Given a subgroup A in G, consider the normalizer. Prove A is normal iff normalizer=G.

**JJMC89** · March 2nd 2011, 07:24 PM

Given a subgroup $A \subseteq G$ , consider the set $N_G(A) = \{g \in G | gAg^{-1} = A\}$ .

Prove that $A$ is normal if and only if $N_G(A) = G$ .

**Drexel28** · March 2nd 2011, 07:32 PM

Originally Posted by JJMC89

Given a subgroup $A \subseteq G$ , consider the set $N_G(A) = \{g \in G | gAg^{-1} = A\}$ .

Prove that $A$ is normal if and only if $N_G(A) = G$ .

This is by definition isn't it......namely $N_G(A)$ is the largest subgroup $H$ (and yes, it is always a subgroup) such that $A\unlhd H$

**JJMC89** · March 2nd 2011, 07:36 PM

Originally Posted by Drexel28

This is by definition isn't it......namely $N_G(A)$ is the largest subgroup $H$ (and yes, it is always a subgroup) such that $A\unlhd H$

The only definition I have for the normalizer is that it is the set $N_G(A) = \{g \in G | gAg^{-1} = A\}$ .

**Drexel28** · March 2nd 2011, 07:54 PM

Originally Posted by JJMC89

The only definition I have for the normalizer is that it is the set $N_G(A) = \{g \in G | gAg^{-1} = A\}$ .

I mean for normality. One has that $\displaystyle N\unlhd G$ if and only if $gNg^{-1}=N$ for all $g\in G$ which is precisely saying that $N_g(N)=N$ .

**JJMC89** · March 2nd 2011, 08:01 PM

Originally Posted by Drexel28

I mean for normality. One has that $\displaystyle N\unlhd G$ if and only if $gNg^{-1}=N$ for all $g\in G$ which is precisely saying that $N_g(N)=N$ .

Oh right.

I've got it now.

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