3k-4 theorem for ordered groups

TL;DR

This paper provides elementary proofs for two recent structure theorems concerning ordered groups, simplifying the understanding of their algebraic properties.

Contribution

It offers accessible, elementary proofs of complex theorems in the theory of ordered groups, making these results more widely understandable.

Findings

Elementary proofs of two structure theorems for ordered groups

Simplification of existing complex proofs

Enhanced understanding of ordered group properties

Abstract

Recently, G. A. Freiman, M. Herzog, P. Longobardi, M. Maj proved two `structure theorems' for ordered groups \cite{FHLM}. We give elementary proof of these two theorems.

Equations

$$|S^2|\ge 2|S|-1.$$

$$x_1{x}_1<x_1{x}_2<\dots <x_1{x}_k<x_2{x}_k<\dots <x_k{x}_k$$

$$\{x_1{x}_1,x_1{x}_2,\dots ,x_1{x}_k,x_2{x}_k,\dots ,x_k{x}_k\}=S^2.$$

$$x_2{x}_1=x_1{x}_2,x_2{x}_2=x_1{x}_3,x_2{x}_3=x_1{x}_4,\dots ,x_2{x}_{k-1}=x_1{x}_k.$$

$$x_{k-1}{x}_k=x_k{x}_{k-1},x_{k-2}{x}_k=x_k{x}_{k-2}=x_{k-1}^2.$$

$$|S^2|\ge |T^2|+|T|+1.$$

$$|S^2|\ge |T^2|+|T|+1+1.$$

Full text

1. Introduction

For any group $G$ (written multiplicatively) and a subset $S$ of $G$ we define $S^{2}=\{ab:a,b\in S\}$. Then, the main theorem of [1] is

Theorem 1.1**.**

[Theorem 1.3, [1]] Let $G$ be an ordered group and $S$ be a finite subset of $G$. If $|S^{2}|\leq 3|S|-3$ then the subgroup generated by $S$ is an abelian subgroup of $G$.

As a corollary to Theorem 1.1, they deduce a $3k-4$ type theorem for ordered groups.

Theorem 1.2**.**

[Corollary 1.4, [1]] Let $G$ be an ordered group and $S$ be a finite subset of $G$ with $|S|=k\geq 3$. If $|S^{2}|\leq 3|S|-4$, then there exist two commuting elements $x,y$ such that $S\subset\{yx^{i}:0\leq i\leq N\}$ for $N=|S^{2}|-|S|$.

We give elementary proofs of Theorem 1.1 and Theorem 1.2. We shall always assume that $G$ is an ordered group and $S$ is a finite subset of $G$ with $k$ elements. We shall write $S=\{x_{1},\ldots,x_{k}\}$ and assume that $x_{1}<\ldots<x_{k}$.

2. Proofs

As in the case of integers, the following inequality holds:

[TABLE]

In equation (1) equality holds only if $S$ is a geometric progression $\{yx^{i}:0\leq i\leq k\}$, with $x,y$ two commuting elements of $G$.

Lemma 1**.**

If $S$ is not a geometric progression then $|S^{2}|\geq 2|S|.$

Proof.

Let $S=\{x_{1}<\ldots<x_{k}\}$. Certainly

[TABLE]

are $2|S|-1$ distinct elements in $S^{2}$. If $|S^{2}|<2|S|$ then

[TABLE]

Now, consider the elements $x_{2}x_{1}<x_{2}x_{2}<\ldots<x_{2}x_{k}$. All these elements are in $S^{2}$ and $x_{1}x_{1}<x_{2}x_{1},\ldots,x_{2}x_{k-1}<x_{2}x_{k}$. Thus we must have

[TABLE]

From the above relations it follows that $x_{1}$ and $x_{2}$ commute and for $i>2$, $x_{i}$ is contained in the subgroup generated by $x_{1},\ldots,x_{i-1}$. Consequently we get that each $x_{i}$ commutes with each $x_{j}$ for $i,j=1,\ldots,k$.

Put $y=x_{1},x=x_{2}x_{1}^{-1}$, then $x$ and $y$ commute and $S=\{y,xy,x^{2}y,\ldots,x^{k-1}y\}$.

Thus, if $S$ is not a geometric progression then $|S^{2}|\geq 2|S|$. ∎

Proof of Theorem 1.2.

We shall use induction on $k$. For $k=3$, we have $|S^{2}|\leq 5$. We have five distinct elements $x_{1}^{2}<x_{1}x_{2}<x_{2}^{2}<x_{2}x_{3}<x_{3}^{2}$ in $S^{2}$. Since $x_{1}x_{3}\in S^{2}$, so $x_{1}x_{3}$ must equal to one of these five elements. Using the order relation, we get $x_{1}x_{3}=x_{2}^{2}$. Similarly, we get $x_{1}x_{2}=x_{2}x_{1}$. Let $y=x_{1}$ and $x=x_{2}x_{1}^{-1}$. Then $x$ and $y$ commute and $S=\{y,yx,yx^{2}\}.$

Now we assume that $k\geq 4$ and the theorem is true for any subset $T$ of $G$ with $|T|\leq k-1$. Put $T=\{x_{1},\ldots,x_{k-1}\}.$

Case (1): $|T^{2}|\leq 3|T|-4$.

By induction hypothesis, there are commuting elements $x,y$ such that $T\subset\{yx^{j}:j=0,\ldots,M\}$ with $M=|T^{2}|-|T|$.

In case $x_{k}T\cap T^{2}=\emptyset$, then, taking $x_{k}^{2}$ in account, we see that $|S^{2}|\geq|T^{2}|+(|T|+1).$ Since $|T^{2}|\geq 2|T|-1$, we immediately obtain $|S^{2}|\geq 3|S|-3,$ which contradicts the hypothesis. Thus, we get $x_{k}T\cap T^{2}\neq\emptyset$. Consequently, there are $yx^{i},yx^{u},yx^{v}\in T$ such that $x_{k}yx^{i}=yx^{u}yx^{v}$. This gives $x_{k}=yx^{(u+v-i)}$ and $S\subset\{yx^{j}:j=0,\ldots,M^{\prime}\}$ with $M^{\prime}=\max\{M,u+v-i\}$. Clearly the map $yx^{j}\mapsto j$ gives a $2-isomorphism$ of $S$ with a subset of $\mathbb{Z}$. From the Freiman’s $3k-4$-theorem for integers, it follows that $M^{\prime}\leq N$, and the theorem is proved.

Case (2): $|T^{2}|\geq 3|T|-3=3|S|-6$. Using the order relation of $G$ we see that the elements $x_{k}^{2}$ and $x_{k}x_{k-1}$ of $S^{2}$ are not in $T^{2}$. Consider the element $x_{k-1}x_{k}$ of $S^{2}$. If $x_{k-1}x_{k}\neq x_{k}x_{k-1}$ then we get $|S^{2}|\geq|T^{2}|+3$, which contradicts the hypothesis. So, we obtain $x_{k-1}x_{k}=x_{k}x_{k-1}$. Next, we consider the element $x_{k-2}x_{k}$ of $S^{2}$. If $x_{k-2}x_{k}\neq x_{k-1}^{2}$, then we already get $|S^{2}|\geq|T^{2}|+3$, leading to a contradiction. Similarly it follows that $x_{k}x_{k-2}=x_{k-1}^{2}$. Thus we have

[TABLE]

Put $y=x_{k},x=x_{k-1}x_{k}^{-1}$. Then $x$ and $y$ commute and $x_{k}=y,x_{k-1}=yx,x_{k-2}=yx^{2}$. Considering the elements $x_{k-3}x_{k},x_{k-4}x_{k},\ldots,x_{1}x_{k}$ successively we see that each of $x_{i}$ is of the form $yx^{t_{i}}$. Clearly $S$ is $2-isomorphic$ to the subset $\{t_{i}:1\leq i\leq k\}$ of $\mathbb{Z}$. Now the theorem follows from the Freiman’s $3k-4$-theorem for integers. ∎

Proof of Theorem 1.1.

We shall use induction on $k$. For $k=1,2$, the theorem holds trivially. Now, let $k\geq 3$ and assume that the theorem is true for any set $T$ with $|T|\leq k-1$. Put $T=\{x_{1},\ldots,x_{k-1}\}.$

Case (1): $|T^{2}|\leq 3|T|-3$.

By induction hypothesis, $T$ generates a commutative subgroup. If $x_{k}T\cap T^{2}\neq\emptyset$ or $Tx_{k}\cap T^{2}\neq\emptyset$ then $x_{k}$ lies in the subgroup generated by $T$. Consequently, $S$ generates a commutative subgroup. So we assume that $x_{k}T\cap T^{2}=\emptyset$ and $Tx_{k}\cap T^{2}=\emptyset$. Using the order relation, we see that $x_{k}^{2}\not\in T^{2}\cup x_{k}T$, so we obtain

[TABLE]

If $T$ is not a geometric progression then, using Lemma 1 in (2), we see that $|S^{2}|\geq 3|S|-2$, which contradicts the hypothesis. Thus, $T$ must be a geometric progression.

Next, observe that if $x_{k}T\neq Tx_{k}$ then we have an element in $Tx_{k}$ which is not in $T^{2}\cup x_{k}T\cup\{x_{k}^{2}\}$. This leads to

[TABLE]

From this one obtains $|S^{2}|\geq 3|S|-2$, which contradicts the hypothesis. Thus, we must have $x_{k}T=Tx_{k}$. Now using the order relation we see that $x_{k}$ commutes with all the elements of $T$ and consequently $S$ generates an abelian group.

Case (2): $|T^{2}|>3|T|-3$.

As in the proof of Theorem 1.2 (the arguments used in Case (2)) we see that either $|S^{2}|\geq|T^{2}|+3$ or $S=\{yx^{t_{i}}:1\leq i\leq k\}$ with commuting elements $x$ and $y$. The former leads to a contradiction and hence we get $S=\{yx^{t_{i}}:1\leq i\leq k\}$ with commuting elements $x$ and $y$. This proves the theorem. ∎

Remark 1**.**

From the proof of Theorem 1.2 it is clear that the subgroup generated by $S$ (with $|S|>2$) is, in fact, generated by $|S|-1$ or less elements.