Order Equal to 252

Let $\mid G \mid = 252 = 2^{2} \times 3^{2} \times 7$.

$$\begin{tabular}{\vert c\vert c\vert l\vert} \hline prime & Sy... ... & 1, 4, 7, 28 \\ \hline 7 & 7 & 1, 36 \\ \hline \end{tabular}$$

Since the index of the normalizer in G of a Syl₇ subgroup must be 36 in a simple group, and all Sylow p-subgroups of single prime power interesct in the identity, then the Syl₃ subgroups must intersect non-trivially or there would be too many elements in the group. Therefore, if $H,H' \in Syl_{3}(G)$ then $\mid H \bigcap H' \mid = 3$. But this subgroup is normal in both H and H' and so the normailzer of $H \bigcap H'$ contains HH'.


We know that $\mid HH' \mid = 27$. Therefore letting $k = \mid N_{G}(H \bigcap H') \mid$, we have the following conditions: $k \ge 27, 9 \mid k$, and $k \mid 252$. So k = 36 and $\mid G:N_{G}(H \bigcap H') \mid \le 4$. But the normalizer in G of the intersection of these Sylow subgroups is a subgroup in G with index 4. $\mid G \mid$ must divide 4! and here is our contradiction. G can not be simple!