## Count the number of sub conjugacy classes comprising a conjugacy class contained in a normal subgroup

Let $G$ be a group, $H \leq G$ a normal subgroup, and $K \subseteq G$ a conjugacy class such that $K \subseteq H$. Prove that $K$ is a union of $[G : HC_G(x)]$ distinct $H$-conjugacy classes, each of equal size. Deduce that a conjugacy class in $S_n$ which consists of even permutations is either a single $A_n$ conjugacy class or a union of two $A_n$-conjugacy classes of equal size.

Considering the induced action of $H$ on $K$ by conjugation, $K$ is a union of $H$-conjugacy classes. Now let $H \cdot a, H \cdot \subseteq K$ be distinct $H$-conjugacy classes. Because $K$ is a $G$-conjugacy class, we have $g \cdot a = b$ for some $g \in G$. ($\cdot$ denotes conjugation.) Now $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$. Thus $H \cdot a$ and $H \cdot b$ are conjugate as subsets of $G$, and we have $|H \cdot a| = |H \cdot b|$. That is, all $H$-conjugacy classes in $K$ have the same cardinality. If we fix $x \in K$, then, $|K| = m \cdot |H \cdot x|$ for some integer $m$, where $m$ is the number of distinct $H$-conjugacy classes in $K$.

Now note that $C_H(x) = H \cap C_G(x) \leq H \leq H C_G(x) \leq G$ and that $C_H(x) \leq C_G(x) \leq G$. We then have the following.

 $[G : H \cap C_G(x)] = [G : H \cap C_G(x)]$ $[G : C_G(x)] \cdot [C_G(x) : H \cap C_G(x)] = [G : H] \cdot [H : C_H(x)]$ $|K| \cdot [C_G(x) : H \cap C_G(x)] = [G : HC_G(x)] \cdot [HC_G(x) : H] \cdot |H \cdot x|$ $|K| = [G : HC_G(x)] \cdot |H \cdot x|$

Since by the Second Isomorphism Theorem, $[HC_G(x) : H] = [C_G(x) : H \cap C_G(x)]$.

Thus $m = [G : HC_G(x)]$ is the number of $H$-conjugacy classes contained in $K$, and all these classes have the same cardinality.

Consider now $A_n \vartriangleleft S_n$. If $K \subseteq A_n$ is a conjugacy class and $\sigma \in K$, then by the above argument $K$ is a union of $[S_n : A_n C_{S_n}(\sigma)]$ distinct $A_n$-conjugacy classes of the same cardinality. Recall that $A_n$ is a maximal subgroup of $S_n$. Since $A_n \leq A_nC_{S_n}(\sigma) \leq S_n$, we have that $A_nC_{S_n}(\sigma)$ is either $A_n$ or $S_n$. Thus the number of $A_n$-conjugacy classes in $K$ is either 1 or 2, and these classes have the same size.