Order Equal to 264

If $\mid G \mid = 264 = 2^{3} \times 3 \times 11$, and G is simple then n₁₁ = 12. Therefore, by Theorem 0.2 G must be isomorphic to a subgroup of S₁₂. But since G is simple, it is also isomorphic to a subgroup of A₁₂. Choose $H \in Syl_{11}(G)$. Since $n_{11} = \mid G:N_{G}(H) \mid$, we have $\mid N_{G}(H) \mid = 22$. Note that $\mid N_{S_{12}}(G) \mid = 11(11-1) \Longrightarrow \mid N_{A_{12}} \mid = 55$ by Theorem 1.5. But 22 does not divide 55, and so $N_{G}(H) \not\le N_{A_{12}}(G)$. Therefore, G is not simple!