# Normalizer as a subgroup. Unique largest subgroup in which A is normal.

Printable View

• March 2nd 2011, 08:34 PM
joestevens
Normalizer as a subgroup. Unique largest subgroup in which A is normal.
Prove that the normalizer of $A \subseteq G$ is always a subgroup of $G$ which contains $A$. Further prove it is the unique largest subgroup in which $A$ is normal.
• March 2nd 2011, 08:57 PM
Drexel28
Quote:

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

What particularly are you having trouble with?
• March 2nd 2011, 09:07 PM
joestevens
Quote:

Originally Posted by Drexel28
What particularly are you having trouble with?

Am I supposed to show $A \subseteq N(A) \subseteq G$ or just that $N(A) \subseteq G$, when $A \subseteq G$?
Where should I start?