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Removed some text that was no longer meaningful (after Benjamin Steinberg's remark)
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Salvo Tringali
Salvo Tringali
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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, say, left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below), i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

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.

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, say, left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below), i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

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.

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.

Embedded BS's remark
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Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64

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, say, left-invertibleleft-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below), i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

It is known from a paper of A. I. Mal'cev [Math. AnnalenAnn. 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.

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, say, left-invertible in $\mathbb B$, i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

It is known from a paper of A. I. Mal'cev [Math. Annalen 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.

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, say, left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below), i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

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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Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64

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, say, left-invertible in $\mathbb B$, i.e. there exists an element $\tilde x \in B$ such that $\tilde x + x = 0$?

It is known from a paper of A. I. Mal'cev [Math. Annalen 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.