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

• Mar 2nd 2011, 07:34 PM
joestevens
Normalizer 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 PM
Drexel28
Quote:

Originally Posted by joestevens
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.

What particularly are you having trouble with?
• Mar 2nd 2011, 08:07 PM
joestevens
Quote:

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

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