Every abelian simple group has prime order

Prove that if G is an abelian simple group then G \cong \mathbb{Z}/(p) for some prime p.


Let G be an abelian simple group.

Suppose G is infinite. If x \in G is a nonidentity element of finite order, then \langle x \rangle < G is a nontrivial normal subgroup, hence G is not simple. If x \in G is an element of infinite order, then \langle x^2 \rangle is a nontrivial normal subgroup, so G is not simple.

Suppose G is finite; say |G| = n. If n is composite, say n = pm for some prime p with m \neq 1, then by Cauchy’s Theorem G contains an element x of order p and \langle x \rangle is a nontrivial normal subgroup. Hence G is not simple. Thus if G is an abelian simple group, then |G| = p is prime. We saw previously that the only such group up to isomorphism is \mathbb{Z}/(p), so that G \cong \mathbb{Z}/(p). Moreover, these groups are indeed simple.

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