I followed the template from my last post to show that 726 isn't simple, could you tell me if it's correct?
Abelian case: $a \in G / \{1\}$. If $\langle a\rangle \neq G$, then we are done. If $\langle a\rangle = G$, then $\langle a^{66}\rangle $ is a proper normal subgroup of $G$. General case: WLOG we can assume $G \neq Z(G)$. $\langle 1\rangle \neq Z(G)$ which is a proper normal subgroup of $G$. Done. Otherwise $|Z(G)|= 1$. $ 726 = 1 + \sum_{**}\frac{|G|}{|C_G(x)|} $ There must be some $a\in G$ such that 11 does not divide $ \frac{|G|}{|C_G(a)|} $ It follows that $\frac{|G|}{|C_G(a)|} = 2 $ or $3$ or $6$ $\Rightarrow [G:C_G(a)] = 2$ or $3$ or $6$ $\Rightarrow 726 \mid2!$ or $726\mid3!$ or $726 \mid6!$.
Therefore group of order 726 is not simple.