Abelian Group is Simple iff Prime

Theorem
Let $G$ be an abelian group.

Then $G$ is simple $G$ is a prime group.

Necessary Condition
Let $G$ be a simple abelian group whose order is $n$.

By definition of simple, $G$ has no normal subgroups.

From Subgroup of Abelian Group is Normal, it follows that $G$ can have no subgroups at all.

From Cauchy's Group Theorem, if $p$ is a prime number which is a divisor of $n$, then $G$ has a subgroup of order $p$.

It follows that if $G$ is simple, there can be no prime number less than $n$ which is a divisor of $n$.

It follows that $n$ is prime.

Sufficient Condition
Suppose $G$ is a prime group.

By Prime Group is Simple, it follows that $G$ is simple.