$\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is
$$
 G/\left(ab\sim ba\ \middle|\ a,b\in G\right)?
$$
Answer: not $G^{\mathrm{ab}}$, [but](https://math.stackexchange.com/a/765583) the set of conjugacy classes of $G$.


When passing to monoids, the situation gets more complicated: the equivalence relations generated by the following declarations are all equivalent for $M$ a group, but not for $M$ a monoid:


(1) *Conjugacy.* $a\sim\_{1}b$ iff there exists an invertible $g\in M$ such that $gag^{-1}=b$;


(2) *Conjugacy, II.* $a\sim\_{2}b$ iff there exists a non-necessarily invertible $m\in M$ such that $ma=bm$;


(3) *"Commutativity".* $ab\sim\_{3}ba$ for each $a,b\in M$;


In an [MO answer](https://mathoverflow.net/a/52110), Tom Leinster explained how to describe $M/\unsim\_{1}$ and $M/\unsim\_{2}$: First, let $[\mathbb{N},M]$ be the category where


* $\mathrm{Obj}([\mathbb{N},M])=M$;
* A morphism $m\longrightarrow m'$ is an element $g$ of $M$ such that $gm=m'g$.


Then we have bijections
$$
\begin{align\*}
 M/\unsim\_{1} &\cong [\mathbb{N},M]/\{\text{isos}\},\\
 M/\unsim\_{2} &\cong \pi\_{0}([\mathbb{N},M]).
\end{align\*}
$$


**Question:** Given a monoid $M$, what is the set $M/\unsim\_{3}=M/\left(ab\sim ba\ \middle|\ a,b\in M\right)$?



Conjugacy classes of monoids II: Abelianising a monoid, wrongly
Defining conjugacy for monoids is a dicey subject because many different notions that are equivalent for groups are different for monoids and it is not clear which of these is interesting. The one you call 3 is probably the most commonly studied one, although it varies depending on the context how useful its.


I am not sure of a category theoretic description of the third relation because I am semigroup theorist and not a category theorist but I will say that if you move to representations or algebras, then it is one of the more natural notions. This is because any trace on the monoid algebra factors through the quotient of $KM$ by this relation. If $K$ is a field, then the equivalence classes of $\sim\_3$ form a basis for the $0$-Hochschild cohomology $HH\_0(KM)$ and that seems to me a good reason already to think about it.


For finite von Neumann regular monoids, one has that two elements are equivalent under $\sim\_3$ if and only if all complex characters of the monoid agree on them. This is not true for finite monoids in general, which have an extra relation that you need to add.


