Upper Bound of Order of Non-Abelian Finite Simple Group

Theorem
Let $G$ be a non-abelian finite simple group.

Let $t \in G$ be a self-inverse element of $G$.

Let $\map {C_G} t$ denote the centralizer of $t$ in $G$.

Let $m = \order {\map {C_G} t}$ be the order of $\map {C_G} t$.

Then:
 * $\order G \le \paren {\dfrac {m \paren {m + 1} } 2}!$