**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.4 Exercise 3.4.1**

Solution: Let $G$ be an abelian simple group.

Suppose $G$ is infinite. If $x \in G$ is a nonidentity element of finite order, then $\langle x \rangle < G$ is a nontrivial normal subgroup, hence $G$ is not simple. If $x \in G$ is an element of infinite order, then $\langle x^2 \rangle$ is a nontrivial normal subgroup, so $G$ is not simple.

Suppose $G$ is finite; say $|G| = n$. If $n$ is composite, say $n = pm$ for some prime $p$ with $m \neq 1$, then by Cauchy’s Theorem $G$ contains an element $x$ of order $p$ and $\langle x \rangle$ is a nontrivial normal subgroup. Hence $G$ is not simple. Thus if $G$ is an abelian simple group, then $|G| = p$ is prime. We saw previously that the only such group up to isomorphism is $\mathbb{Z}/(p)$, so that $G \cong \mathbb{Z}/(p)$. Moreover, these groups are indeed simple.