Order Equal to 315

Let $\mid G \mid = 315 = 3^{3} \times 5 \times 7$. Then by Sylow's Theorem:

$$\begin{tabular}{\vert c\vert c\vert l\vert} \hline prime & Sy... ... 5 & 5 & 1, 21 \\ \hline 7 & 7 & 1, 15 \\ \hline \end{tabular}$$

If G is simple then n₅ = 21, and n₃ = 7. But then $\mid N_{G}(Syl_{3}) \mid = 135 = 5 \times 27$, and since this is a group of order pq it has a normal Syl₅ making $\mid G:N_{G}(Syl_{5}) \mid \le 7$. Therefore, $n_{5} \ne 21 \Longrightarrow n_{5} = 1$ and G is not simple!