Cyclic of prime power order iff not generated by proper subgroups

Statement
The following are equivalent for a nontrivial group:


 * 1) It has a unique maximal subgroup, and this subgroup contains every proper subgroup.
 * 2) It is not generated by its proper subgroups.
 * 3) It is a cyclic group whose order is a power of a prime.

Related facts

 * Cyclic iff not a union of proper subgroups
 * No proper nontrivial subgroup implies cyclic of prime order

Equivalence of (1) and (2)

 * (2) implies (1): The subgroup generated by all proper subgroups is a subgroup containing every proper subgroup. In particular, if it is not the whole group, it is a maximal subgroup. Moreover, since it contains every proper subgroup, it must be the unique maximal subgroup.
 * (1) implies (2): If there exists a maximal subgroup that contains every proper subgroup, then this is precisely the subgroup generated by all the proper subgroups. Thus, the whole group is not generated by its proper subgroups.

Equivalence of (1) and (3)
Given: A group $$G$$ with a maximal subgroup $$M$$ that contains every proper subgroup of $$G$$.

To prove: $$G$$ is a finite cyclic group and its order is a power of a prime.

Proof:


 * 1) $$G$$ is a cyclic group, generated by any element outside $$M$$: Pick any element $$g \in G \setminus M$$. Then, the subgroup generated by $$g$$ cannot be proper, otherwise it would be contained in $$M$$. This forces $$\langle g \rangle = G$$.
 * 2) $$G$$ is not an infinite cyclic group: Any infinite cyclic group is isomorphic to the group of integers $$\mathbb{Z}$$, and $$\mathbb{Z}$$ contains infinitely many maximal subgroups: the subgroup generated by any prime number is maximal.
 * 3) $$G$$ is isomorphic to the group of integers modulo $$n$$, and $$n$$ must be a prime power: If $$G$$ is finite cyclic, it is isomorphic to the group of integers modulo $$n$$ for $$n$$ equal to the order of the group. If $$n$$ has two distinct prime factors, say $$p$$ and $$q$$, then $$G$$ has two distinct maximal subgroups: the subgroup comprising multiples of $$p$$ mod $$n$$ (which has order $$n/p$$) and the subgroup comprising multiples of $$q$$ mod $$n$$ (which has order $$n/q$$). This contradicts the assumption. Thus, $$n$$ has exactly one prime factor, forcing it to be a prime power.