Abelian simple groups have prime order

Statement

Let $G$ be an abelian group. Then $G$ is simple if and only if $G$ is finite of prime order. In that case, $G$ is cyclic, and $G \cong \mathbb{Z}/p\mathbb{Z}$ for a prime number $p$.

Remarks

In an abelian group, each subgroup is normal, so a simple abelian group has no subgroup except $\{e\}$ and itself. Then $G = \langle g \rangle$ for each $g \neq e$, so $G$ is cyclic; $\mathbb{Z}$ is not simple, because $2\mathbb{Z}$ is a proper nontrivial subgroup. A finite cyclic group of order $n$ has a subgroup of order $d$ for each divisor $d$ of $n$, so $n$ must be prime.

Explanations

L1 human

In an abelian group, all subgroups are normal. So a simple abelian group has no subgroups at all except the trivial ones. Only the groups of prime order have this property.