On Some Subgroup Chains Related to Kneser's Theorem

TL;DR

This paper explores subgroup chains related to Kneser's Theorem, extending results from abelian to nonabelian groups using additive number theory and poset properties.

Contribution

It introduces an analogous poset for nonabelian groups and extends key results of Balandraud to this broader context.

Findings

Short proof of Balandraud's results in abelian groups

Extension of subgroup chain properties to nonabelian groups

Use of classical additive number theory tools

Abstract

A recent result of Balandraud shows that for every subset S of an abelian group G, there exists a non trivial subgroup H such that |TS| <= |T|+|S|-2 holds only if the stabilizer of TS contains H. Notice that Kneser's Theorem says only that the stabilizer of TS must be a non-zero subgroup. This strong form of Kneser's theorem follows from some nice properties of a certain poset investigated by Balandraud. We consider an analogous poset for nonabelian groups and, by using classical tools from Additive Number Theory, extend some of the above results. In particular we obtain short proofs of Balandraud's results in the abelian case.

Equations

$$|XY|=|HX|+|HY|-|H|,$$

$$|X\setminus Y|+|Y\setminus X|\ge |H|+|K|-2|H\cap K|.$$

$$u=|MS|-|M|=|(M\cap N)S|-|M\cap N|.$$

$$|NS|-|N|=|(N\cap M)S|-|N\cap M|.$$

$$u=|XS|-|HX|=|HS|-|H|,$$

Full text

This strong form of Kneser’s theorem follows from some nice properties of a certain poset investigated by Balandraud. We consider an analogous poset for nonabelian groups and, by using classical tools from Additive Number Theory, extend some of the above results. In particular we obtain short proofs of Balandraud’s results in the abelian case.

1 Introduction

In order to avoid switching from multiplicative to additive notation, all groups will be written multiplicatively.

Kneser’s addition theorem states that if $S,T$ are finite subsets of an abelian group $G$ then $|ST|\leq|S|+|T|-2$ holds only if $ST$ is periodic (i.e, there is a non trivial subgroup $H$ such that $HST=ST$.) Kneser’s Theorem is a fundamental tool in Additive number Theory. Proofs of this result may be found in [4, 5, 6, 7, 9].

In all previously known proofs of Kneser’s Theorem, the subgroup $H$ depends crucially on both sets $S$ and $T$. With the goal of breaking this double dependence in $S$ and $T$, Balandraud investigated in recent work [1, 2] the properties of a combinatorial poset that we now present.

Let $S$ be a finite subset containing $1$ of a group $G$. Following Balandraud, let us define a cell of $S$ as a finite subset $X$ such that, for all $z\notin X$, it holds that $zS\not\subset XS$. This notion is defined in [1, 2] and it is equivalent to the notion of nonextendible subset used in [3]. Throughout the paper, by a cell we always mean a cell of $S$.

A cell $X$ is called a $u$-cell if $|XS|-|X|=u$. A $u$-cell with minimal cardinality is called a $u$-kernel (of $S$).

Balandraud showed that, for a finite set $S$ in an abelian group $G$, in the poset of $j$–cells containing the unity ordered by inclusion with $1\leq j\leq|S|-2$, the set of kernels form a chain of subgroups. Moreover, if there exists a $u$–cell, then there is a unique $u$–kernel containing the unit element which is contained in all $u$–cells containing the unit element.

One of the consequences of this work is a new proof and the following strengthening of Kneser’s Theorem:

Theorem 1** (Balandraud)**

For any non-empty finite subset $S$ of an abelian group $G$, there exists a finite subgroup $H$ of $G$ such that for any finite subset $T$ of $G$ one of the following conditions hold :

• •

$|TS|\geq|T|+|S|-1$**

• •

$HTS=TS$* and $|TS|\leq|HS|+|HT|-|H|$*

As far as the authors are aware this is a surprising and strong formulation that was not observed before and does not follow straightforwardly from the classical forms of Kneser’s Theorem.

The purpose of the present note is to give a short proof for the nonabelian case that, in the poset of $j$–cells that are subgroups ordered by inclusion with $0\leq j\leq|S|-1$, the set of kernels form a chain of subgroups. Moreover, each $u$-kernel of this poset is unique and contained in all $u$–cells of this poset.

From this statement Kneser’s theorem allows one to deduce Balandraud’s results for the abelian case, and in particular Theorem 1. Kneser’s Theorem has several equivalent forms. We use the following one; see e.g [4, 7]:

Theorem 2** (Kneser [5])**

Let $G$ be an abelian group and $X,Y\subset G$ be finite subsets such that $|XY|\leq|X|+|Y|-2$. Then

[TABLE]

where $H=stab(XY)=\{x:xXY=XY\}.$

Our main tool is the following Theorem of Olson[8, Theorem 2]. We give an equivalent formulation here where we use left–cosets instead of right–cosets.

Theorem 3** (Olson [8])**

Let $X,Y$ be finite subsets of a group $G$, and let $H$ and $K$ be subgroups such that $HX=X$, $KY=Y$ and $KX\neq X$, $HY\neq Y$. Then

[TABLE]

In particular either $|X\setminus Y|\geq|H|-|H\cap K|$ or $|Y\setminus X|\geq|K|-|H\cap K|$.

We shall use the following lemma.

Lemma 4** ([1, 2])**

Let $G$ be a group and $1\in S\subset G$ be a finite subset. Then the intersection of two cells $M_{1},M_{2}$ of $S$ is a cell of $S$.

*Proof.  *  Let $x\notin M_{1}\cap M_{2}$. There is $i$ with $x\notin M_{i}$. Then $xS\not\subset M_{i}S.$ Hence $xS\not\subset(M_{1}\cap M_{2})S$.

We can now state our main result, namely Theorem 5 below.

2 An application of Olson’s Theorem

Balandraud [1, 2] proved that, in the abelian case, the set of kernels containing the unit element and ordered by inclusion is a chain of subgroups. In the non abelian case we can prove only that the set of kernels that are subgroups forms a chain. The abelian case can then be easily recovered, since Kneser’s Theorem implies (as we shall see below) that a kernel containing the unit element is a subgroup.

Theorem 5

Let $S$ be a finite subset containing $1$ of a group $G$. Let $M$ be a $u$–kernel of $S$ which is a subgroup. Let $N$ be a subgroup which is a $v$–cell and suppose $u,v\leq|S|-1$.

• (i)

If either $N$ is a $v$–kernel or $u=v$ then $M\subset N$ or $N\subset M$.

• (ii)

If $N$ is a $v$–kernel and $v\leq u$ then $M\subset N$.

*Proof.  *  Suppose that $M\not\subset N$ and $N\not\subset M$. Note that, since $M$ is a cell, if $NMS=MS$ then $NM=M$, thus $N\subset M$ against our assumption. Hence we may assume $NMS\neq MS$ and similarly $MNS\neq NS$. By Theorem 3 we have one of the two following cases.

**Case ** 1: $|MS|-|(MS)\cap(NS)|=|(MS)\setminus(NS)|\geq|M|-|M\cap N|.$ It follows that $|(M\cap N)S|-|M\cap N|\leq|(MS)\cap(NS)|-|M\cap N|\leq|MS|-|M|.$ On the other hand we have $u=|MS|-|M|<|S|\leq|(M\cap N)S|.$ Since $|MS|-|M|$ is a multiple of $|M\cap N|$ we have

[TABLE]

By Lemma 4, $M\cap N$ is a cell. Since $M$ is a $u$–kernel, we have $M\cap N=M,$ a contradiction.

**Case ** 2: $|NS|-|(NS)\cap(MS)|=|(NS)\setminus(MS)|\geq|N|-|N\cap M|.$ It follows that $|(N\cap M)S|-|N\cap M|\leq|(NS)\cap(MS)|-|N\cap M|\leq|NS|-|N|.$ On the other hand we have $|NS|-|N|<|S|\leq|(N\cap M)S|.$ Since $|NS|-|N|$ is a multiple of $|N\cap M|$ we have

[TABLE]

Assume first $u=v$. Then $u=|MS|-|M|=|NS|-|N|=|(N\cap M)S|-|N\cap M|$. Since $M$ is a $u$–kernel, we have $M\cap N=M,$ a contradiction.

Assume that $N$ is a $v$–kernel. Then (1) implies $N\cap M=N,$ a contradiction. This proves $(i)$.

Assume now that $v\leq u$. Suppose $M\not\subset N$. By $(i)$ we have $N\subset M$, which implies in particular that $|MS|-|M|$ is a multiple of $N$. Therefore, from $u=|MS|-|M|<|S|\leq|NS|$ we have $u=|MS|-|M|\leq|NS|-|N|=v$ which gives $u=v$. But then $M\not\subset N$ and $N\subset M$ imply $|N|<|M|$, and since $N$ is now a $u$–cell, this contradicts $M$ being a $u$–kernel.

We can now deduce Balandraud’s description for kernels and cells :

Corollary 6** (Balandraud [1, 2])**

Let $G$ be an abelian group and $S\subset G$ be a finite subset with $1\in S$. Let $M$ be a $u$–kernel of $S$ containing $1$ with $1\leq u\leq|S|-2$. Then,

• (i)

$M$* is a subgroup.*

• (ii)

Each $u$-cell is $M$–periodic.

• (iii)

Each $v$–kernel with $u<v\leq|S|-2$ is a proper subgroup of $M$.

*Proof.  *  Let $X$ be a $u$-cell with $u\leq|S|-2$. By Kneser’s Theorem, the inequality $|XS|-|X|=u\leq|S|-2$ implies

[TABLE]

where $H$ is the stabilizer of $XS$. Since $X$ is a cell and $HXS=XS$, we have $X=HX$. Note that, since $G$ is abelian, $(\{y\}\cup H)S=HS$ implies $y\in Stab(HS)\subset Stab(XS)$, so that $y\in H$. This observation and (2) imply that $H$ is an $u$–cell. In particular, by taking $X=M$, the period $K$ of $MS$ is a $u$–cell. Since $KMS=MS$ and $M$ is a $u$–cell, we have $K\subset KM\subset M$. Since $M$ is a $u$–kernel we have $M=K$. This proves (i).

Now let $H$ be the stabilizer of $XS$, where $X$ is a $u$–cell. As shown in the preceding paragraph $H$ is also a $u$–cell. By Theorem 5 we have $M\subset H$ and thus $MH=H$. Since $X$ is a cell and $HXS=XS$, we have $X=HX=MHX$. Hence $X\subset MX\subset MHX=X$ implies $X=MX$. This proves (ii).

Finally, by (i), a $v$–kernel $N$ is a subgroup. By Theorem 5 we have $N\subset M$.

From Corollary 6, one can deduce Theorem 1.

Proof of Theorem 1:   We may assume without loss of generality that $1\in S$.

Case 1: There is no $m$–cell for any $1\leq m\leq|S|-2$.

• •

either we have $|TS|\geq|S|+|T|-1$ for any non-empty finite $T$, in which case the theorem clearly holds with $H=\{1\}$.

• •

or there exists some non-empty finite $T$ such that $|TS|\leq|S|+|T|-2$. Without loss of generality, we may also suppose $1\in T$. Now $T$ must be contained in an $m$–cell with $m\leq|S|-2$, but since no such cell exists for $1\leq m$, we have that $T$ itself must be a cell (a [math]-cell) i.e. $|TS|=|T|$. We therefore have $HT=TH=T=TS=HTS$ where $H$ is the (necessarily finite) subgroup generated by $S$. We have just proved that the theorem holds in this case with $H=\langle S\rangle$.

Case 2: There exists an $m$–cell with $1\leq m\leq|S|-2$. We may therefore consider the largest integer $u\leq|S|-2$ for which $S$ admits a $u$–cell. Let $H$ be the $u$–kernel containing $1$. Note that $u\leq|S|-2$ implies that $H$ is different from $\{1\}$. Now let $T$ be any finite non-empty subset such that $|TS|-|T|\leq|S|-2$. We shall prove that $HTS=TS$.

By adding elements to $T$ as long as necessary, we can find a cell $X$ that contains $T$ and such that $XS=TS$. Note that we then have $|XS|-|X|\leq|TS|-|T|\leq|S|-2$, so that $X$ is a $v$–cell for some $v\leq u$. By Corollary 6 (ii) we have $TS=XS=MXS=MTS$ where $M$ is the $v$-kernel containing $1$. By part (i) of Corollary 6, $H$ is a subgroup of $M$ so that $TS=XS=HTS$ as well.

Finally, $|ST|\leq|HS|+|HT|-|H|$ follows from $|ST|$ being a multiple of $|H|$.