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?



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


