## No simple group of order 6545 exists

Let $G$ be a group. Prove that if $|G| = 6545$ then $G$ is not simple.

Note that $6545 = 5 \cdot 7 \cdot 11 \cdot 17$. Note that the divisibility and congruence criteria of Sylow’s Theorem force the following.

• $n_5 \in \{ 1, 11 \}$
• $n_7 \in \{ 1, 85 \}$
• $n_{11} \in \{ 1, 595 \}$
• $n_{17} \in \{ 1, 35 \}$

Note in particular that for each prime $p$ dividing 6545, the Sylow $p$-subgroups of $G$ intersect trivially. If no Sylow subgroup is normal, then $G$ has at least $4 \cdot 11 + 6 \cdot 85 + 10 \cdot 595 + 16 \cdot 35 = 7094$ elements, a contradiction. Thus some Sylow subgroup of $G$ is normal, and hence $G$ is not simple.