Prove that the normalizer of $\displaystyle A \subseteq G$ is always a subgroup of $\displaystyle G$ which contains $\displaystyle A$. Further prove it is the unique largest subgroup in which $\displaystyle A$ is normal.

- Mar 2nd 2011, 07:34 PMjoestevensNormalizer as a subgroup. Unique largest subgroup in which A is normal.
Prove that the normalizer of $\displaystyle A \subseteq G$ is always a subgroup of $\displaystyle G$ which contains $\displaystyle A$. Further prove it is the unique largest subgroup in which $\displaystyle A$ is normal.

- Mar 2nd 2011, 07:57 PMDrexel28
- Mar 2nd 2011, 08:07 PMjoestevens