Order Equal to 336

Let $\mid G \mid = 336 = 2^{4} \times 3 \times 7$. Then,

$$\begin{tabular}{\vert c\vert c\vert l\vert} \hline prime & Sy... ... 4, 16, 28, 112 \\ \hline 7 & 7 & 1, 8 \\ \hline \end{tabular}$$

But by Cayley's Theorem $n_{2} \ne 3$, and $n_{3} \ne 4$. Therefore, n₃ = 7, or 21 and the minimum value of n₃ = 16 for G to be simple.

The key to this order turns out to be the Syl₇ subgroups.


For G simple, then n₇ = 8 and therefore G is isomorphic to a subgroup of S₈ and also of A₈. Also, the index in Gof the normalizer of a Syl₇ subgroup is 8, and so $\mid N_{G}(Syl_{7})\mid = 42$.


However, from Theorem 0.7, $\mid N_{S_{*}}(Syl_{7}) \mid = 7(7-1) = 42$, and Theorem 1.5 assures us that $\mid N_{A_{8}}(Syl_{7}) \mid = 21$. But $\mid N_{G}(Syl_{7})\mid = 42$. The normalizer in G "simply" can't fit into A₈. Therefore, G can not be isomorphic to a subgroup of A₈. This is a contradiction. Therefore, G is not simple!