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?(adsbygoogle = window.adsbygoogle || []).push({});

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# I Proving a property when elements of a group commute

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