on $𝒦$, then $𝒦$ is the orbit of $x$ for the action of $G$ on $𝒦$, so the orbit-stabilizer formula gives

$$\begin{array}{lll}\hfill |G:C_G(x)|=|𝒦|.& & \hfill \text{(2)}\end{array}$$

By part (a), all the orbits ${𝒪}_i$ have the same cardinality, and there are $k$ orbits which form a partition of $𝒦$, therefore

$$\begin{array}{lll}\hfill |𝒦|=k|{𝒪}_1|.& & \hfill \text{(3)}\end{array}$$

Finally, the Second Isomorphism Theorem for the subgroups $H$ and $C_G(x)$ of $G$ (see figure) shows that $C_H(x)⊴C_G(x)$ and $(H{C}_G(x))∕H\simeq C_G(x)∕C_H(x)$, so

$$\begin{array}{lll}\hfill |H{C}_G(x):H|=|C_G(x):C_H(x)|.& & \hfill \text{(4)}\end{array}$$

Put $z=|H{C}_G(x):H|$. The equality (4) shows that

$$\begin{array}{llll}\hfill |H{C}_G(x):C_H(x)|& =|H{C}_G(x):H|\cdot |H:C_H(x)|=z|H:C_H(x)|,& \hfill & \\ \hfill |H{C}_G(x):C_H(x)|& =|H{C}_G(x):C_G(x)|\cdot |C_G(x):C_H(x)|=|H{C}_G(x):C_G(x)|z,& \hfill & \end{array}$$

so

$$\begin{array}{lll}\hfill |H{C}_G(x):C_G(x)|=|H:C_H(x)|.& & \hfill \text{(5)}\end{array}$$

Putting all together, the equalities (1), (2) and (3) give

$$\begin{array}{lll}\hfill |G:C_G(x)|& =k|H:C_H(x)|,& \hfill \text{(6)}\end{array}$$

and $|G:C_G(x)|=|G:H{C}_G(x)|\cdot |H{C}_G(x):C_G(x)|$, so by (5)

$$\begin{array}{lll}\hfill |G:C_G(x)|& =|G:H{C}_G(x)|\cdot |H:C_H(x)|.& \hfill \text{(7)}\end{array}$$

The comparison of (6) and (7) shows that

$$k=|G:H{C}_G(x)|.$$

In conclusion, $𝒦$ is a union of $k$ conjugacy classes of equal size in $H$ (for the conjugation action of $H$ on $𝒦$), where $k=|G:H{C}_G(x)|$.

(b)

We apply this result to $G=S_n$ and $H=A_n⊴S_n$.

Since $H{C}_G(x)$ contains $H=A_n$, where $A_n$ is a maximal subgroup of $S_n$, then

$$H{C}_G(x)=H=A_n\text{or}H{C}_G(x)=G=S_n.$$

Since $k=|G:H{C}_G(x)|$ by part (a),

$$k=1\text{or}k=2.$$

So a conjugacy class in $S_n$ contained in $A_n$ is either a single conjugacy class under the action of $A_n$ or is a union of two classes of the same size in $A_n$.