# Proving a property when elements of a group commute

## Main Question or Discussion Point

By commutative, we know that ##ab = ba## for all a,b in G. Thus, why do we need to prove separately that ##a^n b^m = b^ma^n##? Isn't it the case that ##a^n## and ##b^m## are in fact elements of the group? So shouldn't the fact that they commute automatically be implied?

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blue_leaf77
Homework Helper
Isn't it the case that anana^n and bmbmb^m are in fact elements of the group?
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.

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.
Oh, I see. I neglected to see that the group is not actually abelian

mathman
The proof you want seems trivial.

WWGD
Gold Member
2019 Award
By commutative, we know that ##ab = ba## for all a,b in G. Thus, why do we need to prove separately that ##a^n b^m = b^ma^n##? Isn't it the case that ##a^n## and ##b^m## are in fact elements of the group? So shouldn't the fact that they commute automatically be implied?
Why do you believe you need to prove it separately?

lavinia
Proof: $$a^nb^m=a^{n-1}bab^{m-1}$$, etc.