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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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2 Answers 2

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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 Schlitt 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

A solvable finite simple group is cyclic of prime order. It is not necessary to assume that the given simple group is abelian, solvable is sufficient.

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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 Schlitt Oct 18 '12 at 2:23
    
@lovinglifein2012 Any subgroup of an abelian group is normal, and so the group is generated by any one of its elements if it is simple. Then you are reduced to showing that a cyclic simple group is finite and of prime order. –  i. m. soloveichik Oct 18 '12 at 2:30

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