# Normalizer of a subgroup of a p - group

• Dec 28th 2010, 05:50 AM
sashikanth
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?
• Dec 28th 2010, 06:52 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, 09:54 AM
tonio
Quote:

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