No simple group of order $300$

Solution 1:

OK, the following results lead to a solution to this and similar problems.

Theorem. Let $G$ be a finite nonabelian simple group with a subgroup $H$ of index $n>1$. Then $n \ge 5$, and $|G|$ divides $n!/2$.

Proof. Let $\phi: G \rightarrow S_n$ be the permutation representation of $G$ acting by (left or right depending on whether you use left or right group actions) multiplication on the set of (left or right) cosets of $H$ in $G$. Then $G/{\rm Ker}(\phi) \cong {\rm im}(\phi) \le S_n$. Since $n>1$ and ${\rm im}(\phi)$ is transitive, $|{\rm im}(\phi)| > 1$ and so $G$ simple implies ${\rm Ker}(\phi) = 1$, and hence $G \cong {\rm im}(\phi)$. Now $S_n$ is solvable for $n < 5$, so we must have $n \ge 5$. Furthermore, we must have ${\rm im}(\phi) \le A_n$, since otherwise ${\rm im}(\phi) \cap A_n$ would be a normal subgroup of ${\rm im}(\phi)$ of index 2, and so $G$ would not be simple. Hence $|G|$ divides $|A_n| = n!/2$.

Corollary. Let $G$ be a finite simple group and $n = |{\rm Syl}_p(G)|$ for some prime $p$ dividing $|G|$. Then $n \ge 5$ and $|G|$ divides $n!/2$.

Proof. Let $P \in {\rm Syl}_p(G)$. We cannot have $n=1$ because then $P$ would be normal in $G$. Now apply the theorem to the subgroup $N_G(P)$ of index $n$ in $G$.

Solution 2:

Assume $G$ is simple. Then the existence of 6 Sylow 5-groups implies $G$ embeds in $S_{6}$ (let $G$ act on the Sylow 5-subgroups by conjugation and use the assumption that $G$ is simple). But 300 does not divide 6 factorial. So $G$ is not simple.