I want to solve the following Exercise from Dummit & Foote's Abstract Algebra text:
Assume $H$ is a normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x \in \mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k=|G:HC_G(x)|$. Deduce that a conjugacy class in $S_n$ which consists of even permutations is either a single conjugacy class under the action of $A_n$, or is a union of two classes of the same size in $A_n$. [Let $A=C_G(x)$ and $B=H$ so $A \cap B=C_H(x)$. Draw the lattice diagram associated to the Second Isomorphism Theorem and interpret the appropriate indices. See also Exercise 9, Section 1.]
Here is my solution:
Let $G$ act on the set $\mathcal{K}$ by conjugation (since $\mathcal{K}$ is a conjugacy class this is a well-defined action). The normal subgroup $H \trianglelefteq G$ acts on $\mathcal{K}$ via conjugation as well, let its orbits (which are $H$-conjugacy classes) be $\{\mathcal{O}_i\}_{i \in I}:=\mathcal{F}$. For each $i \in I$, let $o_i \in \mathcal{O}_i$ be a representative. Let $g \in G$ and $\mathcal{O}_i \in \mathcal{F}$ be arbitrary. We claim that $g \mathcal{O}_i g^{-1}=\mathcal{O}_j$ where $\mathcal{O}_j$ is the orbit containing $g o_i g^{-1}$. The proof uses the normality of $H$ like so
\begin{equation} \begin{split} &g \mathcal{O}_ig^{-1}=\{g h o_i h^{-1} g^{-1}:h \in H\}=\{(ghg^{-1})g o_i g^{-1}(ghg^{-1})^{-1}: h \in H\}\\ &=\{h' (go_ig^{-1})h'^{-1}:h' \in H\}=\mathcal{O}_j. \end{split} \end{equation}
Let $o_i,o_j$ be two arbitrary representatives, as $G$ acts transitively on $\mathcal{K}$ we may find $g \in G$ such that $g o_i g^{-1}=o_j$. We have by the above that $g \mathcal{O}_i g^{-1}=\mathcal{O}_j$, so that $G$ acts transitively on $\mathcal{F}$ as well.
Since conjugations are bijections, restricting them to the appropriate orbit shows that all orbits have the same cardinality. (in greater detail, since $G$ acts transitively on $\mathcal{F}$, it can be shown that all orbits have the same cardinality as any fixed one of them.)
In order to finish the exercise, we apply the orbit-stabiliser theorem to the action of $G$ on $\mathcal{F}$. If we choose the orbit $\mathcal{O}_{i_0}$ which contains $x \in \mathcal{K}$ we find that since $G$ acts transitively
\begin{equation} |\mathcal{F}|=|G:G_{\mathcal{O}_{i_0}}|. \end{equation}
Moreover, since
\begin{equation} \begin{split} &g \mathcal{O}_{i_0} g^{-1}=\mathcal{O}_{i_0} \Leftrightarrow \exists h \in H: g x g^{-1}=h x h^{-1} \\ &\Leftrightarrow \exists h \in H: (h^{-1} g) x (h^{-1} g)^{-1}=x \Leftrightarrow \exists h \in H: h^{-1}g \in C_G(x) \\ & \Leftrightarrow g \in HC_G(x), \end{split} \end{equation} we have that the number (or cardinality) of the orbits in $H$ is $|G:HC_G(x)|$ as desired.
Taking $G=S_n,H=A_n$, the index $|G:HC_G(x)|$ can either be 1 (if $HC_G(x) \supsetneq H$) or 2 (if $HC_G(x) = H$), which proves the last part of the exercise.
I believe that my solution is fine (is it really?). The only thing that bothers me is the fact that the authors name the index $|G:HC_G(x)|$ $k$, which makes be believe that it has to be finite. Is it true that $k$ has to be a finite integer? What if $G$ is an infinite group?
P.S. there is an alternative solution here, but I'm still not convinced that $k$ has to be finite.
Thank you!