# Let $|G| = p^n$ where $p$ is prime. Show that the number of non-normal subgroups of $G$ is $pm$. [duplicate]

Let $$|G| = p^n$$ where $$p$$ is prime. Show that the number of non-normal subgroups of $$G$$ is $$pm$$.

I am just studying about group theory so please try to use some simple knowledge.

## marked as duplicate by Dietrich Burde 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(); } ); }); }); Apr 22 '18 at 15:46

If $G$ is trivial group, then there are $0=p\cdot 0$ non-normal subgroups.
Let $G$ be non-trivial group, which contains non-normal subgroups. Then there is an action of $G$ on the set $\mathcal{N}$ of non-normal subgroups by conjugation. That is, if $H\in\mathcal{N}$ then for $g\in G$ we have $H^g =g^{-1}Hg \in \mathcal{N}$. Also we know that for every $H\in\mathcal{N}$ we have $N_G(H)<G$ and the size of the orbit containing $H$ is $|G:N_G(H)|$ is divisible by $p$. So, the cardinality of $\mathcal{N}$ is the sum of sizes of different orbits is also divisible by $p$.