• MostlyHarmless
In summary, the conversation discusses a question about a property of cosets, specifically whether the equation abH=baH implies that ab=ba. The group G modded by its center, Z(G), and its relationship to nontrivial cyclic groups is also mentioned. The conversation concludes with a discussion about using the fact that G/Z(G) is cyclic to write elements of G/Z(G) as powers of a generator. Ultimately, it is determined that the equation does not necessarily imply that ab=ba, but rather it is true if and only if a certain condition is met.

## Homework Statement

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.

## Homework Equations

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.

## The Attempt at a Solution

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.

MostlyHarmless said:
Here is my question:
Does, abH=baH imply that ab=ba
No. What is true is that ##abH = baH## if and only if ##a^{-1}b^{-1}ab \in H##. Certainly this is true if ##ab = ba##, but it can also be true if ##ab \neq ba##, for example if ##H## is (or contains) the commutator subgroup of ##G##: http://en.wikipedia.org/wiki/Commutator_subgroup

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)
So far so good. For brevity, let's write ##Z## instead of ##Z(G)##. Suppose that ##G/Z## is cyclic. Then ##G/Z = \langle xZ\rangle## for some ##x \in G##, in other words, the elements of ##G/Z## are cosets of the form ##x^kZ## where ##x## is some fixed element of ##G## and ##k \in \mathbb{Z}##. In particular, if ##a,b \in G## then there are integers ##m,n## such that ##aZ = x^mZ## and ##bZ = x^nZ##, or equivalently, ##a \in x^mZ## and ##b \in x^nZ##. What does this imply about ##a^{-1}b^{-1}ab##?

Sorry for not responding to this, the proof I turned in had a similar idea, it just wasn't quite right. This was the solution shown to us in class.