# Show H is normal to G iff $N_G(H)=G$

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.

From the definition of $N_G(H)$, we see that if $N_G(H) = G$, that $H \lhd G$, that direction is fairly trivial.
Your proof in the other direction is dangerously close to being circular reasoning. What we CAN say is that if $H \lhd G$, then $G \subseteq N_G(H)$. However, since $N_G(H)$ is also a subset of $G$, the two sets must then be equal.
• By definition: $N_G(H) = \{g \in G| gHg^{-1} = H\}$. The first part of that definition: $N_G(H) = \{g \in G|\dots$ says that $N_G(H)$ is a subset of $G$ (which subset? The elements that satisfy the property $P(g) :=$"$\ gHg^{-1} = H\$"). – David Wheeler May 1 '15 at 0:08
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 $N_G(H)=G$. $\square$