Order Equal to 396

If $\mid G \mid = 396 = 2^{2} \times 3^{2} \times 11$, then from sylow's Theorem:

$$\begin{tabular}{\vert c\vert c\vert l\vert} \hline prime & Sy... ...9 & 1, 4, 22 \\ \hline 11 & 11 & 1, 12 \\ \hline \end{tabular}$$

If G is simple then n₁₁ = 12. Therefore, $\mid N_{G}(Syl_{11}) \mid = 33$, and $G \cong \le A_{12}$.


But then N_G(Syl₁₁) is abelian, being of order pq, and has a normal Syl₁₁ and a normal Syl₃. Therefore, N_G(Syl₁₁) is equal to the product of a Syl₁₁ and a Syl₃ and so is isomorphic to Z₃₃. But A₁₂ has no subgroups isomorphic to Z₃₃. This is a contradiction. Therefore, n₁₁ = 1 and G is not simple!