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It appears that my former officemate Andreas Reinhart (University of Graz, Austria) has disproven the Ankeny–Artin–Chowla conjecture: more precisely, Andreas has found that $$ d := 331914313984493$$ i …

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, …

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 …

Sorry for answering my own question, but YCor's construction in a related thread (here) gave me a lightbulb moment. Hopefully, it's not a broken lightbulb. The answer to the question asked in the OP …

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 …

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ta …

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, …

I shared the link to this thread with several colleagues, inviting them to contribute to the discussion. Notably, I received a response from Melvyn Nathanson himself, most of which is reproduced below …

Given $a, b \in \mathbb Z$, set $[\![a,b]\!] := \{x \in \mathbb Z: a \le x \le b\}$. A basic result in additive number theory goes as follows: If $A$ is a finite subset of $\mathbb N$ with $0 \in A$ …

Let $S$ be a semigroup. An ideal (of $S$) is a subset $I$ of $S$ such that $SI$ and $IS$ are both contained in $I$. The non-empty ideals constitute a subsemigroup, $\mathfrak I(S)$, of the power semig …

Let $S$ be a semigroup and $\widehat{S}$ be its unitization, i.e., the monoid obtained from $S$ by adjoining an identity element if necessary (so that $\widehat{S} = S$ when $S$ is already a monoid). …

EDIT. I've just noticed a thread from 2011 in the "Related" column on the right (click me), where a closely related question is being discussed (the main difference seems to be that, in Question 2 bel …

Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (m …

An ideal of a semigroup $S$ (written multiplicatively) is a set $I \subseteq S$ such that $IS$ and $SI$ are both contained in $I$ (here, $XY$ means, for all $X, Y \subseteq S$, the setwise product of …

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