# If every cyclic subgroup of a group G be normal in G, prove that every subgroup of G is normal in G. [duplicate]

If every cyclic subgroup of a group G be normal in G, prove that every subgroup of G is normal in G.

Attepmt

Let G be a group. Let H be a Normal subgroup of G. Let $K=\langle a \rangle$ be a cyclic subgroup of G generated by $a\in G$. We shall have to show that H is normal in G. That is to prove that for $h\in H$, $ghg^{-1}\in G$ for all $g\in G$.

## marked as duplicate by Dietrich Burde, Stefan4024, Namaste 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(); } ); }); }); Oct 7 '16 at 11:17
• In Clayton solution, math.stackexchange.com/questions/276957/…, why $g\langle k\rangle g^{-1}=\langle k\rangle$ ? – user1942348 Oct 7 '16 at 9:30
• Because the cyclic group $\langle k\rangle$is normal in $G$. This is the definition of being normal. – Dietrich Burde Oct 7 '16 at 9:35
Let $H$ be a subgroup of $G$, $h \in H$ and $g \in G$. You have to show that $g^ {-1}hg \in H$. Since $H$ is a subgroup, the cyclic group $\langle h \rangle$ generated by $h$, is a subgroup of $H$ and hence of $G$. Your premise tells us that $\langle h \rangle$ is normal, so $g^ {-1}hg \in \langle h \rangle$, since of course $h \in \langle h \rangle$. Hence $g^ {-1}hg=h^i$ for some $i$, and $h^i \in H$.