- Prove that every simple abelian group is cyclic.
Solution
Let \(G\) be a nontrivial simple abelian group. Choose \(1\ne g\in G\). Since \(G\) is abelian, every subgroup is normal, so the nontrivial subgroup \(\langle g\rangle\) must equal \(G\). Hence \(G\) is cyclic.
If \(G\) were infinite, then \(G\cong\mathbb Z\), which has the proper nontrivial subgroup \(2\mathbb Z\), contradicting simplicity. Thus \(G\) is finite cyclic, say of order \(n\). If \(n\) were composite, a proper divisor of \(n\) would give a proper nontrivial subgroup of \(G\). Therefore \(n=p\) is prime.
Hence every simple abelian group is isomorphic to \(\mathbb Z/p\mathbb Z\) for some prime \(p\).