Normalizer of a subgroup of a p - group

**sashikanth** · December 28th 2010, 05:50 AM

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?

**DrSteve** · December 28th 2010, 06:52 AM

I don't think I believe that this is true. If G is abelian, then the normalizer of any subgroup is equal to G.

**tonio** · December 28th 2010, 09:54 AM

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: $H<N_G(H)$ for any proper subgroup $H\,\,of\,\,G$

Tonio

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