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

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

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

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

2. Originally Posted by JJMC89
Given a subgroup $\displaystyle A \subseteq G$, consider the set $\displaystyle N_G(A) = \{g \in G | gAg^{-1} = A\}$.

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

3. Originally Posted by Drexel28
This is by definition isn't it......namely $\displaystyle N_G(A)$ is the largest subgroup $\displaystyle H$ (and yes, it is always a subgroup) such that $\displaystyle A\unlhd H$
The only definition I have for the normalizer is that it is the set $\displaystyle N_G(A) = \{g \in G | gAg^{-1} = A\}$.

4. Originally Posted by JJMC89
The only definition I have for the normalizer is that it is the set $\displaystyle N_G(A) = \{g \in G | gAg^{-1} = A\}$.
I mean for normality. One has that $\displaystyle \displaystyle N\unlhd G$ if and only if $\displaystyle gNg^{-1}=N$ for all $\displaystyle g\in G$ which is precisely saying that $\displaystyle N_g(N)=N$.

5. Originally Posted by Drexel28
I mean for normality. One has that $\displaystyle \displaystyle N\unlhd G$ if and only if $\displaystyle gNg^{-1}=N$ for all $\displaystyle g\in G$ which is precisely saying that $\displaystyle N_g(N)=N$.
Oh right.
I've got it now.