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

You're missing something. A is defined using every element from G, so if the identity is an element in G then it's trivially in A.

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?

You just check the axioms.

 

For the existence of an inverse, we use the fact that G is abelian:

 

a^n * (a^-1)^n = (a * a^-1)^n = ...

Don't know whether this will help, but sometimes it saves a bit of time. If G is a group with H a subset of G, then if you can prove the condition [imath]g, h \in H \Rightarrow gh^{-1} \in H[/imath], H is a subgroup. As I said, it's saved me some time in exams, so there we go