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From Rotman "Introduction to the Theory of Groups", Exercise 3.7.

Let $G$ be a finite group, let $H$ be a normal subgroup of prime index, and let $x \in H$ satisfy $C_H(x) < C_G(x)$. If $y \in H$ is conjugate to $x$ in $G$, then $y$ is conjugate to $x$ in $H$.

i don't know how i can use the hypothesis that $C_H(x) < C_G(x)$, please if anyone could help me .

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  • $\begingroup$ Perhaps you can provide some context, for instance what is $C_G(x)$ ? And what have you tried ? $\endgroup$ – Max Jul 4 '17 at 18:55
  • $\begingroup$ $C_{G}(x)$ is the set $\left \{ g \in G \mid gx=xg \right \}$ $\endgroup$ – Abderrahim Abdou Jul 5 '17 at 11:54
  • $\begingroup$ Well then the hypothesis "$C_H(x)$ is a subgroup of $C_G(x)$" is always satisfied, so the proof should rely on the prime index of $H$, and on the fact that it's a normal subgroup. $\endgroup$ – Max Jul 5 '17 at 12:20
  • $\begingroup$ I know this is an old question but I was puzzled by what Rotman meant by $C_H(x) < C_G(x)$ too. He means that $C_H(x)$ is a proper subgroup of $C_G(x)$. $\endgroup$ – F M yesterday
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Because $|G:H|$ is a prime number and $HC_G(x)$ is a subgroup of $G$ (containing $H$), since $H$ is normal, we must have either $G=HC_G(x)$ or $H=HC_G(x)$. The latter case is equivalent to $C_G(x) \subseteq H$, and then we would have $C_H(x)=H \cap C_G(x)=C_G(x)$. Given the fact that $C_H(x) \lt C_G(x)$ this gives a contradiction.
Hence, $G=HC_G(x)$ and this implies, by comparing indices, that $|Cl_G(x)|=|G:C_G(x)|=|H:C_H(x)|=|Cl_H(x)|$. But of course $Cl_H(x) \subseteq Cl_G(x)$, so we must have $Cl_G(x)=Cl_H(x)$. It now follows that if $y \in H$ and $x \sim_G y$ then $x \sim_H y$.

Note I use the notation $Cl_G(x)$ for the conjugacy class of $x$ in $G$. Rotman would write $x^G$.

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