For indirect proof, suppose that G is a simple group of order 315 = (3²)(5)(7). By the Sylow theorems, it follows that there are exactly 7 Sylow 3-subgroups of G. Letting P be a Sylow 3-subgroup, we have [G: N_G(P)] = 7. So letting G act on the left cosets of N_G(P) by left translation induces a permutation representation G --> S₇ with kernel contained in N_G(P). Simplicity of G forces this kernel to be the identity and hence we may assume that G < S₇.

Let Q be a Sylow 5-subgroup of G. Note that Q is also a Sylow 5-subgroup of S₇. So, since 5² does not divide 7!, it follows that

|N_S7(Q)| = (7 - 5)!5(5 - 1) = 40.

Since Q < G < S₇, we have N_G(Q) < N_S7(Q). So |N_G(Q)| divides |N_S7(Q)| = 40.

Now, the Sylow theorems imply that there are exactly 21 Sylow 5-subgroups of G. So

|N_G(Q)| = |G|/21 = 315/21 = 15.

Since 15 does not divide 40, we have a contradiction.