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게시글 주소: https://orbi.kr/00039302089
Q : Let $p$ be a prime and $H<S_p$ s.t. $|H| =p$. If $H<N<S_p$ where $N$ is nilpotent, $H = N$.\\
A : $H$ is a subgroup of order $p$. Then $H$ is cyclic so that $H =\langle\sigma\rangle$ for some $\sigma\in S_p$. Then as there are $(p-1)!$ many $p$-cycle in $S_p$, $(p-1)!= |\sigma| = |H| = |S_p|/|C_{S_p}(\sigma)|$ by the orbit stabilizer theorem so that $|C_{S_p}(\sigma)| = |C_{S_p}(H)| = p$. Hence, $H = C_{S_p}(H)$. Now let $N$ be any nilpotent subgroup of $G$ containing $H$. Then $Z(N)$ is not trivial. Since $H<N$, $Z(N)<Z(H)$. Hence, $Z(N)<C_G(H) = H$. Hence $Z(N) = H$ as the order of $H$ is $p$ and $Z(N)$ is nontrivial. Then by definition, $N<C_G(H) =H$ so that $H =N$.
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