how to show that there is no simple group of order $1755 = 3^3 \cdot 5 \cdot 13$? Thank you very much.
using the sylow theorems we see that:
the number of 5-subgroups must be 351 or 1 (the only divisors of 13*27 which are 1 mod 5).
if the number of 5 subgroups and 13 subgroups were both not equal to 1 then the number of non-identity elements of these subgroups would be: $$(5-1)\cdot(351)+(13-1)\cdot(27)=1728=1755-27$$ (noting that the 5 and 13 subgroups all intersect trivially) so that the remainder of the group would have to consist of exactly the 27-subgroup. hence there is a normal sylow subgroup.