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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.

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    $\begingroup$ 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 $\endgroup$ – Benjamin Steinberg Mar 13 '16 at 16:44
  • $\begingroup$ To my shame, I hadn't thought of it. Let me edit and include your remark. $\endgroup$ – Salvo Tringali Mar 13 '16 at 16:46
  • $\begingroup$ 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. $\endgroup$ – Benjamin Steinberg Mar 13 '16 at 16:49
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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.

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  • $\begingroup$ @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. $\endgroup$ – Benjamin Steinberg Mar 13 '16 at 20:03

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