1. Jan 15, 2015

### MostlyHarmless

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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.

2. Jan 16, 2015

### jbunniii

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

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 $a^{-1}b^{-1}ab$?

3. Jan 21, 2015

### MostlyHarmless

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.