Claim: If G has no subgroups H /= (e), G, then G must be cyclic of prime order.
Proof:
One Line Proof: If the order of G is composite, then it has Sylow Subgroups.
More than one Line Proof: If the order of G is composite then there exists d that divides |G| and 1 < d < |G|. Pick any element g from G. Note that ( g^d \neq e ) otherwise we will find a nontrivial subgroup. Then consider the non-trivial subgroup generated by ( g^d ). As ( g^{d \times \frac{|G|}{d} } = e ) hence we find a non trivial subgroup.
Therefore the order of G cannot be composite.
Rest is left as an exercise.
