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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

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3 Answers 3

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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.

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  • $\begingroup$ Don't you mean $cf=f$ for any $f$? $\endgroup$ Oct 21, 2017 at 19:02
  • $\begingroup$ @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. $\endgroup$ Oct 23, 2018 at 15:40
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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.

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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.

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  • $\begingroup$ Sorry, but what does $<x,x^{-1},y>$ mean $\endgroup$
    – Amr
    Sep 4, 2013 at 16:40
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    $\begingroup$ 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. $\endgroup$
    – Jim
    Sep 4, 2013 at 16:42
  • $\begingroup$ Ahh OK. The free monoid on the symbols $\{x,x^{-1},y\}$ subject to the relations .... $\endgroup$
    – Amr
    Sep 4, 2013 at 16:43
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    $\begingroup$ 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 $xz = zx = 1$. $\endgroup$
    – Jim
    Sep 4, 2013 at 16:46
  • $\begingroup$ Surely this monoid is isomorphic to its opposite, by the homomorphism sending $x$ to $x$, $y$ to $y$, and in general reversing words? $\endgroup$ Sep 25, 2013 at 20:10

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