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.

Converse

Further information: Prime order implies no proper nontrivial subgroup

The converse is clearly true from Lagrange's theorem. Namely, if we have a cyclic group of prime order $p$, any subgroup must have order either 1 or $p$, by Lagrange's theorem. The only possible subgroup of order $p$ is the whole group, and the only possible subgroup of order 1 is the trivial subgroup.

Note that this also shows something stronger: any group of prime order, since it has no proper nontrivial subgroup, must be cyclic of prime order.

Related facts

Corollaries

• Simple Abelian implies cyclic of prime order

Other related facts

• Cyclic of prime power order iff not generated by proper subgroups
• Every subgroup is a direct factor iff trivial or elementary Abelian
• Cyclic iff not a union of proper subgroups

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 $\langle g\rangle =G$, 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 $g\neq e$ in $G$. Then, consider the subgroup generated by $g$. This is a nontrivial subgroup, hence, by assumption, $\langle g\rangle =G$.

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. Then, $G$ is isomorphic to the cyclic group of order $n$, under the identification $a\mapsto g^{a}$. Pick a positive integer $d\neq 1,n$ such that $d|n$. Then the subgroup generated by $g^{d}$ is a proper nontrivial subgroup of $G$ (corresponds to the multiples of $d$ mod $n$). More precisely, it is a subgroup of order $n/d$, because the order of $g^{d}$ is $n/d$. This is again a contradiction.

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