Let $G$ be a p-group. Let $H$ be a proper subgroup of $G$. Show that there exists $g$ $\in$ $G \setminus H$ such that $gHg^{-1}=H$.

Question

Let $G$ be a p-group. Let $H$ be a proper subgroup of $G$. Show that there exists $g$ $\in$ $G \setminus H$ such that $gHg^{-1}=H$.
I tried to use a counting argument. Let's assume by contradiction that it isn't true.
So, for every $g$ $\in$ $G \setminus H$, $gHg^{-1}$ is not $H$. We know that conjugation is also a subgroup of $G$, so $gHg^{-1}$ is a different subgroup of order $p^k$ where $|G|=p^n$ and $n>k$.
Now, we can see that each conjugation as above is different if $g_1 \neq g_2h$, since if $g_1Hg_1^{-1} = g_2Hg_2^{-1}$ then we can get $g_2^{-1}g_1Hg_1^{-1}g_2=H$ and clearly, from what's given, $g_2^{-1}g_1 \in H$.
Consequently, we will have $p^{n-k}*(p^k-1)+p^k+1$ elements in $G$ which clearly is not a contradiction, and I don't know how to continue from here. Any hint will be helpful.

Answer 1

We consider the action of $G$ on the set $S := G/H$ of left cosets of $H$. Restrict this to an action of $H$ on $S: H × S → S$. Note that $H$ is a $p$-group itself and that $S$ has $[G : H]$ elements; this number is a positive power of $p$. Then we know that $|S^H|$ is divisible by $p$. Now $H ∈ S^H$, so $|S^H| \ge 1$, hence in fact $|S^H|\ge p$. Thus there exists some coset, call it $gH$, with $g\not\in H$, such that $gH\in S^H$ . Then for any $h∈H$ we have $hgH=gH$,i.e., $g^{-1}hgH=H$,so that $g^{−1}hg \in H$. Thus $gHg^{-1} = H$.