The discussion revolves around the necessity of proving that elements of a group commute when the group is known to be commutative. Participants explore whether the commutativity of individual elements implies the commutativity of their powers.
Participants express differing views on whether a separate proof is necessary for the commutativity of powers in a commutative group. The discussion remains unresolved regarding the necessity of the proof in the context of non-Abelian groups.
Participants discuss the implications of group properties and the definitions of commutativity, highlighting the potential for misunderstanding when the nature of the group is not clearly defined.
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.Mr Davis 97 said:Isn't it the case that anana^n and bmbmb^m are in fact elements of the group?
Oh, I see. I neglected to see that the group is not actually abelianblue_leaf77 said: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.
Why do you believe you need to prove it separately?Mr Davis 97 said: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?
Mr Davis 97 said: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?