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Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively writ …

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups …

Thanks to Valentin Havlovec (TU Graz, Austria), I've finally got a copy of (i) Tamura and Shafer's paper [Power Semigroups, Math. Japon. 12 (1967), 25-32] and (ii) the note by Shafer mentioned by Benj …

Denote by $\mathcal P(S)$ the semigroup obtained by endowing the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced by $S$: $ …

Update: I had an email exchange with Victor Guba. He has kindly confirmed that there is indeed a typo in (the Russian and English versions of) his paper: a "кусок" (as per his paper) and an "$s$-piece …

I'm confused by the definition of a "cusp" as found in V.S. Guba, Conditions for the embeddability of semigroups into groups, Math. Notes 56 (1994), Nos. 1-2, 763-769 (link). In the words of Ma …

Preliminaries. Let $\mathbb A = (A, +)$ be a possibly non-commutative semigroup. For $X, Y \subseteq A$ we set $$ X - Y := \{a \in A: a + y \in X\text{ for some }y \in Y\}, $$ which is just the usual …

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 $\mathbb B = (B, +)$ s …

Let $\bf H$ be a set, $f: \mathcal P({\bf H}) \rightharpoonup \bf R$ a partial function, and $\mathcal{D}$ the domain of $f$. Next, denote by $\mathcal M(f)$ the set of all (total) functions $\theta: …

An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a monomorphis …

A linearly ordered (shortly, l.o.) monoid is a triple $\mathbb M = (M, \cdot, \le)$ for which $(M, \cdot)$ is a (multiplicatively written) monoid and $\le$ is a total order on $M$ such that $xy < xz$ …

We say that a group $(A, \cdot)$ is bi-orderable if there exists a total order $\preceq$ on $A$ such that $xz \prec yz$ and $zx \prec zy$ for all $x,y,z \in A$ with $x \prec y$. Let $m,n$ be non-zero …

Let $(A, +, \preceq)$ be an ordered group, namely $(A, +)$ is a group and $\preceq$ is a total order on $A$ such that $x + z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in A$ with $x \prec y$ …

Not really an example (see the edit below), but somehow related to Greg Kuperberg's one: Let $\mathbb A = (A, +, \cdot, \preceq)$ be a strictly totally orderable semiring (*) and for a fixed integer $ …

Let $\mathbb G = (G, +)$ be a group. We say that $\mathbb G$ is strictly totally orderable (others would say bi-orderable) if there exists a total order $\preceq$ on $G$ such that $x+z \prec y + z$ an …