Using the fact that G is abelian in this abstract algebra problem

[jdinatale]

I'll post the problem and my attempt at solution all in one picture:

In the red step, I'm using commutative multiplication. Am I allowed to do this? I'm not sure, because the subset of G might not be a subgroup, so I don't know if its necessarily abelian like G is. Or does the fact that G is abelian override this somehow?

 

[deluks917]

Any subgroup of an abelian group is abelian.

 

[jdinatale]

Any subgroup of an abelian group is abelian.

i'm trying to PROVE it's a subgroup. That's the question itself.

 

[micromass]

That looks good. I don't see the confusion. In an abelian group, it always holds that

$$(ab)^n=a^nb^n$$

You don't need any subgroups to prove this.

 

[deluks917]

Sorry any subset of an abelian group is abelian.