Hi Guy's, I know this is not for home work questions however I have had no luck in that section. Have I done enough to show that 10 cannot be a sub-group of order 324 1. The problem statement, all variables and given/known data Let G be a group of order 324. Show that G has subgroups of order 2, 3, 4, 9, 27 and 81, but no subgroups of order 10. 2. Relevant equations Sylow showed that if a prime power divides the order of a finite group G, then G has a subgroup of order . 3. The attempt at a solution I can see that G can have the subgroup 2 because [itex]2^n n=1 = 2[\latex] subgroup 3 because [itex]3^n n=1 = 3[\latex] divides 324 subgroup 4 because [itex]2^n n=2 = 4[\latex] divides 324 subgroup 9 because [itex]3^n n=2 = 9[\latex] divides 324 subgroup 27 because [itex]3^n n=3 = 27[\latex] divides 324 subgroup 81 because [itex]3^n n=4 = 81[\latex] divides 324 I know that 10 does not divide 324 in Z Is that enough to show that the sub group can't be order 10 ?