# Embedding a cancellative monoid into another in such a way that $|X-x|=|X|$, where $X$ is a fixed finite set and $x\in X$

## 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 difference set of the pair $(X, Y)$ in the case when $\mathbb A$ is a group. As is customary in these matters, we write $X-y$ in place of $X -\{y\}$ if $Y = \{y\}$ and there is no likelihood of confusion, and we say that $y$ is a right-subtractive element of $X$ (relative to the semigroup $\mathbb A$, or to the binary operation $+$) if $|X-y| = |X|$.

Given $X \subseteq A$, we denote by $X^{\rm rs}$ the set of right-subtractive elements of $X$. It is seen that $X^{\rm rs}$ contains the set of left-invertible elements of $\mathbb A$ (which is, of course, empty if $\mathbb A$ is not a monoid); in particular, $X^{\rm rs} = X$ if $\mathbb A$ is a group. Moreover, if $\mathbb A$ is cancellative then $|X-y| \le |X|$ for all $y \in A$, and if in addition $X$ is finite then $y$ is a subtractive element of $X$ iff for each $x \in X$ there exists a (provably unique) $a \in A$ such that $a + y = x$.

## Question.

Let $\mathbb B = (B, +_B)$ be a cancellative monoid, and let $X$ be a non-empty finite subset of $B$. Is it always possible to embed $\mathbb B$ into a cancellative monoid $\mathbb A$ in such a way that at least one element of $X$ becomes right-subtractive relative to $\mathbb A$?

## Background.

Yesterday, I asked a question concerning the embeddability of a cancellative monoid into another in such a way that a prescribed element of the former becomes a unit (i.e., invertible) in the latter.

Shortly later, Benjamin Steinberg showed that this isn't possible in general, so proving that a naive attempt I was hoping for to extend the range of applicability of a certain transform introduced by Kemperman in:

On complexes in a semigroup, Indag. Math. 18 (1956) 247-254,

(and now commonly referred to as Kemperman's transform), can't work but for groups.

I've now realized that something that seems very much weaker would suffice, and this is where the question in the above comes from.

• If $X$ is just the identity, then isn't $X$-subtractive the same as left-invertible? – Benjamin Steinberg Mar 14 '16 at 17:47
• Sorry for the delay, I was in a hurry this afternoon and didn't realize that there were at least three things I had to fix in the old formulation. In particular, I would/should have asked if $\mathbb B$ can be embedded into $\mathbb A$ in such a way that at least one $x\in X$ becomes right-subtractive. @BenjaminSteinberg. Yes, and if $\mathbb A$ is a canc. monoid, then the right-subtractive elements of a finite set $X$ that contains the identity of $\mathbb A$ are precisely the left-invertible elements of $\mathbb A$ (hence $X^{\rm rs}=\mathbb A^\times$, from what you made me note yesterday). – Salvo Tringali Mar 14 '16 at 22:18