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Questions about the branch of algebra that deals with groups.

12 votes
1 answer
466 views

Bi-orderability of Baumslag-Solitar group $\langle a,b \mid a^{-1} b^m a = b^n\rangle$ and o...

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 …
Salvo Tringali's user avatar
8 votes
1 answer
226 views

Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topol...

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 …
Salvo Tringali's user avatar
8 votes
2 answers
541 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

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 …
Salvo Tringali's user avatar
8 votes
1 answer
314 views

Does every cancellative duo semigroup embed into a group?

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 …
Salvo Tringali's user avatar
7 votes
2 answers
538 views

A linearly orderable monoid which does not embed into a linearly orderable group

It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable sem …
Salvo Tringali's user avatar
6 votes
1 answer
371 views

Embedding a cancellative monoid into another in such a way that a prescribed element becomes...

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 …
Salvo Tringali's user avatar
5 votes
0 answers
79 views

Some questions about the Lévy monoid of certain densities

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: …
Salvo Tringali's user avatar
4 votes
1 answer
441 views

What is a "cusp" ("кусок") in relation to Guba's embedding theorem?

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 …
Salvo Tringali's user avatar
4 votes
2 answers
383 views

Embedding a linearly ordered free monoid into a linearly ordered group

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 $xy < xz$ …
Salvo Tringali's user avatar
4 votes
1 answer
254 views

On the divisibility of the special linear group of degree $n$ over an algebraically closed f...

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 …
Salvo Tringali's user avatar
4 votes
2 answers
1k views

Every abelian torsion-free group is strictly totally orderable (via the compactness theorem)

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 …
Salvo Tringali's user avatar
3 votes

Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$

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 …
Salvo Tringali's user avatar
2 votes

What's a non-abelian totally ordered group?

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 $ …
Salvo Tringali's user avatar
2 votes
1 answer
218 views

A categorical framework for Freiman s-morphisms

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 …
Salvo Tringali's user avatar
2 votes
1 answer
269 views

Apropos of two groups being globally isomorphic iff they are isomorphic

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 $S$: $ …
Salvo Tringali's user avatar

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