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A monoid is invertible-free if $xy=1$ implies $x=y=1$ for all $x,y$.

Question: Can every cancellative invertible-free monoid be embedded in a group?

I'm fairly sure that a quotient of the free product of such a monoid with its mirror (this is the monoid with the same elements and identity but reversed multiplication, i.e. $x\cdot y=yx$) is the "most general" group in which it can be embedded.

This is the non-commutative version of the construction of the integers from the natural numbers.

Does this appear anywhere in the literature as a problem / proposition / theorem?

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    $\begingroup$ No, see mathoverflow.net/questions/109566 and references therein. $\endgroup$
    – Salvo Tringali
    Dec 4, 2020 at 8:56
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    $\begingroup$ To add detail to @SalvoTringali's answer Malcev gave the first example of a cancelative monoid not embeddable in a group and if is given by a finite presentation where both sides of each relation have length 2. So two equivalent words have the sam length and so if is invertible free $\endgroup$
    – Benjamin Steinberg
    Dec 4, 2020 at 10:17
  • $\begingroup$ Thank you! This is helpful. $\endgroup$
    – David Pokorny
    Dec 4, 2020 at 11:23

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No, it is not true even for finitely generated monoids. Take any semigroup $S$ which is cancellative and does not embed into a group (first examples were constructed by Malcev). Consider the monoid $S^1$ which is $S\sqcup\{1\}$ with $1$ a (new if $S$ is a monoid) neutral element. Then $S^1$ is an invertible-free monoid which does not embed into a group. It is cancellative iff $S$ does not have a neutral element.

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    $\begingroup$ This was the answer given in the comments $\endgroup$
    – Benjamin Steinberg
    Dec 5, 2020 at 2:05
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    $\begingroup$ If you mean your comment, then it is clearly different. $\endgroup$
    – markvs
    Dec 5, 2020 at 2:30
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    $\begingroup$ @BenjaminSteinberg dodd’s answer applies to any cancellative groupoid that does not embed into a group, it does not rely on the specifics of Malcev’s construction. $\endgroup$
    – Emil Jeřábek
    Dec 5, 2020 at 8:00
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    $\begingroup$ @EmilJerabek I guess this depends perspective. My answer is that Malcev's example has no units because his construction doesn't have the empty word in the relations. That is the same as adjoining and identify $\endgroup$
    – Benjamin Steinberg
    Dec 5, 2020 at 12:55
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    $\begingroup$ In any case, MO works best if questions are answered with an answer, rather than a comment. $\endgroup$
    – HJRW
    Dec 7, 2020 at 13:43

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