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

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?

• No, see mathoverflow.net/questions/109566 and references therein. Dec 4, 2020 at 8:56
• 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 Dec 4, 2020 at 10:17
• Thank you! This is helpful. Dec 4, 2020 at 11:23

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.