# Normalizer of a subgroup of a p - group

• Dec 28th 2010, 06:50 AM
sashikanth
Normalizer of a subgroup of a p - group
Let $G$ be a group of order $p^n$ where $p$ is prime, and let $H$ be a proper subgroup of $G$. Prove that $H = N(H)$ where $N(H)$ is the normalizer of $H$.

I have not yet studied theorems on p - groups. All I have done is theorems and concepts of subgroups, cosets and the Lagrange's Theorem. Can this question be solved using these concepts?
• Dec 28th 2010, 07:52 AM
DrSteve
I don't think I believe that this is true. If G is abelian, then the normalizer of any subgroup is equal to G.
• Dec 28th 2010, 10:54 AM
tonio
Quote:

Originally Posted by sashikanth
Let $G$ be a group of order $p^n$ where $p$ is prime, and let $H$ be a proper subgroup of $G$. Prove that $H = N(H)$ where $N(H)$ is the normalizer of $H$.

I have not yet studied theorems on p - groups. All I have done is theorems and concepts of subgroups, cosets and the Lagrange's Theorem. Can this question be solved using these concepts?

The claim is false: any finite p-group is nilpotent and thus fulfills the normalizer condition: $H for any proper subgroup $H\,\,of\,\,G$

Tonio