Let $G$ be a group and $H\leq G$. The normalizer of $H$ in $G$ is $N_G(H)=\{ g\in G |gHg^{-1} =H \}$. Show H is normal to G iff $N_G(H)=G$
I know that $H$ is normal to $N_G (H)$, for the inverse direction, if $N_G (H)=G$, then $H$ is normal to $G$.
Attempt: Since $N_G (H)=G$, and $H$ is normal to $N_G (H)$ by construction,
thus $H$ is normal to $N_G (H)$
For the foreword direction, if $ H$ is normal to $G$, then $N_G(H)=G$
Attempt: Suppose $H$ is normal to $G$, then $gHg^{-1}=H, \forall g \in G$,
and since $H$ is normal to $N_G (H)$, we know that $H = gHg^{-1}, \forall g \in G$, hence, $N_G(H)=G$
I don't think my proof is valid, can anyone show me how to do it?
Thanks.
Best Answer
The statement follows directly from the definition.
If $N_G(H)=G$, then for every $g\in G$ we have $g^{-1}Hg=H$, hence $H$ is normal.
Now if $H$ is normal, then for every $g\in G$ we have $g^{-1}Hg=H$, therefore $N_G(H)=G$. $\square$