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One often reads that divisible groups are important since they help us understanding the structure of abelian groups, for they are all and the only injectives in the usual category of abelian groups ( …

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 $x \prec y$ …

The question is fairly dry: Is there any semigroup analogue of Lagrange's theorem for groups (counting as a generalization of the latter)? Let me guess the answer: Obviously yes. So the real question …

I don't know how interesting my answer can be after a comment by Ben Green, but this would be too long for a comment, and I hope it can be helpful, somehow. Your question is tightly related to the b …

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 $S$: $ …

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 …

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 …

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 …

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 $ …

The title has it all. I'm looking for a proof/disproof of the fact that an algebraically closed field, say $\mathbb K$, has characteristic zero iff the following property (R) holds: For all $n,k \in \ …

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 …

Let $n$ be a positive integer, $p$ a (positive rational) prime, and $\mathbb K$ an algebraically closed field. If ${\rm char}(\mathbb K) = 0$ then ${\rm GL}_n(\mathbb K)$ is divisible (see here). But …

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 $\mathbb A = (A, +_A)$ 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 $\mathfrak A_i$ be groups ($i = 1, 2$), written multiplicatively, and $s$ a non-negative integer (here, as usual, I am abusing notation and denoting the operations of $\mathfrak A_1$ and $\mathfra …