**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.