Monochrome Symmetric Subsets in Colorings of Finite Abelian Groups

Abstract

A subset S of a group G is symmetric if there is an element $g\in G$ such that $g{S}^{-1}g=S$. We study some Ramsey type functions for symmetric subsets in finite Abelian groups.

1. Introduction

Let G be a finite group. Given an element $g\in G$, the symmetry on G with the centre g is the mapping

$${\eta }_g:G\ni x\mapsto g{x}^{-1}g\in G$$

This is an old notion, which can be found in the book [1]. And it is a very natural one, since

$${\eta }_g={\lambda }_g\circ \iota \circ {\lambda }_g^{-1}={\rho }_g\circ \iota \circ {\rho }_g^{-1}$$

where

$${\lambda }_g:G\ni x\mapsto gx\in G,{\rho }_g:G\ni x\mapsto xg\in G,\mathrm{and}\iota :G\ni x\mapsto x^{-1}\in G$$

are the left translation, the right translation, and the inversion, respectively. Indeed, it follows from ${\lambda }_g(x)=gx$ that ${\lambda }_g^{-1}(gx)=x$, so ${\lambda }_g^{-1}(x)=g^{-1}x$. Consequently, ${\lambda }_g^{-1}={\lambda }_{g^{-1}}$. Similarly, ${\rho }_g^{-1}={\rho }_{g^{-1}}$. Then

$$\begin{array}{cc}\hfill {\lambda }_g\circ \iota \circ {\lambda }_g^{-1}(x)& ={\lambda }_g\circ \iota \circ {\lambda }_{g^{-1}}(x)=g{(g^{-1}x)}^{-1}=g{x}^{-1}g\mathrm{and}\hfill \\ \hfill {\rho }_g\circ \iota \circ {\rho }_g^{-1}(x)& ={\rho }_g\circ \iota \circ {\rho }_{g^{-1}}(x)={(x{g}^{-1})}^{-1}g=g{x}^{-1}g\hfill \end{array}$$

A subset $S\subseteq G$ is symmetric if it is invariant with respect to some symmetry on G. Equivalently, S is symmetric if there exists an element $g\in G$ (centre of symmetry) such that $g{S}^{-1}g=S$.

Given $r\in \mathbb{N}$, an r-coloring of G is any mapping $\chi :G\to \{1,\dots ,r\}$.

Definition 1.1

For every finite group and $r\in \mathbb{N}$, define the numbers $s_r(G)$ and ${\sigma }_r(G)$ as follows.

$s_r(G)$ is the greatest number of the form $\frac{k}{\left|G\right|}$, where $k\in \mathbb{N}$ such that for every r-coloring of G there exists a monochrome symmetric subset of cardinality k.

${\sigma }_r(G)$ is the greatest number of the form $\frac{k}{\left|G\right|}$, where $k\in \mathbb{N}$ such that for every r-coloring $\chi$ of G there exists a subset $X\subseteq G$ of cardinality k and element g such that $\chi (x)=\chi (g{x}^{-1}g)$ for all $x\in X$.

It is easy to see that

$$s_r(G)\le {\displaystyle \frac{1}{r}}+{\displaystyle \frac{1}{\left|G\right|}},{\sigma }_r(G)\le 1,s_r(G)\ge {\displaystyle \frac{{\sigma }_r(G)}{r}}$$

For every finite Abelian group G, ${\sigma }_r(G)\ge \frac{1}{r}$, and consequently, $s_r(G)\ge \frac{1}{r^2}$ [2]. In the non-Abelian case this inequality fails [3]. In this note we describe groups with ${\sigma }_r(G)=\frac{1}{r}$, ${\sigma }_r(G)=1$, and $s_2(G)=\frac{1}{4}$. Since the journal [2] is not easy to access and it is in Ukrainian, we give here also a short proof of the inequality from [2].

2. The Inequality

In this section we prove the following theorem.

Theorem 2.1

Let G be a finite group of odd order or any finite Abelian group, and let $r\in \mathbb{N}$. Then

$${\sigma }_r(G)\ge \frac{1}{r}$$

and consequently

$$s_r(G)\ge \frac{1}{r^2}$$

Let G be a finite group. For every r-coloring $\chi :G\to \{1,\dots ,r\}$ and $g\in G$, let

$$S(\chi ,g)=\left|\right\{x\in G:\chi (x)=\chi (g{x}^{-1}g)\left\}\right|$$

and let

$$\sigma (\chi )=\frac{1}{\left|G\right|}\underset{g\in G}{\max }S(\chi ,g)$$

Then

$${\sigma }_r(G)=\underset{\chi :G\to \{1,\dots ,r\}}{\min }\sigma (\chi )$$

For every $a\in G$, let

$$\nu (a)=|\{x\in G:x^2=a\}|$$

Lemma 2.2

For every $\chi :G\to \{1,\dots ,r\}$,

$$\sum _{g\in G}S(\chi ,g)=\sum _{i=1}^r\sum _{(x,y)\in A_i^2}\nu (y{x}^{-1})$$

where $A_i={\chi }^{-1}(i)$.

Proof 

Computing in two ways the number of all triples $(g,x,y)\in G\times G\times G$ such that $g{x}^{-1}g=y$, we obtain

$${\displaystyle \sum _{g\in G}}S(\chi ,g)=\sum _{i=1}^r\sum _{(x,y)\in A_i^2}\left|\right\{g\in G:g{x}^{-1}g=y\left\}\right|$$

It remains to notice that

$$|\{g\in G:g{x}^{-1}g=y\}|=|\{g\in G:g{x}^{-1}g{x}^{-1}=y{x}^{-1}\}|=\nu (y{x}^{-1})$$

Proof of Theorem 2.1

Let $\chi :G\to \{1,\dots ,r\}$ and let $A_i={\chi }^{-1}(i)$. By Lemma 2.2

$$\sum _{g\in G}S(\chi ,g)=\sum _{i=1}^r\sum _{(x,y)\in A_i^2}\nu (y{x}^{-1})$$

If G has odd order, then $\nu (y{x}^{-1})=1$ for any $x,y\in G$. Since the function $x_1^2+\cdots +x_r^2$, where $x_1+\cdots +x_r=C$, attains minimum when $x_1=\cdots =x_r=\frac{C}{r}$,

$$\sum _{g\in G}S(\chi ,g)=\sum _{i=1}^r|A_i{|}^2\ge \underset{r}{\underbrace{{\left(\frac{\left|G\right|}{r}\right)}^2+\cdots +{\left(\frac{\left|G\right|}{r}\right)}^2}}=\frac{{\left|G\right|}^2}{r}$$

If G is Abelian, then $\nu (y{x}^{-1})>0$ if and only if $y{x}^{-1}\in G^2=\{g^2:g\in G\}$ and in this case $\nu (y{x}^{-1})=[G:G^2]$. Let $C_j$ $(1\le j\le k)$ be cosets of G modulo $G^2$, $C_{j,i}=C_j\bigcap A_i$. Then

$$\sum _{g\in G}S(\chi ,g)=\sum _{i=1}^r\sum _{j=1}^k|C_{j,i}{|}^2·k\ge rk{\left(\frac{\left|G\right|}{rk}\right)}^2·k=\frac{{\left|G\right|}^2}{r}$$

Therefore, in each case, there exists an element $g\in G$ such that $S(\chi ,g)\ge \frac{\left|G\right|}{r}$ and so $\sigma (\chi )\ge \frac{1}{r}$.

3. Finite Abelian Groups with ${\sigma }_r(G)=\frac{1}{r}$ and ${\sigma }_r(G)=1$

In this section we describe finite Abelian groups with ${\sigma }_r(G)=\frac{1}{r}$ and ${\sigma }_r(G)=1$.

Theorem 3.1

${\sigma }_r(G)=\frac{1}{r}$ if and only if r divides $\left|2G\right|$.

Proof 

Define the subgroups $2G=\{2x:x\in G\}$ and $B(G)=\{x\in G:2x=0\}$. Denote $\left|2G\right|=m$ and $|B(G)|=k$. Obviously, $\left|G\right|=mk$.

Consider first the case when r does not divide m. Fix any r-coloring $\chi$ of a group G. Let $C_j$ $(1\le j\le k)$ be cosets of G modulo $2G$, $C_{j,i}=C_j\bigcap {\chi }^{-1}(i)$. Then

$$\sum _{i=1}^r|C_{j,i}{|}^2>r{\left(\frac{m}{r}\right)}^2=\frac{m^2}{r}$$

Hence,

$$\sum _{g\in G}S(\chi ,g)=k\sum _{j=1}^k\sum _{i=1}^r|C_{j,i}{|}^2>k^2\frac{m^2}{r}=\frac{{\left|G\right|}^2}{r}$$

Therefore, there exists an element $g\in G$ such that $S(\chi ,g)>\frac{\left|G\right|}{r}$ and so $\sigma (\chi )>\frac{1}{r}$.

Now consider the case where r divides m. By Theorem 2.1, ${\sigma }_r(G)\ge \frac{1}{r}$, so it suffices to construct a coloring $\chi$ with $\sigma (\chi )=\frac{1}{r}$. Pick subgroup H of a group G such that $B(G)\subseteq H$ and $[G:H]=r$. Then $[2G:2H]=r$. Define r-coloring $\chi$ of G as follows:

(1) every coset of G modulo $2H$ is monochrome;

(2) every r cosets of G modulo $2H$ which form a coset of G modulo $2G$ are colored in r different colors.

Then

$$\begin{array}{cc}\hfill \chi (x)=\chi (2g-x)& \iff x-(2g-x)\in 2H\hfill \\ & \iff 2(x-g)\in 2H\hfill \\ & \iff \exists h\in H:2(x-g-h)=0\hfill \\ & \iff \exists h\in H:x-g-h\in B(G)\hfill \\ & \iff x-g\in H+B(G)=H\hfill \\ & \iff x\in g+H.\hfill \end{array}$$

So $S(\chi ,g)=\left|H\right|$ for every $g\in G$. Therefore $\sigma (\chi )=\frac{\left|H\right|}{\left|G\right|}=\frac{1}{[G:H]}=\frac{1}{r}$.

Theorem 3.2

${\sigma }_r(G)=1$ if and only if one of the following cases holds:

(1)

$r=1$;

(2)

$r=2$ and G is a cyclic group of order either 3 or 5;

(3)

G is a Boolean group.

Proof 

Sufficiency is obvious. We need to prove Necessity. Assume on the contrary that neither of cases (1)–(3) holds.

Suppose first that $\left|G\right|$ is even. Then both subgroups $2G$ and the elementary Abelian 2-group $B(G)$ are different from G. Pick $a,b\in G$ such that $a+b\notin 2G$ and $a-b\notin B(G)$. Define $\chi :G\to \{1,2\}$ by

$$\chi (x)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}x\in \{a,b\}\hfill \\ 2\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

Let $g\in G$. Since $a+b\notin 2G$, $2g-a\ne b$. If $2g-a=a$, then $2g-b\ne b$, because $a-b\notin B(G)$. It follows that either $\chi (a)\ne \chi (2g-a)$ or $\chi (b)\ne \chi (2g-b)$, a contradiction.

Now suppose that $\left|G\right|$ is odd. Then $2G=G$. Since $\left|G\right|\ge 7$, we can choose distinct $a,b,c\in G$ such that for any distinct $g,x\in \{a,b,c\}$, $2g-x\notin \{a,b,c\}$.

To see this, pick any distinct $a,b\in G$. There is a unique $g\in G$ such that $b=2g-a$. We then pick $c\in G\setminus \{a,b,g,2a-b,2b-a\}$.

Define $\chi :G\to \{1,2\}$ by

$$\chi (x)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}x\in \{a,b,c\}\hfill \\ 2\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

Let $g\in G$. If $g\notin \{a,b,c\}$ and $2g-a=b$, then $2g-c\notin \{a,b,c\}$. If $g\in \{a,b,c\}$, say $g=a$, then $2g-b\notin \{a,b,c\}$. It follows that there is $x\in \{a,b,c\}$ such that $\chi (x)\ne \chi (2g-x)$, again a contradiction.

Remark 

Theorem 3.2 describes finite Abelian groups where each r-coloring is symmetric. A coloring $\chi$ of G is symmetric if there exists $g\in G$ such that

$$\chi (g{x}^{-1}g)=\chi (x)\mathrm{for}\mathrm{all}x\in G$$

Obviously, the number of all r-colorings of a group G of order n equals $r^n$. To find the number of all symmetric r-colorings of a group G is a quite complicated exercise involving Möbious inversion on the lattice of subgroups. Precise formula for the number of all symmetric r-colorings of finite Abelian group was established in [4], and the corresponding formula for every finite group has been found only recently [5] (for the quaternion group, see [6]).

4. Finite Abelian Groups with $s_2(G)=\frac{1}{4}$

In this section we prove the following.

Theorem 4.1

Let G be a finite Abelian group and $n\in \mathbb{N}$.

(1)

If G contains subgroup ${\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4$, then $s_{2^n}(G)=\frac{1}{4^n}$;

(2)

If G does not contain subgroup ${\mathbb{Z}}_4$, then $s_2(G)>\frac{1}{4}$.

We first prove some auxiliary statements.

Lemma 4.2

$s_{2^n}({\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4)=\frac{1}{4^n}$.

Proof 

Define the coloring $\chi :{\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4\to {\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_2$ by

$${(\chi (x))}_i=\left\{\begin{array}{cc}0\hfill & \mathrm{if}{(x)}_i\in \{0,1\}\hfill \\ 1\hfill & \mathrm{if}{(x)}_i\in \{2,3\}\hfill \end{array}\right.$$

Fix $g\in {\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4$. If $\chi (x)=\chi (2g-x)$, then ${(\chi (x))}_i={(\chi (2g-x))}_i$. It remains to notice that $s_2({\mathbb{Z}}_4)=\frac{1}{4}$.

For every group G and a coloring $\chi$, let $s(\chi )$ denote the cardinality of the largest monochrome symmetric subset of G divided by $\left|G\right|$.

Lemma 4.3

Let G be a finite group, let $f:G\to H$ be a surjective homomorphism and let $\chi$ be a coloring of H. Define coloring $\phi$ of G by $\phi =\chi \circ f$. Then $s(\phi )=s(\chi )$.

Proof 

Let S be a monochrome subset of G symmetric with respect to $g\in G$. By definition of $\phi$ it follows that $\phi (x)=\phi (g{x}^{-1}g)$ if and only if $\chi (f(x))=\chi (f(g)f{(x)}^{-1}f(g))$. So, $f(S)$ is a monochrome subset of H symmetric with respect to $f(g)$. Since $\left|S\right|\le \left|\ker f\right|·|f(S)|$,

$${\displaystyle \frac{\left|S\right|}{\left|G\right|}}\le {\displaystyle \frac{\left|\ker f\right|·|f(S)|}{\left|G\right|}}={\displaystyle \frac{f(S)}{\left|H\right|}}$$

Thus $s(\phi )\le s(\chi )$.

Conversely, let S be a monochrome subset of H symmetric with respect to $h\in H$. Then $f^{-1}(S)$ is a monochrome subset of G symmetric with respect to any $g\in f^{-1}(h)$. Since $|f^{-1}(S)|=|\ker f|·|f(S)|$,

$${\displaystyle \frac{|f^{-1}(S)|}{\left|G\right|}}={\displaystyle \frac{\left|S\right|}{\left|H\right|}}$$

Thus $s(\phi )\ge s(\chi )$.

Corollary 4.4

Let G be a finite group and let H be a homomorphic image of G. Then $s_r(G)\le s_r(H)$.

Proof of Theorem 4.1

(1) By Theorem 2.1, we have that $s_{2^n}(G)\ge \frac{1}{4^n}$. If G contains subgroup ${\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4$, then there exists a homomorphism from G onto ${\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4$. Thus, by Corollary 4.4, $s_{2^n}(G)\le s_{2^n}({\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4)$. By Lemma 4.2, $s_{2^n}({\displaystyle \underset{n}{⨁}}{\mathbb{Z}}_4)=\frac{1}{4^n}$. Thus $s_{2^n}(G)=\frac{1}{4^n}$.

(2) Suppose that G does not contain ${\mathbb{Z}}_4$. Then $G=H\times B$ for some subgroup H of odd order and Boolean group B (which can be trivial). Let $\chi$ be an arbitrary 2-coloring of G. For every $b\in B$ define the coloring ${\chi }_b$ on H by ${\chi }_b(x)=\chi (x,b)$. Then

$$\sum _{h\in H}S(\chi ,h)=\sum _{h\in H}\sum _{b\in B}S({\chi }_b,h)=\sum _{b\in B}\sum _{h\in H}S({\chi }_b,h)$$

Since H has odd order, we obtain that

$$\sum _{h\in H}S({\chi }_b,h)>\frac{{\left|H\right|}^2}{2}$$

Then

$$\sum _{h\in H}S(\chi ,h)=\sum _{h\in H}\sum _{b\in B}S({\chi }_b,h)>\left|B\right|·\frac{{\left|H\right|}^2}{2}$$

It follows that there exists $h\in H$ such that

$$S(\chi ,h)>{\displaystyle \frac{\left|B\right|·\left|H\right|}{2}}={\displaystyle \frac{\left|G\right|}{2}}$$

Consequently,

$$\sigma (\chi )>\frac{1}{2}$$

and hence

$$s_2(G)>\frac{1}{4}$$

Theorem 4.1 implies the following two criteria.

Corollary 4.5

For every finite Abelian group G, $s_2(G)=\frac{1}{4}$ if and only if G contains subgroup ${\mathbb{Z}}_4$.

Corollary 4.6

$s_2({\mathbb{Z}}_n)=\frac{1}{4}$ if and only if $4|n$.

Below we give the corresponding coloring.

We conclude the paper with the following table (

Table 1. Ramsey functions $s_2({\mathbb{Z}}_n)$ and ${\sigma }_2({\mathbb{Z}}_n)$ for $n\le 8$.

n	k	$s_2({\mathbb{Z}}_n)=\frac{k}{n}$	m	${\sigma }_2({\mathbb{Z}}_n)=\frac{m}{n}$	χ
1	1	1	1	1	●
2	1	0.5	2	1
3	2	0.66666666	3	1
4	1	0.25	2	0.5
5	3	0.6	5	1
6	2	0.33333333	4	0.66666666
7	3	0.42857142	5	0.71428571
8	2	0.25	4	0.5

).