Let $G$ a finite group and $H$ subgroup of index $2$. Let $x\in H$ so that the number of conjugates of $x$ in $G$ is $n$. Show that the the number of conjugates of $x$ in $H$ is $n$ or $n/2$.
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Since $[G:H]=2$, $H$ is maximal and normal. Now, we know that the number of conjugates of $x$ is $[G:C_G(x)]$ in $G$ and $[H:C_H(x)]$ in $H$.
Now, $C_H(x)=H\cap C_G(x)$, and there is a bijection $$H/C_H(x)\to HC_G(x)/C_G(x).$$ Indeed, let $f:H\to HC_G(x)/C_G(x)$ be the map $f(h)=hC_G(x)$. Then $f(h)=f(h')$ if, and only if $h^{-1}h'\in C_G(x)$. This means that $f^{-1}(hC_G(x))=h(C_G(x)\cap H)=hC_H(x)$. It now follows that the induced map $\tilde{f}:H/C_H(x)\to HC_G(x)/C_G(x)$ is a bijection and $$[H:C_H(x)]=[HC_G(x):C_G(x)].$$
The two possibilities are $C_G(x)\subset H$ or not. If not, $HC_G(x)=G$, so both orbits have size $n$. Otherwise, $HC_G(x)=H$ and the $H$-orbit has size $n/2$.
David Hill
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why if not $C_G(x)\nsubseteq H $ then $HC_g(X)=G$? – Roiner Segura Cubero Oct 19 '15 at 00:14
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@RoinerSeguraCubero Because H is maximal and $HC_G(x)$ properly contains $H$. – David Hill Oct 19 '15 at 00:50
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$H\lhd G$, $(C_G(x)\cap H)\lhd C_G(x)$, $C_G(x)<G$, by the second isomorphism theorem not is $$C_G(x)/C_H(x)\cong HC_G(x)/H?$$ – Roiner Segura Cubero Oct 19 '15 at 04:33
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@RoinerSeguraCubero You are right. I've corrected the argument. – David Hill Oct 19 '15 at 15:12