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

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

jdjd.jpg




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?
 
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Any subgroup of an abelian group is abelian.
 
deluks917 said:
Any subgroup of an abelian group is abelian.

i'm trying to PROVE it's a subgroup. That's the question itself.
 
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.
 
Sorry any subset of an abelian group is abelian.
 
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