# Show the centralizer of H in G is a subgroup of the normalizer of H in G.

$G$ is a group and $H$ is a subgroup of $G$. Prove that $C_G(H)<N_G(H)$.

Missed the discussion on normalizer and centralizer. I just have the definitions: $$N_G(H)=\{g\in G\mid gHg^{-1}=H\}$$ $$C_G(H)=\{g\in G\mid gh=hg, \forall h\in H\}$$

## 1 Answer

If $x\in C_G(H)$ then $xH=Hx$.

So $xHx^{-1}=H$.

Hence $x\in N_G(H)$.

• Well, that is simple. I think I was looking way too into this question. – maidel b Nov 26 '15 at 0:04
• no problem my friend – janmarqz Nov 26 '15 at 0:05