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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.

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marked as duplicate by Dietrich Burde abstract-algebra Apr 22 '18 at 15:46

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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 $p$.

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