# A question on prime power order group.

After some trying on this problem I could not solve it:

For a group of order $p^n$, $p$ prime, prove that for any subgroup $H\ne G$, $\exists x\in G, x \notin H$ such that $xHx^{-1}=H$.

Can someone give a small hint about to how to approach this problem.

p.s.-this is not a homework problem.

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Have you tried letting $G$ act on the set of groups of order $|H|=p^k$, then counting orbits? – Alexander Gruber Oct 5 '12 at 17:24
Another way to prove this is to use the fact that the centre of a finite $p$-group is nontrivial and use induction. By using induction on the nilpotency class, you can prove this property for all proper subgroups of a (possibly infinite) nilpotent group. – Derek Holt Oct 6 '12 at 11:07

There is nothing to do if $H = 1,$ so suppose not. Let $H$ act by right translation on the right cosets of $H$ in $G.$ The number of such cosets is $[G:H]$, a power of $p$ greater than $1$. On the other hand, the order of $H$ is a power of $p$ greater than $1$. Now $H$ fixes the right coset $He = H$ in this action. Use the orbit counting theorem to show that $H$ must fix at least one other right coset, say $Hx$. Then we have $Hxh = Hx$ for all $h \in H$. This is one choice of $x$ for the problem (some details have been omitted because you asked for hints).
Suppose $H$ has order $p^k$. Consider the double coset representation $G = \cup HxH$ with representatives $x_1, \dots, x_t$. Without loss of generality we can assume $x_1 \in H$ and $x_i \not\in H$ for all $i \geq 2$. Thus
$$|G| = |H| + \sum_{i = 2}^t \frac{|H|^2}{|H \cap x_iHx_i^{-1}|}$$
If $x_iHx_i^{-1} \neq H$ for all $i \geq 2$, the order of $H$ would be divisible by $p^{k+1}$ which is absurd.