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

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March 2nd 2011, 07:34 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, 07:57 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, 08:07 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?

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