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게시글 주소: https://orbi.kr/00039282148
Q : If $G$ is a non-cyclic group of order $3^n$, $n>0$ then it has at least $6$ normal subgroups.\\
A : If $G$ is abelian then by the fundamental theorem of finite abelian group, $\Bbb Z/3\times\Bbb Z/3<G$. Since $\BBb Z/3\times\Bbb Z/3$ has $6$ subgroups, those subgroups gives the $6$ normal subgroups.\\
If $G$ is not abelian then $G/Z(G)$ is not cyclic. If $G/Z(G)$ is abelian then again by the fundamental theorem of finite abelian group, $\Bbb Z/3\times\Bbb Z/3<G/Z(G)$ so that $G/Z(G)$ contains $6$ normal subgroups. By the correspondence theorem, $G$ contains $6$ normal subgroups. If $G/Z(G)$ is not abelian then $G/Z(G)$ is a non-cyclic group of order $3^k<3^n$. Then by induction on $n$ and the correspondence theorem, $G$ contains $6$ normal subgroups.
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