# Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible

Let $\mathbb A = (A, +_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ such that $\mathbb A$ is a submonoid of $\mathbb B$ and $x$ is left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below)?

It is known from a paper of A. I. Mal'cev [Math. Ann. 113 (1937), No. 1, 686-691] that there exist (finitely generated) cancellative monoids that do not embed into a group, but the question here is about an embeddability condition that looks very much weaker.

• Why do you say left invertible? In a cancellative monoid left invertible elements are right invertible. If yx =1 then xyx=x and so xy =1 by cancellation Mar 13, 2016 at 16:44
• To my shame, I hadn't thought of it. Let me edit and include your remark. Mar 13, 2016 at 16:46
• It seems to me the answer should be no. Take a fg cancellative monoid $M$ not embeddable in a group. Enumerate the elements and invert them one by one. Then the original monoid would embed in the unit group of the direct limit. Mar 13, 2016 at 16:49

## 1 Answer

The answer is no.

In a cancellative monoid left invertible elements are right invertible. If yx =1 then xyx=x and so xy =1 by cancellation. So you are asking to invert a given element in some cancellative monoid.

Take a fg cancellative monoid M not embeddable in a group. Invert the generators one by one to embed M in the group of units of a monoid and get a contradiction.

• @EmilJerabek, No. But the OP is asking whether it can always be done and I am arguing this is the same as embedding in a group. Mar 13, 2016 at 20:03