Ramsey-product subsets of a group

TL;DR

This paper introduces the concept of Ramsey-product subsets in infinite groups, showing they form a filter and relate to a specific subsemigroup of ultrafilters, advancing the understanding of algebraic and combinatorial properties of groups.

Contribution

It defines Ramsey-product subsets in groups, proves they form a filter, and links them to a subsemigroup of ultrafilters, providing new insights into group combinatorics.

Findings

The family of all Ramsey-product subsets forms a filter.

Ramsey-product subsets define a subsemigroup of ultrafilters.

The structure of these subsets relates to algebraic properties of the group.

Abstract

We say that a subset $S$ of an infinite group $G$ is a Ramsey-product subset if, for any infinite subsets $X$, $Y$ of $G$, there exist $x \in X$ and $y\in Y$ such that $x y \in S$ and $ y x \in S$ . We show that the family $\varphi$ of all Ramsey-product subsets of $G$ is a filter and $\varphi$ defines the subsemigroup $ \overline{G^*G^*}$ of the semigroup $G^*$ of all free ultrafilters on $G$.

Full text

All groups under consideration are supposed to be infinite; a countable set means a countably infinite set.

We say that a subset $S$ of a group $G$ is

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a *Ramsey-product subset * if, for any infinite subsets $X,Y$ of $G$, there exist $x\in X$ and $y\in Y$ such that $xy\in S$ and $yx\in S$;

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a *Ramsey-square subset * if, for every infinite subset $X$ of $G$ there exist distinct elements $x,y\in X$ such that $xy\in S$, $yx\in S$;

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a *Ramsey-quotient subset * if, for every infinite subset $X$ of $G$, there exist distinct elements $x,y\in X$ such that $x^{-1}y\in S$ and $y^{-1}x\in S$.

We show that above defined subsets of $G$ arise naturally in context of the Stone-$\check{C}$ech compactification $\beta G$ of the group $G$ endowed with the discrete topology. We identify $\beta G$ with the set of all ultrafilters on $G$. Then the family $\{\bar{A}:A\subseteq G\}$, where $\bar{A}=\{p\in\beta G:A\in p\}$, forms a base for the topology of $\beta G$. Given a filter $\varphi$ on $G$, we denote $\bar{\varphi}=\cap\{\bar{A}:A\in\varphi\}$, so $\varphi$ defines the closed subset $\bar{\varphi}$ of $\beta G$, and every non-empty closed subset of $\beta G$ can be defined in this way .

We use the standard extension [3, Section 4.1] of the multiplication on $G$ to the semigroup multiplication on $\beta G$. Given two ultrafilters $p,q\in\beta G$, we choose $P\in p$ and, for each $x\in P$, pick $Q_{x}\in q$. Then $\bigcup_{x\in P}xQ_{x}\in pq$ and the family of these subsets forms a base of the product $pq$. We note that the set $G^{*}$ of all free ultrafilters of $G$ is a closed subsemigroup of $\beta G$, and the closure $\overline{G^{*}G^{*}}$ of $G^{*}G^{*}$ is a subsemigroup of $G^{*}$. An ultrafilter $p\in G^{*}\setminus\overline{G^{*}G^{*}}$ is called strongly prime.

By [5, Propositions 3, 4], the family $\psi$ of all Ramsey-quotient subsets of a group $G$ is a filter and $\overline{\psi}$ is the smallest closed subset of $G^{*}$ containing all ultrafilters of the form $q^{-1}q$, $q\in G^{*}$, where $q^{-1}=\{Q^{-1}:Q\in q\}$. Analogously, the closure of $\{pp:p\in G^{*}\}$ is defined by the filter of all Ramsey-square subsets of $G$. In both cases, we used the classical Ramsey theorem [2, p. 16].

In this note, the key technical part plays the following version of Ramsey theorem (Theorem 6 from [2, p.98]).

Let $\chi$ be a finite coloring $\chi:\omega\times\omega\rightarrow[r]$. Then there exists an infinite set $A=\{a_{i}\}_{<\omega}$ and colors $c_{L}$, $c_{G}$, $c_{E}$ (not necessarily distinct) such that $\chi(a_{i},a_{j})=c_{L}$ if $i<j$ , $\chi(a_{i},a_{j})=c_{G}$ if $i>j$ and $\chi(a_{i},a_{j})=c_{E}$ if $i=j$.

Lemma 1. If $S$ is a Ramsey-product subsets of a group $G$ then, for any countable subsets $X,Y$ of $G$, there exist disjoint countable subsets $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$ such that $X^{\prime}Y^{\prime}\subseteq S$, $Y^{\prime}X^{\prime}\subseteq S$.

Proof. Passing to subsets of $X$ and $Y$, we may suppose that $X\cap Y=\emptyset$. We enumerate $X=\{x_{n}:n<\omega\}$, $Y=\{y_{n}:n<\omega\}$ and define a $\{0,1\}$-coloring $\chi$ of $\omega\times\omega$ by the rule: $\chi((n,m))=\chi((m,n))=1$ if $m>n$, $x_{n}y_{m}\in S$, $y_{m}x_{n}\in S$, and $\chi((m,n))=0$ otherwise. By Theorem 6 from [2, p.98], there exists a countable subset $A$ of $\omega$ such that $\chi$ is monochrome on all $(n,m)$ , such that $n\in A$, $m\in A$, $n\neq m$. We partition $A$ into two infinite subsets $A=B\cup C$ and denote $X^{\prime}=\{x_{n}:n\in B\}$, $Y^{\prime}=\{y_{n}:n\in C\}$. Since $S$ is a Ramsey-product set, there are $x^{\prime}\in X^{\prime}$, $y^{\prime}\in Y^{\prime}$ such that $x^{\prime}y^{\prime}\in S$, $y^{\prime}x^{\prime}\in S$. By the choice of $A$, we have $xy\in S$, $yx\in S$ for all $x\in X^{\prime}$, $y\in Y^{\prime}$. $\ \ \ \ \Box$

Lemma 2. For a free ultrafilter $r$ on a group $G$, $r\in\overline{G^{*}G^{*}}$ if and only if, for every subset $R\in r$, there exist two countable subsets $\{x_{n}:n<\omega\}$ and $\{y_{n}:n<\omega\}$ of $G$ such that $\{x_{n}y_{m}:n\leq m<\omega\}\subseteq R.$

Proof. We assume that $r\in\overline{G^{*}G^{*}}$ , take an arbitrary $R\in r$ and choose $p,q\in G*$ such that $R\in pq$. We choose $P\in p$ and $Q_{x}$, $x\in P$ such that $\bigcup_{x\in P}xQ_{x}\subseteq R$. We take an arbitrary countable subset $\{x_{n}:n<\omega\}$ of $P$ and choose an injective sequence $(y_{n})_{n\in\omega}$ in $G$ such that $y_{n}\in Q_{x_{0}}\cap\ldots\cap Q_{x_{n}}$, $n<\omega$. Then $\{x_{n}y_{m}:n\leq m<\omega\}\subseteq R$.

On the other hand, let $R\in r$ and $\{x_{n}:n\in\omega\}$, $\{y_{n}:n\in\omega\}$ are chosen so that $\{x_{n}y_{m}:n\leq m<\omega\}\subseteq R$. We take an arbitrary free ultrafilters $p,q$ such that $\{x_{n}:n\in\omega\}\in P$, $\{y_{n}:n\in\omega\}\in q$. Then $\{x_{n}y_{m}:n\leq m<\omega\}\in pq$, so $R\in pq$ and $r\in\overline{G^{*}G^{*}}$. $\ \ \ \ \Box$

We recall [1] that a subset $A$ of a group $G$ is sparse if, for every infinite subset $X$ of $G$, there exists a non-empty finite subset $F\subset X$ such that $\bigcap_{x\in F}xA$ is finite. By [1, Theorem 9], an ultrafilter $r\in G^{*}$ is strongly prime if and only if there exists a sparse subset $A$ of $G$ such that $A\in r$. For sparse subsets see also [4], [6].

Teorem. For every group $G$, the following statements hold:

$(i)$* the family $\varphi$ of all Ramsey-product subsets of $G$ is a filter;*

$(ii)$* $\overline{\varphi}=\overline{G^{*}G^{*}}$;*

$(iii)$* $A\in\varphi$ if and only if $G\setminus A$ is sparse.*

Proof. $(i)$ We take two Ramsey-product subsets $S$ and $T$ of $G$ and show that $S\cap T\in\varphi.$ Let $X,Y$ be infinite subsets of $G$. By Lemma 1, there exist disjoint countable subsets $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$ such that $xy\in S$, $yx\in S$ for all $x\in X^{\prime}$, $y\in Y^{\prime}$. Since $T$ is a Ramsey-product subsets, there exist $x^{\prime}\in X^{\prime}$, $y^{\prime}\in Y^{\prime}$ such that $x^{\prime}y^{\prime}\in T$, $y^{\prime}x^{\prime}\in T$. Then $x^{\prime}y^{\prime}\in S\cap T$, $y^{\prime}x^{\prime}\in S\cap T$, so $S\cap T\in\varphi$.

$(ii)$ To show that $\overline{G^{*}G^{*}}\subseteq\overline{\varphi}$, we suppose the contrary, choose $p\in\overline{G^{*}G^{*}}\setminus\overline{\varphi}$ and take $P\in p$ such that $G\setminus P\in\varphi$. By Lemma 2, there are countable sets $\{x_{n}:n<\omega\}$ and $\{y_{n}:n<\omega\}$ such that $\{x_{n}y_{m}:n\leq m<\omega\}\subseteq P$. We use Lemma 1 to choose disjoint countable subsets $X^{\prime}\subseteq\{x_{n}:n<\omega\}$, $Y^{\prime}\subseteq\{y_{n}:n<\omega\}$ such that $X^{\prime}Y^{\prime}\subseteq G\setminus P$, $\ \ Y^{\prime}X^{\prime}\subseteq G\setminus P$. To get a contradiction, we take an arbitrary $x_{n}\in X^{\prime}$ and pick $y_{m}\in Y^{\prime}$ such that $m>n$.

Now we take an arbitrary $p\in\overline{\varphi}$ and prove that $p\in\overline{G^{*}G^{*}}$. Given any $P\in p$, by Lemma 1, there exist two disjoint countable subsets $X,Y$ of $G$ such that, for any countable subsets $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq X$, either $X^{\prime}Y^{\prime}\cap P\neq\emptyset$ or $Y^{\prime}X^{\prime}\cap P\neq\emptyset$ . We enumerate $X=\{x_{n}:n<\omega\}$, $Y=\{y_{n}:n<\omega\}$ and define a $\{0,1\}$-coloring $\chi$ of $\omega\times\omega$ by the rule: $\chi(n,m)=1$ if and only if $n<m$ and $x_{n}y_{m}\in P$ or $n>m$ and $y_{m}x_{n}\in P$. Applying Theorem 6 from [2, p. 98], we get an infinite subset $W$ of $\omega$ such that the restriction of $\chi$ to $W_{1}=\{(n,m):n<m,\ \ n\in W,\ \ m\in W\}$ is monochrome, and the restriction of $\chi$ to $W_{2}=\{(n,m):n>m,\ \ n\in W,\ \ m\in W\}$ is monochrome. By the choice of $X$ and $Y$, either $\chi|_{W_{1}}\equiv 1$ or $\chi|_{W_{2}}\equiv 1$. In the first case, we choose two injective sequences $(x_{n}^{\prime})_{n<\omega}$ in $X$ and $(y_{n}^{\prime})_{n<\omega}$ in $Y$ such that $\{x_{n}^{\prime}y_{m}^{\prime}:n<m<\omega\}\subseteq P$ . In the second case, we choose two injective sequences $(x_{n}^{\prime})_{n<\omega}$ in $X$ and $(y_{n}^{\prime})_{n<\omega}$ in $Y$ such that $\{y_{n}^{\prime}x_{m}^{\prime}:n<m<\omega\}\subseteq P$ . Applying Lemma 2, we conclude that $p\in\overline{G^{*}G^{*}}$.

$(iii)$ We apply $(ii)$ and Theorem 9 from [1]. $\ \ \ \ \Box$

We recall that a subset $S$ of a group $G$ is *large (extralarge) * if there is a finite subset $K$ of $G$ such that $G=KS$ ($S\cap L$ is large for every large subset $L$ of $G$). By Theorem 4.1 from [4] , the complement of sparse subset is extralarge. Applying $(iii)$, we see that each Ramsey-product subset is extralarge.

By the statement $(ii)$, every Ramsey-product subset of $G$ is a member of every idempotent of the semigroup $G^{*}$, in particular, every idempotent from the minimal ideal of $\beta G$, so $S$ is combinatorially very rich (see [3, Section 14]).

If $G$ is an amenable group and $A$ is a sparse subset of $G$, by Theorem 5.1 from [4], $\mu(A)=0$ for every left invariant Banach measure $\mu$ on $G$. Applying (iii), we see that $\mu(S)=1$ for every Ramsey-product subset $S$ of $G$ and every left invariant Banach measure $\mu$ on $G$.

We conclude the note with the following formal generalization. We fix two infinite subset $X,Y$ of a group $G$ and say that a subset $S$ of $G$ is a *Ramsey $(X,Y)$-product subset * if, for any infinite subsets $X^{\prime}\subseteq X$ and $Y^{\prime}\subseteq Y$, there exist $x\in X^{\prime}$ and $y\in Y^{\prime}$ such that $xy\in S$, $yx\in S$. Then the family $\varphi_{X,Y}$ of all Ramsey $(X,Y)$-product subsets of $G$ is a filter and $\overline{\varphi}_{X,Y}=\overline{X^{*}Y^{*}}\ \cup\ \overline{Y^{*}X^{*}}$.