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.

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