# Normalizer of subgroup of p-group

H is a proper subgroup of p-group G. Show that normalizer of H, N(H) is strictly larger than H, and that H is contained in a normal subgroup index p.

Here's what I've got so far:

• If H is normal, N(H) is all of G and we are done.
• If H is not normal, then suppose for the sake of contradiction that N(H)=H. Then there is no element outside of H that fixes H by conjugation. But the center Z(G) of G does fix H, so Z must be in H.

Don't know if I'm going on the right path or not, but either way can't really think my way out of this one... Any help appreaciated.

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You are on the right track. Now look at the subgroup $H/Z$ of $G/Z$. By induction, its normalizer is strictly larger than $H/Z$. Say it contains the residue class $\overline x$ of $x \in G$ where $\overline x \not\in H/Z$. Now show that $x$ also normalizes $H$ in $G$ to get a contradiction.

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Could you clarify: 1) By induction from what? I thought we were assuming that N(H) is not larger than H. 2) By residue class do you just mean the elements outside of H/Z? – Benjamin Lu Nov 10 '12 at 21:19
1) The induction is on the order of $G$. Since $G$ is a $p$-group its center $Z$ is non-trivial, so $G/Z$ is strictly smaller than $G$. 2) By the residue class of $x$ I mean its image in the quotient group $G/Z$, so the coset $xZ$ if you like. – marlu Nov 10 '12 at 21:28
Very clear. I will look into this, thank you sir. – Benjamin Lu Nov 10 '12 at 21:31
What about the second part, with H being contained in a normal subgroup index p? Any hints there? – Benjamin Lu Nov 11 '12 at 0:31
@BenjaminLu You know that taking normalizers strictly enlarges the subgroup. Suppose you repeat this a lot of times. What would happen? – Miha Habič Nov 11 '12 at 1:36

Any subgroup of index $p$ in a $p$-group is normal.
If you have that $K$ is a subgroup of index $p$ in $G$, you've proven that $N_G(K)$ is strictly greater than $K$, so since there can be no subgroups properly between $K$ and $G$, $N_G(K)$ must be $G$. In other words, $K$ is normal. So, it suffices to prove that every proper subgroup $H$ of a $p$-group $G$ is contained in subgroup of index $p$. Are you allowed to use maximal subgroups?
 Oh I also proved the lemma that index p subgroup of finite group is normal, but it's the last part as you say that's giving me trouble. I'm afraid I have not heard of maximal subgroups, but if the alternative is too tedious I suppose I can look it up. – Benjamin Lu Nov 11 '12 at 2:04 A maximal subgroup of $G$ is a subgroup which is not properly contained in any proper subgroup of $G$. You can argue easily that every subgroup is contained in a maximal subgroup. Now, we can use Cauchy's to argue that such subgroups must have index $p$. As Ali said, every maximal subgroup is normal in a $p$-group, so if $K$ is maximal, we can pick a $p$-element of $G/K$. By the correspondence theorem the subgroup generated by this element is the image of a subgroup $H$ of $G$ containing $K$. So if $[G:K]>p$, we have that $K$ is proper in $H$, a contradiction. – Alexander Gruber♦ Nov 11 '12 at 2:44