Every Simple Abelian group is cyclic of prime order?

This was in a claim in my class notes in a proof that every Solvable Simple group is of prime order.

I was able to verify it in the case where $G$ is finite, which I think might be a missing hypothesis in the statement of the theorem.

Can anyone confirm that this is a necessary assumption to use this fact?

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As you noticed a solvable, simple group is necessarily abelian, since the derived subgroup is trivial. Now, let $a$ be any nonidentity element. The subgroup generated by $\{a\}$ is a normal subgroup, equal to our entire group by the hypothesis that it is simple.

If it were an infinite cyclic group (isomorphic to $\Bbb{Z}$), then there are certainly nontrivial proper subgroups, e.g. the subgroup generated by $a^2$.

Otherwise, it is isomorphic to $\Bbb{Z}_n$, and for each nontrivial divisor of $n$ corresponds a nontrivial proper subgroup. Therefore, $a$ has prime order.

It follows that the given group is prime cyclic.

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Thanks very much! –  Kyle Oct 18 '12 at 2:24
@lovinglifein2012 I am glad I can help you verify a very useful fact. –  peoplepower Oct 18 '12 at 2:35
To clarify, the proof first shows that since $G$ is Simple and Solvable it must be abelian. The claim is then that every abelian simple group must be of prime order and cyclic. I'm asking if the assumption that $G$ be finite is necessary. –  Kyle Oct 18 '12 at 2:23