Let $G$ be a finite group and $H$ a subgroup of $G$ with index $2$, that is, $[G: H]=2$. Recall that $$C_H(x)=H\cap C_G(x), $$ where $C_G(x)=\{g\in G: gx=xg\}$.
How can I use the second isomorphism theorem to show that if $x\in H$ then: $$|x^G|=|x^H|\quad \textrm{or}\ |x^H|=\frac{1}{2}|x^G|?$$
Sketch:
Some facts:
(i) Since $[G: H]=2$ we have $H$ is a normal subgroup of $H$ (that is a condition to use the second isomorphism theorem).
(ii) I also know $|x^G|=[G: C_G(x)]$ as well as $|x^H|=[H: C_H(x)]$.
I guess I should split the problem into two cases if $x=e$ or $x\neq e$. If $x=e$ then $|x^G|=|x^H|$ clearly holds. As to $x\neq e$ I don't know what happens.
Obs: If the second identity holds we have: \begin{align*} \displaystyle |x^H|=\frac{1}{2}|x^G|&\Leftrightarrow [H: C_H(x)]=\frac{1}{2}[G: C_G(x)]\\ &\Leftrightarrow \frac{|H|}{|C_H(x)|}=\frac{1}{2}\frac{|G|}{|C_G(x)|}\\ &\Leftrightarrow \frac{|G|}{|H|}=2\frac{|C_G(x)|}{|C_H(x)|}\\ &\Leftrightarrow |C_G(x)|=|C_H(x)|. \end{align*}
So it would suffice showing $|C_G(x)|=|C_H(x)|$, but it is strange for if $|C_G(x)|=|C_H(x)|$ then $C_G(x)=C_H(x)$ once $C_H(x)\subseteq C_G(x)$.
it is enough to show that $C_G(x)=C_H(x)$ or $|C_G(x):C_H(x)|=2$.
$|C_G(x):C_H(x)|=|C_G(x)/C_G(x)\cap H|=|H.C_G(x):H|$ if $H$ contains $C_G(x)$ then the index is 1 otherwise $H.C_G(x)=G$ so the result is $2$.