# Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$. Prove $H$ is a normal subgroup of $G$ [duplicate]

Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$.

Prove $H$ is a normal subgroup of $G$

Guessinng that exists $a\in G$ where $\langle a\rangle=N$ of some order.

$H \subset N$, $H$ is a subgroup of $N$.

$gH =Hg \equiv g h_1 =h_2 g \equiv g a^{k_1} =a^{k_2}g$.

*Not sure if on the right track just spit balling *

## marked as duplicate by Dietrich Burde, Michael Burr, Juniven, 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(); } ); }); }); Feb 28 '17 at 1:46

• In a cyclic group of order $n$, for each $k$ dividing $n$, there is a unique subgroup of that order.
• Let $g\in G$. $gHg^{-1}$ is a cyclic group of the same order as $H$ (and $H$ is a cyclic group as it is a subgroup of a cyclic group).
• $gHg^{-1}$ is a subgroup of $N$ by the normality of $K$ ($gHg^{-1}$ is a subset of $N$ by normality of $N$ and $gHg^{-1}$ is a group, so its a subgroup of $K$).
Can you use these three facts to prove that $gHg^{-1}=H$?