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게시글 주소: https://orbi.kr/00039350206
Q : Prove that there are no simple group of order $1755$.\\
A : $1755 = 3^3\cdot 5\cdot 13$. $n_3\equiv 1\bmod 3,n_3\mid 65$ so $n_3 = 13$. Then $G$ acts transitively to the collection of Sylow $3$-subgroups. Then there is a nontrivial homomorphism $\varphi:G\to S_{13}$. Clearly, $\varphi$ is injective so we may assume $G<S_{13}$. Let $P$ be a Sylow $13$-subgroup of $G$. Then $P$ is generated by $13$-cycle. $n_{13}\equiv 1\bmod 13, n_{13}\mid |G|$ so $n_{13} = 27$. Hence, $[G:N_G(P)] = 27$ so $|N_G(P)| = 65$. Since $P<G<S_{13}$, $N_G(P)<N_{S_{13}}(P)$ so that $|N_G(P)|\mid |N_{S_{13}}(P)| = 13\cdot 12$. But $65$ does not divide $13\cdot 12$ which is a contradiction.\\
simple group 아닌거 보이는건 실로우만 갖고 하는건 많이 봤는데 이렇게 하는건 또 처음보네
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