# 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?

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).

• Thanks! This is very helpful.
– CWcx
Commented Jul 10, 2012 at 18:53
• Nice and simple. +1 Commented Jul 10, 2012 at 19:17
• Note in fact that what is really proved above is that if $H$ is a non-trivial $p$-subgroup of a finite group $G$ (not necessaily a $p$-group itself), and $[G:H]$ is divisible by $p,$ then $N_{G}(H) >H.$ Commented Jul 11, 2012 at 9:10

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

\begin{align} \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\,\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) \end{align}

and from this it follows at once that $$H\lneq N_G(H)$$

Consider the set of conjugates of $$H$$ in $$G$$, i.e. let $$X = \{H^{g} \ | \ g \in G \}$$ We have $$|X| = |G : N_{G}(H)|$$. If $$|X| = 1$$, there is nothing to show. So, assume $$|X| \ne 1$$. Then $$p\ | \ |X|$$.

Now $$H$$ acts on $$X$$ by conjugation. $$H$$ itself is a fixpoint for this action. Now as in Geoff Robinsons's answer we conclude that there is another orbit of size prime to $$p$$, and therefore another orbit of size $$1$$. Thus we have the existence of $$g \in G \setminus N_{G}(H)$$ with $$H \le N_{G}(H^{g})$$. Since obviously $$H^{g} \le N_{G}(H^{g})$$ we conclude that $$H^{g} < N_{G}(H^{g})$$. But $$N_{G}(H^{g}) = g^{-1} N_{G}(H) \ g$$, so $$g^{-1} H \ g < g^{-1} N_{G}(H) \ g$$, and therefore $$H < N_{G}(H)$$.