$G$ a finite $p-$group, $H < G$ (i.e, $H$ a subgroup of $G$, but $H \neq G$), prove that $H \neq N_G(H)$

Question

I have one problem, that I think it's pretty hard to solve. The problem reads:

Let $G$ be a finite $p-$group, and $H < G$ (i.e $H \le G$, and $H \neq G$). Prove that $H \neq N_G(H)$.

Here are my thoughts on this problem:

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1. I know that if $G$ is a finite $p-$group, then its center is non-trivial, i.e $\mathcal{Z}(G) \neq \{e\}$.

2. Then, I start off by Proof by Contradiction (note that we already have $H \lhd N_G(H)$), i.e, assume that $H = N_G(H)$, then for every $g \notin H$, there exists $h_g\in H$, such that $g^{-1}h_gg \notin H$. And from here, I come up with the fact that, in order for $H$ to satisfy $H = N_G(H)$, we must have $\mathcal{Z}(G) \lhd H$.

However, the two facts above don't seem to help much. :(

Can you guys give me a little push on this problem?

Thanks so much,

And have a good day,

Answer 1

Hint:

In fact ,a $p-$ group is a nilpotent group.

You can prove these step by step:

1. A group $G$ is nilpotent.

2. Every subgroup of $G$ is subsnormal.

3. If $H<G$ (strictly) then $H<N_G(H)$ (strictly).

We have that $1\longrightarrow 2$, $2\longrightarrow 3$ and if $G$ is finite $3\longrightarrow 1$.

Answer 2

I don't know if you are familiar with Group actions, but this claim is a special case of the following general case:

Theorem: If $H\le G$ is a $p-$ subgroup of a finite group such that $p\mid [G:H]$ then $p\mid [N_G(H):H]$.

Indeed, if $|G|=p^m$ for a some $m$, then for $H$ we have an strict inclusion there.i.e; $H<N_G(H)$. It is good to know that $S_3$ has some subgroups that violets the claim. This is because of the order of $S_3$. It is not a finite $p-$ group. The following codes in GAP, investigate the case in $S_3$:

gap> e:=Elements(AllSubgroups(SymmetricGroup(3)));;
     Filtered(e,t-> Normalizer(SymmetricGroup(3),t)=t);
Group([ (2,3) ]),
Group([ (1,2,3), (2,3) ]),
Group([ (1,2) ]),
Group([ (1,3) ])