A Semigroup Is Finite Iff It Is Chain-Finite and Antichain-Finite

Abstract

A subset A of a semigroup S is called a chain (antichain) if $ab\in \{a,b\}$ ($ab\notin \{a,b\}$) for any (distinct) elements $a,b\in A$. A semigroup S is called periodic if for every element $x\in S$ there exists $n\in \mathbb{N}$ such that $x^n$ is an idempotent. A semigroup S is called (anti)chain-finite if S contains no infinite (anti)chains. We prove that each antichain-finite semigroup S is periodic and for every idempotent e of S the set $\sqrt[\infty ]{e}=\{x\in S:\exists n\in \mathbb{N}(x^n=e)\}$ is finite. This property of antichain-finite semigroups is used to prove that a semigroup is finite if and only if it is chain-finite and antichain-finite. Furthermore, we present an example of an antichain-finite semilattice that is not a union of finitely many chains.

1. Introduction

It is well-known that a partially ordered set X is finite iff all chains and antichains in X are finite. The notions of chain and antichain are well-known in the theory of order (see, e.g., ([1] (O-1.6)) or [2]). In this paper we present a similar characterization of finite semigroups in terms of finite chains and antichains.

Let us recall that a magma is a set S endowed with a binary operation $S\times S\to S$, $⟨x,y⟩\mapsto xy$. If the binary operation is associative, then the magma S is called a semigroup. A semilattice is a commutative semigroup whose elements are idempotents. Each semilattice S carries a natural partial order ≤ defined by $x\le y$ iff $xy=yx=x$. Observe that two elements $x,y$ of a semilattice are comparable with respect to the partial order ≤ if and only if $xy\in \{x,y\}$. This observation motivates the following algebraic definition of chains and antichains in any magma.

A subset A of a magma S is defined to be

• a chain if $xy\in \{x,y\}$ for any elements $x,y\in A$;
• an antichain if $xy\notin \{x,y\}$ for any distinct elements $x,y\in A$.

The definition implies that each chain consists of idempotents.

A magma S is defined to be (anti)chain-finite if it contains no infinite (anti)chains.

Let us note that chain-finite semilattices play an important role in the theory of complete topological semigroups. In [3], Stepp showed that for each homomorphism $f:X\to Y$ from a chain-finite semilattice X to a Hausdorff topological semigroup Y, the image $f\left[X\right]$ is closed in Y. Banakh and Bardyla [4] extended the result of Stepp to the following characterization:

Theorem 1.

For a semilattice X the following conditions are equivalent:

• X is chain-finite;
• X is closed in each Hausdorff topological semigroup containing X as a discrete subsemigroup;
• For each homomorphism $f:X\to Y$ into a Hausdorff topological semigroup Y, the image $f\left[X\right]$ is closed;

For other completeness properties of chain-finite semilattices see [4,5,6]. Antichain-finite posets and semilattices were investigated by Yokoyama [7].

The principal result of this note is the following theorem characterizing finite semigroups.

Theorem 2.

A semigroup S is finite if and only if it is chain-finite and antichain-finite.

A crucial step in the proof of this theorem is the following proposition describing the (periodic) structure of antichain-finite semigroups.

A semigroup S is called periodic if for every $x\in S$ there exists $n\in \mathbb{N}$ such that $x^n$ is an idempotent of S. In this case

$$S=\bigcup _{e\in E\left(S\right)}\sqrt[\infty ]{e},$$

where $E\left(S\right)=\{x\in S:xx=x\}$ is the set of idempotents of S and

$$\sqrt[\infty ]{e}=\{x\in S:\exists n\in \mathbb{N}(x^n=e)\}$$

for $e\in E\left(S\right)$.

Proposition 1.

Each antichain-finite semigroup S is periodic and for every $e\in E\left(S\right)$ the set $\sqrt[\infty ]{e}$ is finite.

Theorem 2 and Proposition 1 will be proved in the next section.

Remark 1.

Theorem 2 does not generalize to magmas. To see this, consider the set of positive integers $\mathbb{N}$ endowed with the following binary operation: $nm=n$ if $n<m$ and $nm=1$ if $n\ge m$. This magma is infinite but each nonempty chain in the magma is of the form $\{1,n\}$ for some $n\in \mathbb{N}$, and each nonempty antichain in this magma is a singleton.

Next we present a simple example of an antichain-finite semilattice which is not a union of finitely many chains.

Example 1.

Consider the set

$$S=\{⟨2n-1,0⟩:n\in \mathbb{N}\}\cup \{⟨2n,m⟩:n,m\in \mathbb{N},m\le 2n\}$$

endowed with the semilattice binary operation

$$⟨x,i⟩·⟨y,j⟩=\left\{\begin{array}{cc}⟨x,i⟩\hfill & if x=y and i=j;\hfill \\ ⟨x-1,0⟩\hfill & if x=y and i\ne j;\hfill \\ ⟨x,i⟩\hfill & if x<y;\hfill \\ ⟨y,j⟩\hfill & if y<x.\hfill \end{array}\right.$$

It is straightforward to check that the semilattice S has the following properties:

1.

S is antichain-finite;

2.

S has arbitrarily long finite antichains;

3.

S is not a union of finitely many chains;

4.

The subsemilattice $L=\{⟨2n-1,0⟩:n\in \mathbb{N}\}$ of S is a chain;

5.

S admits a homomorphism $r:S\to L$ such that $r^{-1}\left(⟨x,0⟩\right)=\{⟨y,i⟩\in S:y\in \{x,x+1\}\}$ is finite for every element $⟨x,0⟩\in L$.

Example 1 motivates the following question.

Question 1.

Let S be an antichain-finite semilattice. Is there a finite-to-one homomorphism $r:S\to Y$ to a semilattice Y which is a finite union of chains?

A function $f:X\to Y$ is called finite-to-one if for every $y\in Y$ the preimage $f^{-1}\left(y\right)$ is finite.

2. Proofs of the Main Results

In this section, we prove some lemmas implying Theorem 2 and Proposition 1. More precisely, Proposition 1 follows from Lemmas 1 and 4; Theorem 2 follows from Lemma 5.

The following lemma exploit ideas of Theorem 1.9 from [8].

Lemma 1.

Every antichain-finite semigroup S is periodic.

Proof. 

Given any element $x\in S$ we should find a natural number $n\in \mathbb{N}$ such that $x^n$ is an idempotent. First we show that $x^n=x^m$ for some $n\ne m$. Assuming that $x^n\ne x^m$ for any distinct numbers $n,m$, we conclude that the set $A=\{x^n:n\in \mathbb{N}\}$ is infinite and for any $n,m\in \mathbb{N}$ we have $x^n{x}^m=x^{n+m}\notin \{x^n,x^m\}$, which means that A is an infinite antichain in S. However, such an antichain cannot exist as S is antichain-finite. This contradiction shows that $x^n=x^m$ for some numbers $n<m$ and then for the number $k=m-n$ we have $x^{n+k}=x^m=x^n$. By induction we can prove that $x^{n+pk}=x^n$ for every $p\in \mathbb{N}$. Choose any numbers $r,p\in \mathbb{N}$ such that $r+n=pk$ and observe that

$$x^{r+n}{x}^{r+n}=x^{r+n}{x}^{pk}=x^r{x}^{n+pk}=x^r{x}^n=x^{r+n},$$

which means that $x^{r+n}$ is an idempotent and hence S is periodic. □

An element $1\in S$ is called an identity of S if $x1=x=1x$ for all $x\in S$. For a semigroup S let $S^1=S\cup \left\{1\right\}$ where 1 is an element such that $x1=x=1x$ for every $x\in S^1$. If S contains an identity, then we will assume that 1 is the identity of S and hence $S^1=S$.

For a set $A\subseteq S$ and element $x\in S$ we put

$$xA=\{xa:a\in A\}\mathrm{and}Ax=\{ax:a\in A\}.$$

For any element x of a semigroup S, the set

$$H_x=\{y\in S:y{S}^1=x{S}^1\wedge S^{1}y=S^{1}x\}$$

is called the $\mathcal{H}$-class of x. By Lemma I.7.9 [9], for every idempotent e its $\mathcal{H}$-class $H_e$ coincides with the maximal subgroup of S that contains the idempotent e.

Lemma 2.

If a semigroup S is antichain-finite, then for every idempotent e of S its $\mathcal{H}$-class $H_e$ is finite.

Proof. 

Observe that the set $H_e\setminus \left\{e\right\}$ is an antichain (this follows from the fact that the left and right shifts in the group $H_e$ are injective). Since S is antichain-finite, the antichain $H_e\setminus \left\{e\right\}$ is finite and so is the set $H_e$. □

Lemma 3.

If a semigroup S is antichain-finite, then for every idempotent e in S we have

$$(H_e·\sqrt[\infty ]{e})\cup (\sqrt[\infty ]{e}·H_e)\subseteq H.$$

Proof. 

Given any elements $x\in \sqrt[\infty ]{e}$ and $y\in H_e$, we have to show that $xy\in H_e$ and $yx\in H_e$. Since $x\in \sqrt[\infty ]{e}$, there exists a number $n\in \mathbb{N}$ such that $x^n=e$. Then $x^{n+1}{S}^1=ex{S}^1\subseteq e{S}^1$ and $e{S}^1=x^{2n}{S}^1\subseteq x^{n+1}{S}^1$, and hence $e{S}^1=x^{n+1}S$. By analogy we can prove that $S^{1}e=S^1{x}^{n+1}$. Therefore, $x^{n+1}\in H_e$.

Then $xy=x\left(ey\right)=\left(xe\right)y=\left(x{x}^n\right)y=x^{n+1}y\in H_e$ and $yx=\left(ye\right)x=y\left(ex\right)=y{x}^{n+1}\in H_e$. □

For each $k\in \mathbb{N}$ by ${\left[\mathbb{N}\right]}^k$ we denote the set of all k-element subsets of $\mathbb{N}$. The proofs of the next two lemmas essentially use the classical Ramsey Theorem, so let us recall its formulation, see ([10] (p. 16)) for more details.

Theorem 3

(Ramsey). For any $n,k\in \mathbb{N}$ and map $\chi :{\left[\mathbb{N}\right]}^k\to n=\{0,\dots ,n-1\}$ there exists an infinite subset $I\subseteq \mathbb{N}$ such that $\chi \left[{\left[I\right]}^k\right]=\left\{c\right\}$ for some number $c\in n$.

Lemma 4.

If a semigroup S is antichain-finite, then for every idempotent $e\in E\left(S\right)$ the set $\sqrt[\infty ]{e}$ is finite.

Proof. 

By Lemma 2, the $\mathcal{H}$-class $H_e$ is finite. Assuming that $\sqrt[\infty ]{e}$ is infinite, we can choose a sequence ${\left(x_n\right)}_{n\in \omega }$ of pairwise distinct points of the infinite set $\sqrt[\infty ]{e}\setminus H_e$.

Let $P=\{⟨n,m⟩\in \omega \times \omega :n<m\}$ and $\chi :P\to 5=\{0,1,2,3,4\}$ be the function defined by

$$\chi (n,m)=\left\{\begin{array}{cc}0\hfill & \mathrm{if} x_n{x}_m=x_n;\hfill \\ 1\hfill & \mathrm{if} x_m{x}_n=x_n;\hfill \\ 2\hfill & \mathrm{if} x_n{x}_m=x_m;\hfill \\ 3\hfill & \mathrm{if} x_m{x}_n=x_m;\hfill \\ 4\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

By the Ramsey Theorem 3, there exists an infinite subset $\Omega \subseteq \omega$ such that $\chi [P\cap (\Omega \times \Omega \left)\right]=\left\{c\right\}$ for some $c\in \{0,1,2,3,4\}$.

If $c=0$, then $x_n{x}_m=x_n$ for any numbers $n<m$ in $\Omega$. Fix any two numbers $n<m$ in $\Omega$. By induction we can prove that $x_n{x}_m^p=x_n$ for every $p\in \mathbb{N}$. Since $x_m\in \sqrt[\infty ]{e}$, there exists $p\in \mathbb{N}$ such that $x_m^p=e$. Then $x_n=x_n{x}_m^p=x_{n}e\in H_e$ by Lemma 3. However, this contradicts the choice of $x_n$.

By analogy we can derive a contradiction in cases $c\in \{1,2,3\}$.

If $c=4$, then the set $A={\left\{x_n\right\}}_{n\in \Omega }$ is an infinite antichain in S, which is not possible as the semigroup S is antichain-finite.

Therefore, in all five cases we obtain a contradiction, which implies that the set $\sqrt[\infty ]{e}$ is finite. □

Our final lemma implies the non-trivial “if” part of Theorem 2.

Lemma 5.

A semigroup S is finite if it is chain-finite and antichain-finite.

Proof. 

Assume that S is both chain-finite and antichain-finite. By Lemma 1, the semigroup S is periodic and hence $S={\bigcup }_{e\in E\left(S\right)}\sqrt[\infty ]{e}$. By Lemma 4, for every idempotent $e\in E\left(S\right)$ the set $\sqrt[\infty ]{e}$ is finite. Now it suffices to prove that the set $E\left(S\right)$ is finite.

To derive a contradiction, assume that $E\left(S\right)$ is infinite and choose a sequence of pairwise distinct idempotents ${\left(e_n\right)}_{n\in \omega }$ in S. Let $P=\{⟨n,m⟩\in \omega \times \omega :n<m\}$ and $\chi :P\to \{0,1,2,3,4,5\}$ be the function defined by the formula

$$\chi (n,m)=\left\{\begin{array}{cc}0\hfill & \mathrm{if} e_n{e}_m\in \{e_n,e_m\} \mathrm{and} e_m{e}_n\in \{e_n,e_m\};\hfill \\ 1\hfill & \mathrm{if} e_n{e}_m=e_n \mathrm{and} e_m{e}_n\notin \{e_n,e_m\};\hfill \\ 2\hfill & \mathrm{if} e_n{e}_m=e_m \mathrm{and} e_m{e}_n\notin \{e_n,e_m\};\hfill \\ 3\hfill & \mathrm{if} e_n{e}_m\notin \{e_n,e_m\} \mathrm{and} e_m{e}_n=e_n;\hfill \\ 4\hfill & \mathrm{if}{e}_n{e}_m\notin \{e_n,e_m\} \mathrm{and} e_m{e}_n=e_m;\hfill \\ 5\hfill & \mathrm{if} e_n{e}_m\notin \{e_n,e_m\} \mathrm{and} e_m{e}_n\notin \{e_n,e_m\}.\hfill \end{array}\right.$$

The Ramsey Theorem 3 yields an infinite subset $\Omega \subseteq \omega$ such that $\chi [P\cap (\Omega \times \Omega \left)\right]=\left\{c\right\}$ for some $c\in \{0,1,2,3,4,5\}$.

Depending on the value of c, we shall consider six cases.

If $c=0$ (resp. $c=5$), then ${\left\{e_n\right\}}_{n\in \omega }$ is an infinite (anti)chain in S, which is forbidden by our assumption.

Next, assume that $c=1$. Then $e_n{e}_m=e_n$ and $e_m{e}_n\notin \{e_n,e_m\}$ for any numbers $n<m$ in $\Omega$. For any number $k\in \Omega$, consider the set $Z_k=\{e_n{e}_k:k<n\in \Omega \}$. Observe that for any $e_n{e}_k,e_m{e}_k\in Z_k$ we have

$$\left(e_n{e}_k\right)\left(e_m{e}_k\right)=e_n\left(e_k{e}_m\right)e_k=e_n{e}_k{e}_k=e_n{e}_k,$$

which means that $Z_k$ is a chain. Since S is chain-finite, the chain $Z_k$ is finite.

By induction we can construct a sequence of points ${\left(z_k\right)}_{k\in \omega }\in {\prod }_{k\in \omega }{Z}_k$ and a decreasing sequence of infinite sets ${\left({\Omega }_k\right)}_{k\in \omega }$ such that ${\Omega }_0\subseteq \Omega$ and for every $k\in \omega$ and $n\in {\Omega }_k$ we have $e_n{e}_k=z_k$ and $n>k$. Choose an increasing sequence of numbers ${\left(k_i\right)}_{i\in \omega }$ such that $k_0\in {\Omega }_0$ and $k_i\in {\Omega }_{k_{i-1}}$ for every $i\in \mathbb{N}$. We claim that the set $Z=\{z_{k_i}:i\in \omega \}$ is a chain. Take any numbers $i,j\in \omega$ and choose any number $n\in {\Omega }_{k_i}\cap {\Omega }_{k_j}$.

If $i\le j$, then

$$z_{k_i}{z}_{k_j}=\left(e_n{e}_{k_i}\right)\left(e_n{e}_{k_j}\right)=e_n\left(e_{k_i}{e}_n\right)e_{k_j}=e_n{e}_{k_i}{e}_{k_j}=e_n{e}_{k_i}=z_{k_i}.$$

If $i>j$, then $k_i\in {\Omega }_{k_{i-1}}\subseteq {\Omega }_{k_j}$ and hence

$$z_{k_i}{z}_{k_j}=\left(e_n{e}_{k_i}\right)\left(e_n{e}_{k_j}\right)=e_n\left(e_{k_i}{e}_n\right)e_{k_j}=e_n\left(e_{k_i}{e}_{k_j}\right)=e_n{z}_{k_j}=e_n\left(e_n{e}_{k_j}\right)=e_n{e}_{k_j}=z_{k_j}.$$

In both cases we obtain that $z_{k_i}{z}_{k_j}\in \{z_{k_i},z_{k_j}\}$, which means that the set $Z=\{z_{k_i}:i\in \omega \}$ is a chain. Since S is chain-finite, the set Z is finite. Consequently, there exists $z\in Z$ such that the set $\Lambda =\{i\in \omega :z_{k_i}=z\}$ is infinite. Choose any numbers $i<j$ in the set $\Lambda$ and then choose any number $n\in {\Omega }_{k_j}\subseteq {\Omega }_{k_i}$. Observe that $k_j\in {\Omega }_{k_{j-1}}\subseteq {\Omega }_{k_i}$ and hence $e_{k_j}{e}_{k_i}=z_{k_i}=z$. Then

$$e_{k_j}=e_{k_j}{e}_{k_j}=\left(e_{k_j}{e}_n\right)e_{k_j}=e_{k_j}\left(e_n{e}_{k_j}\right)=e_{k_j}{z}_{k_j}=e_{k_j}z=e_{k_j}{z}_{k_i}=e_{k_j}\left(e_{k_j}{e}_{k_i}\right)=e_{k_j}{e}_{k_i}\notin \{e_{k_j},e_{k_j}\}$$

as $c=1$.

By analogy we can prove that the assumption $c\in \{2,3,4\}$ also leads to a contradiction. □