Given:
$\bullet$ A finite group $G$, an index 2 subgroup $H$, an element $a \in H$
$\bullet$ $[a]_H$ and $[a]_G$ are the conjugacy class in $H$ of $a$ and the conjugacy class in $G$ of $a$, respectively
To prove:
$[a]_H = [a]_G$ or $[a]_H$ is half the size of $[a]_G$, depending on whether or not the centralizer $Z_G(a)$ is contained in $H$.
Attempt:
$H$ being of index 2 means that $H$ is normal, which means that $H \cdot Z_G(a) $ is a subgroup of $G$. By the Second Isomorphism Theorem, $(H \cdot Z_G(a)) \, / \,H \cong Z_G(a) \, / \, (H \cap Z_G(a))$. If $Z_G(a)$ is contained in $H$, then $H \cap Z_G(a) = Z_G(a)$ and the group on the right-hand side of the isomorphism is trivial, and therefore the group on the left-hand side is trivial as well, and so $|H \cdot Z_G(a))| = |H|$.
Where do I go from here? Can I say that $H$ must be equal to $H \cdot Z_G(a))$? Even if I could, I don't know how to continue.
Best Answer
Step 1. From the orbit stablizer theorem:
$$|G| = |[a]_G| \cdot |Z_G(a)|, \quad |H| = |[a]_H| \cdot |Z_H(a)|.$$
Hence
$$\frac{|[a]_G|}{|[a]_H|} = \frac{|G|}{|H|} \cdot \left(\frac{|Z_G(a)|}{|Z_H(a)|}\right)^{-1} = 2 \cdot \left(\frac{|Z_G(a)|}{|Z_H(a)|}\right)^{-1}$$
Step 2. We have $Z_H(a) = Z_G(a) \cap H$.
Step 3. There is a homomorphism:
$$\varphi: Z_G(a) \hookrightarrow G \rightarrow G/H.$$
The kernel is precisely $Z_G(a) \cap H = Z_H(a)$. Thus, by the first isomorphism theorem,
$$Z_G(a)/Z_H(a) \simeq \mathrm{im}(\varphi) \subset G/H,$$
Since $G/H$ has order $2$, it follows that $Z_G(a)/Z_H(a)$ is either trivial or a group of order 2.
Case 1. $Z_G(a) \subset H$. This means that $Z_H(a) = Z_G(a)$, and thus $Z_G(a)/Z_H(a)$ is trivial, and so
$$\frac{|[a]_G|}{|[a]_H|} = 2.$$
Case 2. $Z_G(a) \not\subset H$. This means that $Z_H(a) \ne Z_G(a)$, and thus $Z_G(a)/Z_H(a)$ is non-trivial, and so (by the above) of order $2$, and
$$\frac{|[a]_G|}{|[a]_H|} = \frac{2}{2} = 1.$$