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게시글 주소: https://orbi.kr/00039284715
Q : Let $N<G$ be a cyclic normal subgroup of order $n$ and $G/N$ be a cyclic group of order $m$ s.t. $\gcd(m,\varphi(n))=1$. Then $G$ is abelian.\\
A : Since $N$ is a normal subgroup, $G$ acts on $N$ by conjugation. Then there is an induced homomorphism $f:G\to Aut(N)\simeq(\Bbb Z/n)^\times$. Since $N$ is abelian, $N<\ker f$. Since $|G/N|=m$ and $|Aut(N)| =\varphi(n)$ are coprime, $\ker f = G$. Hence, $N<Z(G)$. Now as $G/N$ is cyclic, if we let $gN$ be the generator of $G/N$ and $x\in N$, then $gx = xg$. Hence $G$ is abelian.
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