Intereting Posts

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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 $n/2$.

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