# Is every monoid isomorphic to its opposite

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

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.

• Don't you mean $cf=f$ for any $f$? Oct 21, 2017 at 19:02
• @ManoPlizzi: No, that would be the condition that $c$ is a left unit, which doesn’t suffice for this example since $\mathrm{End}(X)$ has a unique left unit (the identity function) which is also a unique right unit. Oct 23, 2018 at 15:40

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.

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.

• Sorry, but what does $<x,x^{-1},y>$ mean
– Amr
Sep 4, 2013 at 16:40
• It means that the elements of the monoid are words in the letters $x, x^{-1}, y$ subject to the relations $xx^{-1} = x^{-1}x = 1$. Composition is concatenation of words.
– Jim
Sep 4, 2013 at 16:42
• Ahh OK. The free monoid on the symbols $\{x,x^{-1},y\}$ subject to the relations ....
– Amr
Sep 4, 2013 at 16:43
• Yep, writing $\langle\text{letters} \mid \text{relations}\rangle$ is pretty common, as is writing $x, x^{-1}$ as letters to mean that $x, z$ are letters and there is a relation $xz = zx = 1$.
– Jim
Sep 4, 2013 at 16:46
• Surely this monoid is isomorphic to its opposite, by the homomorphism sending $x$ to $x$, $y$ to $y$, and in general reversing words? Sep 25, 2013 at 20:10