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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.