By the fundamental theorem the problem is trivial:

(means the direct product taken 5 times)

Note . There is a Sylow -subgroup. If is the number of such Sylow subgroups then and so . Since there is only one Sylow -subgroup it must remain invariant under conjugation, i.e. it is a normal subgroup. And hence this group cannot be simple.2. Prove that no group of order 56 is simple. Hint: use the Sylow Theorems.

(By the way a famous conjecture of Burnside says no group of even order > 2 can be simple was proven in 1980's. So you can use it if you want complete overkill).

Using the same argument as about there is only one Sylow 5-subgroup and only one Sylow 7-subgroup . Form the group product , this is a subgroup since and because since .3. Let G be a group of order 35: a) Prove that G has only one 5-Sylow and one 7-Sylow that is, G(5) and G(7) respectively. b) Deduce that G is cyclic.