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I'm working on a rather tough question (well, at least for myself):

Show that no commutative group of composite (or infinite) order is simple.

Any ideas on how I should approach this? All I know is that groups that contain no proper normal subgroups are simple groups. How can I use this to generalize it to any commutative group? Any hints/suggestions are welcome, since that is all I am able to come up with, at the moment at least.

Kyogre
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  • In the finite case: all you need to show is that a group of composite order has a proper nontrivlal subgroup, for that subgroup will be automatically normal, and the Sylow theorem does precisely that. – Mariano Suárez-Álvarez May 08 '17 at 05:25
  • That's the problem I have: since I have not covered (or even mentioned) any of Sylow's Theorems, I am not allowed to use information outside from lecture/textbook. Of course, if this is the only way to approach such a problem, we could make this an exception. – Kyogre May 08 '17 at 05:27
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    The Sylow theorems are not necessary. Let $G$ be a nontrivial abelian group, and let $x \in G$ be a nonidentity element. If $H := \langle x \rangle \neq G$, then $H$ is a nontrivial subgroup of $G$ which is necessarily normal, since $G$ is abelian. If $H = G$, then $G$ is cyclic, and thus has a proper subgroup for every nontrivial divisor of $|G|$ (if $|G| = \infty$, then $x^{k}$ generates a proper subgroup of $G$ for every $k \in \mathbb{N}$). Hence, if $G$ is abelian and simple, then $G$ is cyclic of prime order. This is the contrapositive of the given statement. – Alex Wertheim May 08 '17 at 06:38
  • Apologies for the late reply, but after thoroughly looking at your proof, it would be better of as an answer IMOH. I would have never thought to argue with a proof by contrapositive. – Kyogre May 09 '17 at 02:03

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