No proper nontrivial subgroup implies cyclic of prime order

Statement

If a nontrivial group has no proper nontrivial subgroup, then it is a cyclic group of prime order. In other words, it is generated by a single element whose order is a prime number.

Proof

Given: A nontrivial group $G$, such that the only subgroups of $G$ are the trivial subgroup, and $G$ itself

To prove: There exists an element $g\in G$ such that $⟨g⟩=G$, and the order of $g$ is a prime number. In particular, $G=\{e,g,g^2,\dots ,g^{p-1}\}$ with $g^p=e$

Proof: Since $G$ is nontrivial, there exists $g\ne e$ in $G$. Then, consider the subgroup generated by $g$. This is a nontrivial subgroup, hence, by assumption, $⟨g⟩=G$.

Now there are two possibilities. First, that $g$ has infinite order: no positive power of $g$ is trivial. In this case, the group $G$ is isomorphic to (i.e., can be identified with) the group $\mathbb{Z}$, under the identification $n\mapsto g^n$. In particular, the subgroup generated by $g^2$, which corresponds to the even integers, is a proper nontrivial subgroup, leading to a contradiction.

Hence, $g$ cannot have infinite order, so let $n$ be the order of $g$. Then, $g^n=e$. Suppose that $n$ is composite. Pick a positive integer $d\ne 1,n$ such that $d|n$. This can be identified with the cyclic group of order $n</math.undertheidentification<math>a\mapsto g^a$. Then the subgroup generated by $g^d$ is a proper nontrivial subgroup of $G$ (corresponds to the multiples of $d$ mod $n$), again a contradiction.

Hence, $n$ must be a prime, so $G$ is cyclic of prime order (as desired).