TITLE: Proof that all abelian simple groups are cyclic groups of prime order
QUESTION [22 upvotes]: 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.

REPLY [8 votes]: Let $G$ be an abelian simple group with $|G| > 1$. We want to prove that $G$ is a cyclic group of prime order.
First note that because $G$ is abelian, then every subgroup of $G$ is normal.
Let $x \in G$ and $x \neq 1$, so $\langle x \rangle \leq G$. But because $G$ is simple, $\langle x \rangle$ must be the $G$ itself as there couldn't be any other normal subgroup of $G$, or simply $\langle x \rangle = G$.
If $|G|$ is not finite, then we could have e.g. $\langle x^2 \rangle \neq \langle x \rangle$ as another proper normal subgroup of $G$ which is a contradiction becase G is simple. (It's a proper normal subgroup because $|x^2| < |x|$.)
Hence $G = \langle x \rangle$ must be a finite group.
Let $p = |G|$. Assume $p = mn$ where $m > 1$ and $n > 1$. If that's the case, then we could have $\langle x^m \rangle$ or $\langle x^n \rangle$ as another proper normal subgroup of $G$, which is again a contradiction.
Thus $G = \langle x \rangle$ and $|G| = |x| = p$ where $p$ is a prime number.