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Question about Cosets: Does abH=baH imply ab=ba? | Group Theory Homework
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[QUOTE="MostlyHarmless, post: 4977227, member: 415423"] [h2]Homework Statement [/h2] I really just need clarification about a property of cosets. I can't find anything explicitly stating one way or the other, and it could be because I'm wrong, or because it's deemed trivially true. [h2]Homework Equations[/h2] Left Coset: (aH)(bH)=abH, where a,b are elements of a group G, and H is a normal subgroup of G. G/Z(G):= {gZ(G) | g ##\epsilon##G}, where Z(G) is the center of the group, G. [h2]The Attempt at a Solution[/h2] Here is my question: Does, abH=baH imply that ab=ba The framing of the problem is that I want to show that a group G modded by its center, Z(G), that is, G/Z(G), cannot be a nontrivial cyclic group. My approach is to show that G is abelian and therefor G=Z(G). Moreover, I'm using the fact that G/Z(G) is cyclic to write elements of G/Z(G) as powers of a generator. That is, letting two elements a,b be in G; I get aZ(G), bZ(G) are elements of G/Z(G). Then I proceed similarly to a proof that cyclic implies abelian. If needed I can spell out the proof, but at the moment I'm more concerned with the bit about this paragraph. [/QUOTE]
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Question about Cosets: Does abH=baH imply ab=ba? | Group Theory Homework
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