# Thread: Normalizer of a subgroup of a p - group

1. ## Normalizer of a subgroup of a p - group

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

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?

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

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

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

Tonio