Abelian Group is Simple iff Prime

Theorem
Let $G$ be a non-trivial abelian group.

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

Proof
First we note that the trivial group is (trivially) simple, but not prime because $1$ is not prime.

Hence the specification of $G$ as being non-trivial.

Necessary Condition
Let $G$ be a simple non-trivial abelian group.

Suppose that $G$ is an infinite group.

Then by Infinite Group has Infinite Number of Subgroups $G$ has an infinite number of distinct subgroups.

But then by Subgroup of Abelian Group is Normal, $G$ cannot be simple.

So by contradiction $G$ must be finite.

Let $n$ be the order of $G$.

By definition of simple, $G$ has no non-trivial proper normal subgroups.

From Subgroup of Abelian Group is Normal, it follows that $G$ can have no non-trivial proper 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.

Because $G$ is prime, its order is greater than $1$.

Thus $G$ is non-trivial.