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Yes but each of ##a^n## and ##b^m## might equal another element of the group and they need not commute if the group is non-Abelian.Isn't it the case that anana^n and bmbmb^m are in fact elements of the group?

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Oh, I see. I neglected to see that the group is not actually abelianYes but each of ##a^n## and ##b^m## might equal another element of the group and they need not commute if the group is non-Abelian.

Why do you believe you need to prove it separately?

If you know that the group is commutative then there is nothing to prove. ##a^nb^m = b^ma^n## by definition. If you do not know that the group is commutative then you need a proof.

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