Equivalence relation to make a group commutative

Question

A while ago I was wondering if there is a "natural" way to make a commutative group out of an arbitrary one. I played with the idea a bit and here is what I came up with.

Define a binary relation $\sim$ on a group $G$ such that $x \sim y$ if $\exists a, b \in G$ such that $ab=x$ and $ba=y$.

It's fairly easy to show this is an equivalence relation:

• Reflexivity: $xe = ex = x$, so $x\sim x$.
• Symmetry: obvious.
• Transitivity: Suppose $x \sim y$ and $y \sim z$. Then $\exists a,b,c,d$ such that $ab=x$, $ba=cd=y$, and $dc=z$. Let $e = ad^{-1}$ and $f=db$. Then $ef = ad^{-1}db = ab = x$, and $fe = dbad^{-1} = dyd^{-1} = dcdd^{-1} = dc = z$. So $x \sim z$. $\square$

Question: Let $[x]$ denote the equivalence class of $G/\mathord\sim$ containing $x$. Suppose we let $[x][y] = [xy]$. Is this operation well-defined?

If so, then $G/\mathord\sim$ is clearly a group under this operation. Furthermore, it is commutative, since $[x][y] = [xy] = [yx] = [y][x]$ (where the middle equality is by definition of $\sim$).

Answer 1

No, it is not (except in trivial cases). Note that the only element equivalent to $e$ is $e$ itself.
So if $y \in [x]$ but $y \ne x$, $y x^{-1} \in [x] [x^{-1}]$ but is not in $[x x^{-1}] = \{e\}$.

Answer 2

Note that the equivalence relation is identical to conjugacy classes. If

$$\exists a: axa^{-1}=y$$ then $$a^{-1}axa^{-1}a=x,\;(a)(a^{-1}axa^{-1})=axa^{-1}=y$$ or on a better day just $$a^{-1}(ax)=x,\;(ax)a^{-1}=y$$ and conversely $$\exists a,b: ab=x,\;ba=y\implies aya^{-1}=x$$