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.

marked as duplicate by Dietrich Burde, user26857 abstract-algebra Nov 30 '16 at 13:54

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up vote 8 down vote accepted

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 $k$ ...

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    @bsil: I think Mark is trying to remark you if $k\in H$ then $\langle k\rangle$ is normal as your assumption and so... – mrs Jan 13 '13 at 14:52
  • I've got that. Thank you. – Basil R Jan 13 '13 at 14:56

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.

  • Nice explanation Clayton. Indeed $gkg^{-1}=k^m$ is a key for the OP. +1 – mrs Jan 13 '13 at 15:19

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