Prove that if is an abelian simple group then for some prime .
Let be an abelian simple group.
Suppose is infinite. If is a nonidentity element of finite order, then is a nontrivial normal subgroup, hence is not simple. If is an element of infinite order, then is a nontrivial normal subgroup, so is not simple.
Suppose is finite; say . If is composite, say for some prime with , then by Cauchy’s Theorem contains an element of order and is a nontrivial normal subgroup. Hence is not simple. Thus if is an abelian simple group, then is prime. We saw previously that the only such group up to isomorphism is , so that . Moreover, these groups are indeed simple.