TITLE: Is a monoid commutative if $(ab)^2=a^2b^2$?
QUESTION [6 upvotes]: Let M be a monoid. Suppose that:
$(ab)^2=a^2b^2$
for any elements a,b in M. Is M commutative? The result is obviously true for groups, but I can't find a counterexample for monoids. And without cancellation law, I can't seem to show that it's true either.

REPLY [3 votes]: Here's a less ad hoc way of getting a counterexample.  Consider the free monoid $F$ on $x$ and $y$.  Every element of $F$ is a word in $x$ and $y$.  Let $S\subset F$ be the set of words of length greater than $2$, and let $M$ be the quotient of $F$ by the equivalence relation that identifies all elements of $S$ together (it is easy to see this is a congruence relation).  Then $(ab)^2=a^2b^2$ for all $a,b\in M$, since any instance of this with $a,b\neq 1$ must have both sides be words of length greater than $2$.  However, it is not commutative because $xy\neq yx$.  Explicitly $M=\{1,x,y,xy,yx,\infty\}$ where any product that would give a word of length greater than $2$ is $\infty$.
The motivating idea here is that if there is any counterexample, then the universal example would be a counterexample.  That is, in $F$ modulo the congruence relation that identifies $(ab)^2$ and $a^2b^2$ for all $a,b\in F$, we would need $x$ and $y$ to commute.  Now, this equivalence relation is complicated and hard to think about, but we can notice that it only ever identifies words that have length at least $4$ (except in the trivial cases where $a$ or $b$ is $1$).  So we see we can use a larger equivalence relation that is easily verified to be a congruence, but which still does not identify $xy$ and $yx$.