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I searched the Web to find an conterexample for this statement: If every cyclic subgroup of group $G$ is normal in $G$ then every subgroup of $G$ is normal in $G$.

But couln't find any. It seems it is a right proposition. Please give me a hint to start.

Basil R
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  • How is this a duplicate of that question? That question is about all subgroups of a particular cyclic normal subgroup, and not about all subgroups of the group itself (and we have no information about any other subgroup). Here, it is given that all cyclic subgroups are normal, and we want to prove that then all subgroups of the group are normal. – M. Vinay Jul 23 '19 at 14:23

2 Answers2

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Hint: Suppose you have a subgroup $H$ of $G$ and you want to prove that $H$ is normal. Take an element $k \in H$ and consider the cyclic subgroup $K \subset H$ generated by $k$ ...

Mark Bennet
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Taking Mark Bennett's hint above, suppose that $H$ is a subgroup $G$ whose cyclic groups are normal. We want to show that $ghg^{-1}=h'$ for some $h,h'\in H$. To that end, let $k\in H$. Then $\langle k\rangle$ is a cyclic group, hence it is normal. Then for every $g\in G$, we have $g\langle k\rangle g^{-1}=\langle k\rangle$. In particular, $gkg^{-1}=k^m$ for some $m\geq 1$. Since $k\in H$ and $H$ is a subgroup, it is closed with respect to the operation, i.e., $$\underbrace{k\cdot k\cdots k}_{m\text{ times}}\in H.$$ This implies $H$ is normal since $k\in H$ was arbitrary.

Clayton
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  • Nice explanation Clayton. Indeed $gkg^{-1}=k^m$ is a key for the OP. +1 – Mikasa Jan 13 '13 at 15:19
  • since subgroup of cyclic group is cyclic, and every cyclic group is Abelian, and subgroup of Abelian group is Normal. Is this not suffice? – SSA Nov 07 '21 at 05:06