mt87

how does the fact that the group is albelian help? I know that every cyclic group is abelian but in this case it is the subgroup that is cyclin. Does this proof need the basic prove-the-axioms-of-a-group approach or is it more complex?

If G is an abelian group and n>1 an integer, let A={a^n such that a E G}. Prove that A is a subgroup of G. isn't the identity of A a^0 which does not fall under n>1