Is every monoid isomorphic to its opposite

Question

This may be a trivial question. Every group is isomorphic to its opposite using the isomorphism that sends $x$ to $x^{-1}$. Now does this hold even if the condition of existence of inverses is dropped. More precisely:

Is every monoid isomorphic to its opposite ?

I am expecting the existence of a counterexample, but I don't have any for now.

Thank you

Answer 1

No. Let $X$ be any finite set with at least two elements, and consider the monoid $\mathrm{End}(X)$, of functions $X \to X$, under composition.

Each constant function $c$ is a left-absorbing element, i.e. $cf = c$ for any $f$. However, there are no right-absorbing elements in the monoid. So $\mathrm{End}(X)$ cannot be isomorphic to its opposite, since any such isomorphism would have to interchange left- and right-absorbing elements.

Answer 2

Take the monoid $M = \langle x, x^{-1}, y\rangle$. The invertible elements are powers of $x$ so any isomorphism $M \simeq M^\mathrm{op}$ must send $x \mapsto x^a$ for some integer $a$. To get surjectivity $y$ cannot be sent to a power of $x$, thus the images of $x$ and $y$ aren't going to commute, so the map can't respect the multiplication of both monoids.

Answer 3

No. Take a semigroup of left zeroes ($ab=a$ for all $a,b\in S$). Antiisomorphic to $S$ (but not isomorphic) is a semigroup of right zeroes ($ab=b$). If you want to get a monoid, join to both of them an identity.