Problem: Let $G$ be a finite group and $H\leq G$ of index 2. Show that if $x\in H$ has $m$ conjugates in $G$, then $x$ has either $m$ or $m/2$ conjugates in $H$.
My Solution (or attempt): The size of the conjugacy class of $x$ in $G$ is known to be $[G:C_G(x)]$. Since $|G/H|=2$ (and say $G=H\cup gH$), $H$ is normal and it's easy to show that $C_H(x)$ is normal in $C_G(x)$, then
$$m=[G:C_G(x)] = \frac{[G:C_H(x)]}{[C_G(x):C_H(x)]} = \frac{[G:H][H:C_H(x)]}{[C_G(x):C_H(x)]} =2\frac{[H:C_H(x)]}{[C_G(x):C_H(x)]}.$$
Now, suppose that $C_G(x)=C_H(x)$, that is, all the conjugacy classes are also in $H$, then $[C_G(x):C_H(x)] =1$ and $[H:C_H(x)]=m/2$. Suppose that there is an element $a\in C_G(x)$ such that $a\not\in C_H(x)$, that is, $a\not\in H$, then $a\in gH$, furthermore, if $b\in C_G(x)$ and $b\not\in gH$ then $b\in H$ and $b\in C_H(x)$, therefore if $C_G(x)\neq C_H(x)$, $[C_G(x):C_H(x)]=2$ and $[H:C_H(x)]=m$.
Now, this makes ABSOLUTELY NO SENSE, if $C_G(x)=C_H(x)$ it means that all the conjugacy classes of $x$ in $G$ are made by elements in $H$ and we should have that $[H:C_H(x)]=m$ aswell, but no. Surely I've made a mistake somewhere but I can't find it
Best Answer
Why would the centralizers being equal mean all conjugates are obtainable by conjugating by elements of $H$?
It's the opposite: if $gxg^{-1}=hxh^{-1}$, then $h^{-1}g$ centralizes $x$, hence lies in $C_G(x)=C_H(x)\subseteq H$, hence $h^{-1}g\in H$, so $g\in H$. In particular, any $g\in G-H$ gives you a conjugate that you cannot obtain via conjugation by elements of $H$.