TITLE: Equivalence relation to make a group commutative
QUESTION [6 upvotes]: 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 $\sim$).

REPLY [4 votes]: 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
$$\exists a,b: ab=x,\;ba=y\implies aya^{-1}=x$$