Deduce that if $G$ is a finite $p$-group, the number of subgroups of $G$ that are not normal is divisible by $p$ Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$
Question: Deduce that if $G$ is a finite $p$-group, for some prime $p$, the number of subgroups of $G$ that are not normal is divisible by $p$.
Comments: 


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*The subgroups of $G$ that are normal have the property that $g\cdot H=gHg^{-1}=H$

*The question is equivalent to showing $p$ divides $\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$

*$p$-group: $\forall g\in G,|g|=p^k$ for $k\in\mathbb{N}^+$

*I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution.

*My problems I have considered and not been able to answer are: what is the order of $|G|$? I think it must be of order that is divisible by $p$, hence the question becomes show $p$ divides the order of $\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$. How do you compute the order of this set?

*It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is.

Let $X$ be the set of not-normal subgroups of $G$. A conjugate of a not-normal subgroup is not-normal again, so $G$ acts on $X$ by conjugation and (by construction) there are no trivial orbits, so the size of every orbit is a positive power of $p$.
