Skip to main content
MathOverflow

Return to Question

replaced http://mathoverflow.net/ with https://mathoverflow.net/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

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 showedshowed 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.

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.

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.

Changed a word in the title
Link
Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64

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

Fixed a couple of issues in the formulation of the question
Source Link
Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64

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 $B$, and $y$ an element of $X$. Is it always possible to embed $\mathbb B$ into a cancellative semigroupmonoid $\mathbb A$ so that $a$ becomesin such a left-subtractiveway 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.

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 $y$ 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, $X$ a non-empty finite subset of $B$, and $y$ an element of $X$. Is it always possible to embed $\mathbb B$ into a cancellative semigroup $\mathbb A$ so that $a$ becomes a left-subtractive element of $X$ 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.

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.

added 10 characters in body
Source Link
Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64
Source Link
Salvo Tringali
Salvo Tringali
  • 10.5k
  • 2
  • 29
  • 64