Is it true that any group of order 30 has a normal (hence unique) Sylow-5 subgroup? I know that that the only possibilities for n(5) are 1 or 6. Now suppose there are 6 sylow 5 subgroups in G. This would yield (5-1)6=24 distinct elements of order 5 in G. Now there is only 30-24=6 elements left in G and one of these is the identity. This means that there must be 1 sylow 3 subgroup in G which has 2 distinct elements of order 2 now there is only 4 elements left in G one of them being the identity so there must be 3 sylow 2 subgroups of G in this case each having 1 distinct element of order 2 I don't see where is the contradiction here. The only thing I know is that there must be a cyclic normal subgroup of order 15 in G.