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Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!

Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.

Edit: Thanks, I see that Feit-Thompson is too much.

Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| > 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p > 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.

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Using Feit-Thompson here is ludicrous overkill. –  Chris Eagle Aug 23 '12 at 20:52
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And unless you can check that the proof does not become circular by using it, invalid :-) –  Mariano Suárez-Alvarez Aug 23 '12 at 20:54
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Why don't you try showing that an abelian group of composite order is not simple. You could use cauchy's theorem or some Sylow theory. –  JSchlather Aug 23 '12 at 20:56
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Using Cauchy or Sylow is slightly less ludicrous overkill. –  Chris Eagle Aug 23 '12 at 21:00
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3 Answers

Definitely you're swatting a fly with a nuclear weapon. The Feit-Thompson theorem is not easy to prove, to put it mildly. But it's pretty easy to prove that all abelian simple groups are cyclic groups of prime order.

Suppose it's not cyclic. It's not generated by any one element. Consider $a\ne\mathrm{identity}$. Then some $b$ is not in the subgroup generated by $a$. Think about the subgroups generated by these two elements and see if you can find a nontrivial normal subgroup.

Now suppose the order is $n=k\ell$ and $k,\ell>1$. Let $a\ne\mathrm{identity}$. Think about the group generated by $a^k$ and see if you can find a nontrivial normal subgroup.

And the case of infinite order is not hard.

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Actually, I didn't even need to mention $b$: Consider $a\ne\mathrm{identity}$. Since $a$ doesn't generate the whole group, what can you say about the subgroup that it generates? –  Michael Hardy Aug 23 '12 at 23:32
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Er okay: suppose $G$ is simple, abelian. Then any $g \in G$ has to generate the whole thing, i.e. we have $\mathbb{Z} \twoheadrightarrow G$. $\mathbb{Z}$ is not simple (consider say $2\mathbb{Z}$), so $G \simeq \mathbb{Z}/n \mathbb{Z}$ for some $n$. If $n$ is not prime, take any $p \mid n$, we have $\langle p\rangle \subseteq \mathbb{Z}/n\mathbb{Z}$ is a proper subgroup.

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I changed $<p>$ to $\langle p\rangle$. The TeX code is \langle p\rangle. –  Michael Hardy Aug 23 '12 at 21:03
    
Daww thanks man! –  uncookedfalcon Aug 23 '12 at 21:03
    
I like this solution, because it very quickly makes concrete exactly which groups you have to consider. –  Dustan Levenstein Aug 23 '12 at 21:29
    
haha thanks man –  uncookedfalcon Aug 23 '12 at 21:39
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Every subgroup of an abelian group is normal.Thus, a simple abelian group must have no proper nontrivial subgroup. We also note that no proper nontrivial subgroup implies cyclic of prime order.

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