# If every cyclic subgroup of $G$ is normal so is every subgroup? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 30 '16 at 13:54

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 $k$ ...
• @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
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