proper subgroups of finite p-groups are properly contained in the normalizer I am trying to prove the following, 

Let $G$ be a finite $p$-group and let $H$ be a proper subgroup. Then there exists a subgroup $H'$ such that 
  $$
H\lneq H'\leq G
$$
  and $H\triangleleft H'$. 

Obviously, the natural choice for $H'$ would be the normalizer $N_G(H)$ of $H$ in $G$. However, one needs to prove then that $H\lneq N_G(H)$ in finite $p$-groups. I am aware of a proof of this fact by induction on the order of $G$. However, I was wondering if there was another proof which only used group actions? 
 A: In fact, for a finite group $\,G\,$, to be nilpotent is equivalent to the normalizer condition, namely: every proper subgroup $\,H\leq G\,$ is properly contained in its normalizer, and here enters what Geoff mentioned at the end of his answer. 
Since a (non-trivial) finite $\,p-\,$group is trivially nilpotent (as its center is always non-trivial), we have 
$$1=:Z_0\leq Z_1\leq\ldots\leq Z_n=G\,\,,\,Z_i:=Z\left(G/Z_{i-1}\right)$$ the upper central series. 
It follows, among other things, that $\,[G,Z_i]\leq Z_{i-1}\,\,,\,\forall i=1,2,...,n$
So, if $\,H\lneq G\,$ then there exists $\,0\leq i\leq n\,\,s.t.\,\,Z_{i-1}\leq H\lneq Z_i\,$ , so that
$$\exists\,z\in Z_i-H\Longrightarrow [G:z]\in [G:Z_i]\leq [G:Z_{i-1}]\leq Z_{i-1}\leq H\Longrightarrow$$
$$\Longrightarrow\,\forall\,g\in G\,\,,\,[g,z]:=g^{-1}z^{-1}gz\in H\Longrightarrow h^{-1}z^{-1}hz\in H\,\,,\,\,\forall\,h\in H\Longrightarrow z\in N_G(H)$$
and from this it follows at once that $\,H\lneq N_G(H)\,$
A: Yes, you can do it with group actions. Of course we only need to worry about the case that $H$ is a non-identity proper subgroup. Let $H$ act on the right cosets of $H$ in $G$ by right translation. Since H is proper, the number of such cosets is divisible by $p.$ At least one of these is fixed by $H,$ namely the coset $H.$ There must be another orbit of size prime to $p,$ but since orbit sizes in this situation are powers of $p,$ the orbit size must be $1$. Hence there is some $g \in G \backslash H $ such that $Hgh = Hg$ for all $h \in H.$ Then $gHg^{-1} \leq H,$ so that $gHg^{-1} = H$ as both these subgroups have the same order. Hence $g \in N_{G}(H) \backslash H$ and $N_{G}(H) > H.$
(Another standard proof not by induction is to use the upper central series).
