Let $G$ be a finite simple non abelian group and $\{1\}\neq H\leq G$ be such that $\vert C_{G}(H)\vert=\vert G:H \vert$. Show that $H=G$.
I can find only $[G,G]=G$ and $Z(G)=\{1\}$ and if $H$ is a proper subgroup of $G$, then $G \hookrightarrow S_{\vert G:H\vert}$. Can't bring $C_G(H)$ into the picture. Can anyone please provide a hint to think about it?
See I.M. Isaacs, Finite Group Theory, chapter 1G.
If $G$ is abelian, $|G| = |H||C_G(H)| = |H||G|$, so $H = 1$. So we may assume $G$ to be non-abelian. Let $M$ be the Chermak-Delgado subgroup of $G$. Then $M \lt G$ since $M$ is abelian and $G$ is not. And, $M = 1$, since $M$ is normal and $G$ is simple. Consequently $m_G(M) = |G| = m_G(H)$, so $H \in \mathcal{L}(G)$. Assuming $H \lt G$, we will show $H = 1$. Indeed, by Problem 1G.2 there is a normal subgroup $N$ of $G$ such that $H \subseteq N \lt G$. But $G$ is simple, so $N = 1$. Hence $H = 1$.