2 Preliminaries

We begin by establishing some notation. For a semigroup S and a nonempty subset $X \subseteq S$ , we use $\langle X \rangle $ to denote the subsemigroup generated by the set X. For a congruence $\sim $ on S, we denote the congruence class of the element x by $[x]_\sim $ . A congruence that has only finitely many congruence classes shall be known as a finite index congruence. Given an ideal I of a semigroup S, the Rees congruence $\sim _I$ on S is given by

$$ \begin{align*}a\,\sim_I\,b\iff a, b\in I\text{ or }a=b.\end{align*} $$

The quotient $S/\sim _I$ is called the Rees quotient of S by I, and is denoted by $S/I$ . We denote the Rees congruence class of an element $x\in S$ by $[x]_I$ .

In the introduction we framed separability in the language of general algebra. From now on we focus on separability for semigroups. For clarity we now formally restate our separability conditions in the semigroup setting, and also reformulate them in terms of congruences.

• • A semigroup S is said to be residually finite if for any two distinct elements $s,t \in S$ there exist a finite semigroup U and homomorphism $f: S \to U$ such that $f(s) \neq f(t)$ . This is equivalent to saying there exists a congruence $\sim $ of finite index such that $[s]_{\sim }\neq [t]_{\sim }$ .

• • A semigroup S is weakly subsemigroup separable if for any finitely generated subsemigroup T and any $s \in S\!\setminus \!T$ there exist a finite semigroup U and homomorphism $f: S \to U$ such that $f(s) \notin f(T)$ . This is equivalent to saying there exists a congruence $\sim $ of finite index such that, for all $ t \in T$ , $[s]_{\sim } \neq [t]_ {\sim }$ .

• • A semigroup S is strongly subsemigroup separable if for any subsemigroup T and any $s \in S \!\setminus \! T$ there exist a finite semigroup U and homomorphism $f: S \to U$ such that $f(s) \notin f(T)$ . Again, this is equivalent to saying there exists a congruence $\sim $ of finite index such that, for all $ t \in T$ , $[s]_{\sim } \neq [t]_ {\sim }$ .

• • A semigroup S is completely separable if for any $s \in S$ there exist a finite semigroup U and homomorphism $f:S \to U$ such that $f(s) \notin f(S\!\setminus \! \{s\})$ . This is equivalent to saying there exists a congruence $\sim $ of finite index such that $[s]_{\sim }=\{s\}$ .

For a semigroup S, the profinite topology on S is defined as the topology on S with a basis consisting of all congruence classes of finite index congruences on S. Any semigroup equipped with the profinite topology is a topological semigroup. We can now express the above properties in topological terms.

• • A semigroup is residually finite if and only if every singleton is closed in the profinite topology. This is equivalent to saying that the intersection of all finite index congruences is the equality relation.

• • A semigroup is weakly subsemigroup separable if and only if all finitely generated subsemigroups are closed in the profinite topology.

• • A semigroup is strongly subsemigroup separable if and only if all subsemigroups are closed in the profinite topology.

• • A semigroup is completely separable if and only if the profinite topology is discrete.

We state how these properties relate to one another in the following result.

In general, the converse of each implication from Proposition 2.1 is not true. This is demonstrated in Examples 2.5, 5.10 and 5.11. For every finite semigroup the profinite topology is discrete, so we can immediately observe that every finite semigroup is completely separable and hence that all four separability properties are finiteness conditions. The following proposition shows that subsemigroups inherit each of these separability properties.

In Section 3 we consider the conditions under which certain groups have our separability properties. We consider these groups as semigroups, so it is worth noting how the separability conditions vary between the class of groups and the class of semigroups.

A group G is residually finite within the class of groups if and only if it is residually finite within the class of semigroups. This is because every homomorphic image of a group is itself a group. Similarly, we can see that a group G is completely separable within the class of groups if and only if it is completely separable within the class of semigroups. Furthermore, we can fully classify completely separable groups.

We note that there exist infinite groups that are strongly subsemigroup separable (and hence weakly subsemigroup separable), as Theorem 2.8 demonstrates.

It is also easy to see that if a group G is strongly subsemigroup separable, then it must also be strongly subgroup separable. This is because every subgroup is also a subsemigroup. However, if a group is strongly subgroup separable, then it need not be strongly subsemigroup separable (or even weakly subsemigroup separable), as the following example demonstrates.

As a consequence of Proposition 2.3 and Example 2.5, we see that if a group G is weakly subsemigroup separable then it cannot contain a copy of $\mathbb {Z}$ . In other words, for G to be weakly subsemigroup separable it is necessary for it to be torsion. As every subsemigroup of a torsion group is in fact a subgroup, we observe the following result.

We will now briefly discuss the situation for abelian groups. It is known that a group is residually finite if and only if it is isomorphic to a subdirect product of finite groups; see [Reference Cohn4, Corollary 7.2]. It follows that an abelian group is residually finite if and only if it is isomorphic to a subdirect product of finite cyclic groups.

In order to discuss weak subsemigroup separability in abelian groups, we first give the following lemma. A semigroup $A$ is called locally finite if every finitely generated subsemigroup of $A$ is finite.

For an abelian group to be weakly subsemigroup separable, it is necessary for it to be residually finite. It is also necessary for it to be torsion, as noted above. Since torsion abelian groups are locally finite, being residually finite and torsion are sufficient conditions for an abelian group to be weakly subsemigroup separable by Lemma 2.7.

In [Reference Golubov7] Golubov was able to characterise when commutative semigroups are strongly subsemigroup separable. We can apply his result to abelian groups. For an abelian group A and a prime p, recall that the p-primary component of A is the set

$$ \begin{align*}A_p=\{a \in A \mid o(a)=p^n \text{ for some } n \in \mathbb{N}_0\},\end{align*} $$

where $o(a)$ is the order of the element a. We say that $A_p$ has finite exponent if there exists $n \in \mathbb {N}$ such that $o(a) \leq p^n$ for all $a \in A_p$ . Then Golubov’s result [Reference Golubov7, Theorem 2] tells us that an abelian group A is strongly subsemigroup separable if and only if A is torsion and, for each prime p, the primary p-component has finite exponent.

Finally, Lemma 2.4 tells us that a group is completely separable if and only if it is finite. We summarise these observations.

In the remainder of this section we record some structural results for semigroups that aid us in our investigations. For a semigroup S, let $S^1$ denote S with an identity adjoined if necessary. Green’s relation $\mathcal {H}$ on S is defined by

$$ \begin{align*}\mathcal{H}=\{(x,y) \mid xS^1=yS^1 \text{ and } S^1x=S^1y\}.\end{align*} $$

The $\mathcal {H}$ relation is an equivalence relation, and in commutative semigroups it is also a congruence. For an element $x \in S$ , we denote the $\mathcal {H}$ -class of x by $H_x$ . For an $\mathcal {H}$ -class H, the following are equivalent: H is a maximal subgroup; H contains an idempotent; and the intersection $H \cap H^2$ is nonempty [Reference Howie16, Corollary 2.2.6]. (If $n \geq 2$ , we write $H^n$ for the set of products $h_1 h_2 \cdots h_n$ with each $h_i \in H$ .)

In fact, for an $\mathcal {H}$ -class H, if, for some $n \geq 2$ , $H \cap H^n \neq \emptyset $ then H is a maximal subgroup. This is folklore but we provide a proof for completeness. Suppose $h_1,h_2,\ldots ,h_n,h \in H$ satisfy $h_1h_2 \cdots h_n=h.$ Since $h_1, h_2, h_{n-1}, h_{n}$ and h are pairwise $\mathcal {H}$ -related, there exist $s, t, u, v \in S^1$ such that

$$ \begin{align*} hs=h_1,h_nt=h_2,uh=h_n,vh_1=h_{n-1}.\end{align*} $$

Then

$$ \begin{align*}h\cdot sh_n=h_1h_n, \: h_1h_n \cdot th_3h_4 \cdots h_n=h,\end{align*} $$

$$ \begin{align*} h_1u \cdot h=h_1h_n, \: h_1h_2\cdots h_{n-2}v \cdot h_1h_n=h.\end{align*} $$

Hence $(h,h_1h_n) \in \mathcal {H}$ . Then $H \cap H^2 \neq \emptyset $ and it follows that H is a group.

It is possible to associate a group to an arbitrary $\mathcal {H}$ -class H as follows. The right stabiliser of H in S is

$$ \begin{align*}\operatorname{\mathrm{Stab}}(H)=\{s \in S^1 \mid Hs=H\}.\end{align*} $$

Clearly $\operatorname {\mathrm {Stab}}(H)$ is a submonoid of $S^1$ . We define a congruence $\sigma _H$ on $\operatorname {\mathrm {Stab}}(H)$ , called the Schützenberger congruence of H, by

$$ \begin{align*}(x,y) \in \sigma_H \iff hx=hy\quad \text{for all }h \in H.\end{align*} $$

Then $\Gamma (H)=\operatorname {\mathrm {Stab}}(H)/\sigma _H$ is a group, known as the Schützenberger group of H. It is known that:

• • $\Gamma (H)$ acts regularly (that is, transitively and freely) on H;

• • $|\Gamma (H)|=|H|$ ;

• • if H is a group then $\Gamma (H) \cong H$ .

One could similarly define a group $\Gamma _l(H)$ by considering the left stabiliser of H, but it turns out that $\Gamma _l(H) \cong \Gamma (H)$ . For more on Schützenberger groups and proofs of the above claims, see [Reference Lallement20, Section 2.3].

In a commutative semigroup S, inclusion among principal ideals induces a partial ordering on $\mathcal {H}$ classes: $H_x \leq H_y \text { if } xS^1 \subseteq yS^1.$ It is easy to see that for any $a,x \in S$ we have $H_{xa} \leq H_x.$ From this it follows that there can be at most one minimal $\mathcal {H}$ -class and therefore if such an $\mathcal {H}$ -class exists, it is the least $\mathcal {H}$ -class under this partial ordering. We refer to such an $\mathcal {H}$ -class as the minimum $\mathcal {H}$ -class.

For Section 4 it will be important to be familiar with the basic structure theory of commutative semigroups. An archimedean semigroup is a commutative semigroup S such that for each $a,b \in S$ there exists $n>0$ for which $H_{a^n} \leq H_b$ . We say that a semigroup S is a semilattice of semigroups if, for some semilattice Y, we can write S as a disjoint union of subsemigroups $S =\bigcup _{\alpha \in Y}S_\alpha $ such that $S_{\alpha }S_{\beta } \subseteq S_{\alpha \beta }$ for all $\alpha ,\beta \in Y$ . In this case we write $S=\mathcal {S}(Y,\{S_{\alpha }\}_{\alpha \in Y})$ . This leads us to the following structural theorem for commutative semigroups.

It follows from the definition that an archimedean semigroup can contain at most one idempotent. Hence a finitely generated commutative semigroup contains only finitely many idempotents.

We call a semigroup with a zero nilpotent if every element has a power equal to 0. An ideal extension of a semigroup S by a semigroup Q is a semigroup E such that S is an ideal of E and the Rees quotient $E/S$ is isomorphic to Q.

The following result provides a characterisation of archimedean semigroups with idempotents.

In general, the structure of archimedean semigroups is complex. For more about the decomposition of commutative semigroups into archimedean subsemigroups, see [Reference Grillet14, Ch. 4].

3 Schützenberger groups

It is a natural question to ask, for each of the three generalisations of residual finiteness, whether that property is inherited by Schützenberger groups. This is motivated by the following result.

In this section we show that the properties of complete separability and strong subsemigroup separability are inherited by Schützenberger groups. By way of contrast, it is not true that every Schützenberger group of a weakly subsemigroup separable semigroup is weakly subsemigroup separable. However, we are able to give a sufficient condition on the stabiliser of an $\mathcal {H}$ -class of a weakly subsemigroup separable semigroup to ensure that the corresponding Schützenberger group is also weakly subsemigroup separable. We begin by showing that the nongroup $\mathcal {H}$ -classes of a strongly subsemigroup separable semigroup are finite.

The analogue of Corollary 3.3 for weak subsemigroup separability does not hold in general, as demonstrated in Example 5.10. However, it does hold for commutative semigroups. We deduce this from the following result.

Although Schützenberger groups of weakly subsemigroup separable semigroups need not in general be weakly subsemigroup separable, it is an intriguing open question whether they must be weakly subgroup separable.

In the final part of this section we provide some partial solutions to this problem, one of which is utilised in the proof of Proposition 5.8.

Proof. If H is a group, then $\Gamma (H)\cong H$ is a subgroup of S and hence weakly subsemigroup separable by Proposition 2.3. It then certainly satisfies the separability property with respect to the collection of all finitely generated abelian subgroups.

Suppose that H is not a group. Let G be a finitely generated abelian subgroup of $\Gamma (H)$ and let $b\in \Gamma (H)\!\setminus \!G.$ Now, G is generated (as a group) by some set $\{a_1, \ldots , a_n\}\cup G_0,$ where each $a_i$ is nontorsion and $G_0$ is the finite torsion subgroup of G. For $s \in \operatorname {\mathrm {Stab}}(H)$ , we just write $[s]$ for $[s]_{\sigma _H}.$ Let U denote the subsemigroup

$$ \begin{align*}\{u\in\text{Stab}(H) \mid [u]\in G\}\end{align*} $$

of $\text {Stab}(H).$ For each $i\in \{1, \ldots , n\},$ select $x_i, y_i\in U$ such that $[x_i]=a_i$ and $[y_i]=a_i^{-1}.$ Also, fix an element $h\in H.$

The remainder of this proof is organised as follows.

1. (1) We show that for each $i\in \{1, \ldots , n\},$ there exist $\alpha (i), \beta (i)\in \mathbb {N}$ such that $x_i^{\alpha (i)}h=hy_i^{\beta (i)}.$

2. (2) We build a finitely generated subsemigroup T of S such that $T\cap H=$ $\{hu \mid u\in U\}.$

3. (3) We find a finite group K and a homomorphism $\theta : \Gamma (H)\to K$ such that $\theta (b)\notin \theta (G).$

(1) Let $i\in \{1, \ldots , n\},$ and write $x=x_i$ , $y=y_i.$ We claim that $hy\in \langle h, x\rangle .$ Since S is weakly subsemigroup separable, it suffices to show that $hy$ cannot be separated from $\langle h, x\rangle $ by a finite index congruence. Suppose that $\sim $ is a finite index congruence on S. Then there exist $k, \ell \in \mathbb {N}$ with $k<\ell $ such that $hx^k \sim hx^{\ell }.$ The elements $[x]$ and $[y]$ are inverses of each other in the group $\Gamma (H)$ and so, by the definition of $\sigma _H$ , $hy=hx^ky^{k+1}$ . Hence,

$$ \begin{align*}hy=hx^ky^{k+1}\sim hx^{\ell}y^{k+1}=hx^{\ell-k-1}\in\langle h, x\rangle,\end{align*} $$

which completes the proof of the claim. Since $x\in \text {Stab}(H)$ and H is not a group, we cannot have $hy\in \langle x\rangle .$ Post-multiplying $hy$ by an appropriate power of $y,$ we deduce that $hy^j=uh$ for some $j\in \mathbb {N}$ and $u\in \langle h, x\rangle .$

Now we claim that $hx^j\in \langle h, u\rangle $ . Again, it suffices to show that $hx^j$ cannot be separated from $\langle h, u \rangle $ in a finite index congruence. Suppose that $\sim $ is a finite index congruence on $S.$ Then there exist $k, \ell \in \mathbb {N}$ with $k<\ell $ such that $hy^{jk} \sim hy^{j\ell }.$ Recalling that $[x]$ and $[y]$ are mutually inverse and that $hy^j=uh$ , we deduce

$$ \begin{align*}hx^j=hy^{jk}x^{j(k+1)} \sim hy^{j\ell}x^{j(k+1)}=hy^{j(\ell-k-1)}=u^{\ell-k-1}h\in\langle h, u\rangle,\end{align*} $$

establishing the claim. Since $uh\in H,$ we cannot have $hx^j\in \langle u\rangle ,$ for then $(hx^j)h\in H\cap H^2,$ contradicting that H is not a group. It follows that $hx^j=w_1hw_2$ for some $w_1\in \langle u\rangle ^1$ and $w_2\in \langle h, u\rangle ^1\subseteq \langle h, x\rangle ^1.$ Now $w_1h=hy^{jk}$ for some $k\in \mathbb {N}_0$ ; so $hx^j\neq w_1h$ as $[x]$ has infinite order in $\Gamma (H),$ and hence $w_2\neq 1.$ We claim we cannot write $w_2$ as $sht$ for some $s\in \langle x\rangle ^1$ and $t\in \langle h, x\rangle ^1.$ Indeed, if we could, then since $w_1h\in H$ and $x\in \text {Stab}(H),$ we would have $hx^j\in H\cap H^n$ for some $n\geq 2,$ contradicting that H is not a group. Hence the claim holds. We must then have $w_2\in \langle u\rangle .$ But $u\in \langle h, x\rangle ,$ so we conclude that $u=x^m$ for some $m\in \mathbb {N}.$ Now set $\alpha (i)=m$ and $\beta (i)=j.$ We fix $\alpha (i)$ and $\beta (i)$ for the remainder of this proof.

(2) For each $i\in \{1, \ldots , n\}$ let $m_i=\max (\alpha (i), \beta (i)).$ For each $g\in G_0,$ select $u_g\in U$ such that $[u_g]=g.$ We define a finite set

$$ \begin{align*}W=\{x_1^{j_1}\cdots x_n^{j_n} \mid 0\leq j_i\leq m_i-1\text{ for }1\leq i\leq n\}\cup\{u_g \mid g\in G_0\} \subseteq U.\end{align*} $$

Let $X=\{x_i^{\alpha (i)} \mid 1\leq i\leq n\},$ and let T be the subsemigroup of S generated by

$$ \begin{align*}Z=X\cup\{hw \mid w\in W\}.\end{align*} $$

Note that $U \cap Z =X$ and $H \cap Z=\{hw \mid w\in W\}$ . We prove that $T\cap H=\{hu \mid u\in U\}.$

First, let $h^{\prime }\in T\cap H.$ Then $h^{\prime }=z_1\cdots z_k$ for some $z_j\in Z.$ If every $z_j\in X,$ then $h^{\prime }\in \text {Stab}(H),$ contradicting that H is not a group. Therefore, there exists the minimal j such that $z_j=hw$ for some $w\in W.$ Then for each $i<j$ we have $z_i\in X$ ; and hence $z_ih\in hU$ , as $xh \in hU$ for all $x \in X$ by (1). So, we deduce that $h^{\prime }=hw^{\prime }z_{j+1}\cdots z_k$ for some $w^{\prime }\in U.$ Let $u=w^{\prime }z_{j+1}\cdots z_k.$ We show that $u \in U$ . Suppose that $H \cap \{z_{j+1}, z_{j+2}, \ldots , z_k\}=\{z_{i_1}, \ldots , z_{i_m}\}$ where $j+1\leq i_1<\cdots <i_m\leq k.$ For each $\ell \in \{1, \ldots, m-1\}$ let us set $h_{\ell} = z_{i_\ell} \cdots z_{i_{\ell+1}-1}$ , and also $h_m = z_{i_m} \cdots z_k$. Then $h_{\ell }\in H$ for each $\ell \in \{1, \ldots , m\}.$ But then

$$ \begin{align*}h^{\prime}=(hw^{\prime}z_{j+1}\cdots z_{i_1-1})h_1\cdots h_m\in H\cap H^{m+1},\end{align*} $$

which contradicts that H is not a group. Therefore, we must have $z_i\in X$ for every $i\in \{j+1, \ldots , k\}.$ It follows that $u\in U$ and hence $h^{\prime }\in \{hu \mid u\in U\}.$

For the reverse containment, let $u\in U.$ Since G is abelian, we have $[u]=a_1^{k_1}\cdots a_n^{k_n}c$ for some $k_i\in \mathbb {Z}$ and $c\in G_0.$ Consider $i\in \{1, \ldots , n\}.$ If $k_i\geq 0,$ let $p_i\in \mathbb {N}_0$ and $r_i\in \{0, \ldots , \alpha (i)-1\}$ be such that $k_i=p_i\alpha (i)+r_i,$ and let $q_i=s_i=0.$ If $k_i<0,$ let $q_i\in \mathbb {N}$ and $s_i\in \{0, \ldots , \beta (i)-1\}$ be such that $k_i=-q_i\beta (i)+s_i,$ and let $p_i=r_i=0.$ Now let $t_i=\max (r_i, s_i).$ It follows that

$$ \begin{align*}hu=\bigl(x_1^{\alpha(1)}\bigr)^{q_1}\cdots\bigl(x_n^{\alpha(n)}\bigr)^{q_n}\bigl(hx_1^{t_1}\cdots x_n^{t_n}u_c\bigr)\bigl(x_1^{\alpha(1)}\bigr)^{p_1}\cdots\bigl(x_n^{\alpha(n)}\bigr)^{p_n};\end{align*} $$

so $hu\in T\cap H,$ and hence $T\cap H=\{hu \mid u\in U\}.$

(3) Choose $v\in \text {Stab}(H)$ such that $[v]=b.$ Then $hv\notin T$ by (2). Since S is weakly subsemigroup separable, there exist a finite semigroup P and a homomorphism $\phi : S\to P$ such that $\phi (hv)\notin \phi (T).$ Let $H_{\phi (h)}$ denote the $\mathcal {H}$ -class of $\phi (h),$ and let K be the finite group $\Gamma (H_{\phi (H)})$ . As in the proof of Lemma 3.5, the map $\theta : \Gamma (H)\to K,$ given by $\theta ([t])=[\phi (t)],$ is a homomorphism such that $\theta (b)\notin \theta (G),$ as required.

Groups satisfying the separability property with respect to the collection of all cyclic subgroups are known in the literature as cyclic subgroup separable groups or $\Pi _c$ groups. Such groups have received considerable attention: see, for instance, [Reference Burillo and Martino2, Reference Kim17, Reference Sokolov28, Reference Wong and Wong29].

From Proposition 3.8 we immediately deduce the following corollary.

In Section 6 we return to Schützenberger groups and consider the following question: if all the Schützenberger groups of a semigroup have a separability property, does the semigroup itself have that property?

4 Finitely generated commutative semigroups

In this section we give a characterisation of our separability properties in finitely generated commutative semigroups. We first note that finitely generated abelian groups are strongly subgroup separable. In fact, Mal’cev proved, more generally, that every polycyclic-by-finite group is strongly subgroup separable [Reference Mal’cev24]. However, when it comes to finitely generated commutative semigroups, Example 2.5 shows that there exist finitely generated commutative semigroups that are not even weakly subsemigroup separable. The question of when a finitely generated commutative semigroup is completely separable or strongly subsemigroup separable was considered by Kublanovskiĭ and Lesohin in [Reference Kublanovskiĭ and Lesohin18]. We briefly outline their set-up and results, without giving proofs.

Let S be a finitely generated commutative semigroup with finite generating set A. For $s \in S$ , let $C_s=A \cap \operatorname {\mathrm {Stab}}(H_s)$ . Then $C_s$ is finite and can be empty. We denote $|C_s|$ by $k_s$ . Then $\langle C_s \rangle ^1=\operatorname {\mathrm {Stab}}(H_s)$ . Consider the free commutative monoid $\mathbb {N}_0^{k_s}$ on $k_s$ generators. There is a canonical homomorphism $\phi : \mathbb {N}_0^{k_s} \to \operatorname {\mathrm {Stab}}(H_s).$ We note that $\operatorname {\mathrm {Stab}}(s)$ , the point stabiliser of s, is a submonoid of $\operatorname {\mathrm {Stab}}(H_s)$ . Let $W_s=\phi ^{-1}(\operatorname {\mathrm {Stab}}(s)) \leq \mathbb {N}_0^{k_s}$ be the pre-image of $\operatorname {\mathrm {Stab}}(s)$ . We can view $\mathbb {N}_0^{k_s}$ as a submonoid of the free abelian group $\mathbb {Z}^{k_s}$ . Consider the subgroup $G_s\leq \mathbb {Z}^{k_s} $ generated by $W_s$ . As $G_s$ is a subgroup of $\mathbb {Z}^{k_s}$ , we have $G_s \cong \mathbb {Z}^{m_s}$ for some $m_s \leq k_s$ . Using these parameters, Kublanovskiĭ and Lesohin were able to characterise when S is strongly subsemigroup separable as follows.

From the proof of this result they obtained two corollaries, summarised as follows.

We enhance the result of Corollary 4.2 by showing that for a finitely generated commutative semigroup, weak subsemigroup separability is also equivalent to complete separability. We also provide a new characterisation in terms of $\mathcal {H}$ -classes. This characterisation removes the need for the parameters $k_s$ and $m_s$ of Theorem 4.1. The proof we provide is independent of the work of Kublanovskiĭ and Lesohin, although the reader may note parallels between the methods used.

We now state the main result of this paper.

In order to prove Theorem 4.3, we first establish some notation and prove some preliminary results.

For an $\mathcal {H}$ -class H of a finitely generated commutative semigroup S define

$$ \begin{align*}I(H)=\bigcup\{H_s \mid s \in S, H_s \ngeq H\}.\end{align*} $$

Note that $I(H)$ is nonempty if and only if H is not the minimum $\mathcal {H}$ -class, in which case $I(H)$ is an ideal. In the proofs of the results of this section, when dealing with a nonminimal $\mathcal {H}$ -class H we shall often pass to the Rees quotient $S/I(H)$ . The following observations are needed to justify this strategy.

Let H be a nonminimal $\mathcal {H}$ -class, and denote $I(H)$ by I. Recall that $S/I$ consists of a zero element, namely I, and singleton sets $\{s\}$ where $s\in S\!\setminus \!I$ . For an $\mathcal {H}$ -class $H^{\prime }$ of S, let $H^{\prime }_I=\{[h]_I \mid h \in H^{\prime }\}$ denote the image of $H^{\prime }$ in $S/I$ . The following statements can easily be verified.

• • For an $\mathcal {H}$ -class $H^{\prime }$ of S, the set $H^{\prime }_I$ is an $\mathcal {H}$ -class of $S/I$ .

• • For an $\mathcal {H}$ -class $H^{\prime }$ of S, we have that $H^{\prime } \geq H$ if and only if $H^{\prime }_I \geq H_I$ .

• • The set $H_I$ is the minimum nonzero $\mathcal {H}$ -class in $S/I$ .

• • For $x \in S$ , we have that $x \in \operatorname {\mathrm {Stab}}(H)$ if and only $[x]_{I} \in \operatorname {\mathrm {Stab}}(H_{I})$ .

Lemma 4.4. Let S be a finitely generated commutative semigroup and let H be an $\mathcal {H}$ -class. Fix $s \in S$ with $H_s \geq H$ . Then

$$ \begin{align*}\operatorname{\mathrm{Stab}}(H)=\{x \in S^1 \mid H_{sx^n} \geq H \text{ for all } n \in \mathbb{N}\}.\end{align*} $$

Proof. First we assume that H is the minimum $\mathcal {H}$ -class in S. In this case $S^1=\operatorname {\mathrm {Stab}}(H)$ and $H_{sx^n} \geq H$ for all $s \in S$ , $x \in S^1$ and $n \in \mathbb {N}$ . The result follows.

Now assume that H is a nonminimal $\mathcal {H}$ -class. By noticing that $I(H)\cap \operatorname {\mathrm {Stab}}(H)=\emptyset $ and

$$ \begin{align*}I(H) \cap \{x \in S^1 \mid H_{sx^n} \geq H \text{ for all } n \in \mathbb{N}\} =\emptyset,\end{align*} $$

and factoring out by $I(H)$ , we may assume that H is the minimum nonzero $\mathcal {H}$ -class.

Let $x \in \operatorname {\mathrm {Stab}}(H)$ . For a contradiction assume that some $n \in \mathbb {N}$ has $H_{sx^n}\ngeq H$ . Then, as H is the minimum nonzero $\mathcal {H}$ -class, we have $sx^n=0$ . As $H_s \geq H$ , there exists $t \in S^1$ such that $st \in H$ . Then

$$ \begin{align*}0=sx^nt=stx^n \in Hx^n=H,\end{align*} $$

which is a contradiction.

Now assume that $H_{sx^n} \geq H$ for all $n \in \mathbb {N}$ . Fix $h \in H$ . Assume for a contradiction that $x \notin \operatorname {\mathrm {Stab}}(H)$ . Then $hx=0$ . As S is finitely generated and commutative, it is residually finite; see [Reference Lallement19, Theorem 3]. Let $\sim $ be an arbitrary finite index congruence on S. Then there exist $m,n \in \mathbb {N}$ , with $m < n$ , such that $sx^m \sim sx^n$ . As $H_{sx^m} \geq H$ , there exists $t \in S^1$ such that $sx^mt=h$ . Then

$$ \begin{align*}h=sx^mt \sim sx^nt=sx^mtx^{n-m}=hx^{n-m}=0.\end{align*} $$

As $\sim $ is arbitrary, we have shown we cannot separate h and 0 in a finite quotient. This contradicts S being residually finite.

Now we are ready to prove Theorem 4.3.

Proof of Theorem 4.3. From Proposition 2.1 we know that (1) implies (2) and (2) implies (3). We only need to show that (3) implies (4) and that (4) implies (1).

(3) $\Rightarrow $ (4) For a contradiction assume that S is weakly subsemigroup separable but has an infinite $\mathcal {H}$ -class H. Then the Schützenberger group $\Gamma (H)$ must also be infinite. As $H_{xa} \leq H_x$ for all $x,a \in S$ , the complement of $\operatorname {\mathrm {Stab}}(H)$ is an ideal of S. It follows that if X is a finite generating set for S, then $\operatorname {\mathrm {Stab}}(H)$ is generated by $\operatorname {\mathrm {Stab}}(H) \cap X$ and so is also finitely generated. As $\Gamma (H)$ is a quotient of a finitely generated commutative monoid, it is a finitely generated abelian group. As $\Gamma (H)$ is infinite, it must contain a subgroup isomorphic to $\mathbb {Z}$ . So $\Gamma (H)$ is not weakly subsemigroup separable by Example 2.5 and Proposition 2.3. However, this contradicts Corollary 3.6, so there cannot be an infinite $\mathcal {H}$ -class.

(4) $\Rightarrow $ (1) Let $h \in S$ . We show that there exists a finite index congruence on S such that the congruence class of h is a singleton. For $s \in S$ , let $A_s$ denote the archimedean component of s. Further, let $H$ denote the $\mathcal{H}$ -class of $h$ , let $\sigma$ denote the Schützenberger congruence $\sigma_H$ , and let $I=I(H)$ .

We consider two cases.

Case 1. The first case is that H is not a group. In particular, H is not the minimum $\mathcal {H}$ -class. Factoring out I, we may assume that H is the minimum nonzero $\mathcal {H}$ -class in S. As H is not a group, $H^2=\{0\}$ . So $A_h=A_0$ and $A_h$ is a nilpotent semigroup. It follows from Corollary 4.5 that $S^1$ is the disjoint union of $\operatorname {\mathrm {Stab}}(H)$ and $A_h$ . Consider any finite generating set for S, and write it as $X\cup Y$ , where $X\subseteq \operatorname {\mathrm {Stab}}(H)$ and $Y\subseteq A_h$ . Then $\langle X \rangle \subseteq \operatorname {\mathrm {Stab}}(H)$ and, as $A_h$ is nonempty, Y must also be nonempty. Note that $U= \langle Y \rangle $ is finite as $A_h$ is nilpotent. We may assume that X is nonempty, for otherwise $S=U$ is a finite semigroup and hence certainly completely separable.

Let $X=\{x_1, x_2, \ldots , x_m\}$ and let $\overline {X}=\{\overline {x}_1, \overline {x}_2, \ldots , \overline {x}_m\}$ be disjoint from X. Let $\operatorname {\mathrm {FC}}_{\overline {X}}$ denote the free commutative monoid on $\overline {X}$ . Let $\phi : \operatorname {\mathrm {FC}}_{\overline {X}} \to \operatorname {\mathrm {Stab}}(H)$ be the unique extension to a homomorphism of the map given by $\overline {x_i} \to x_i$ . For $u \in U$ define

$$ \begin{align*}I_u=\{w \in \operatorname{\mathrm{FC}}_{\overline{X}} \mid u\phi(w)\in H\}.\end{align*} $$

Suppose that $w\in I_u$ and $z\in \operatorname {\mathrm {FC}}_{\overline {X}}$ . Then since $u\phi (w)\in H$ and $\phi (z)\in \text {Stab}(H)$ , we deduce that $u\phi (wz)=(u\phi (w))\phi (z)\in H$ . Thus, if $I_u$ is nonempty, then it is an ideal of $\operatorname {\mathrm {FC}}_{\overline {X}}$ .

The monoid $\operatorname {\mathrm {FC}}_{\overline {X}}$ is isomorphic to $\mathbb {N}_0^{|X|}$ . Ideals of $\mathbb {N}_0^{|X|}$ are upward closed sets under the componentwise ordering on tuples. It is well known that this partially ordered set has no infinite antichains (Dickson’s lemma). From this follows the well-known fact that every ideal of $\operatorname {\mathrm {FC}}_{\overline {X}}$ is finitely generated as an ideal.

Let $U^{\prime }=\{u\in U \mid I_u\neq \emptyset \}$ . For each $u\in U^{\prime }$ let $Z_u$ be a finite generating set for $I_u$ , and let

$$ \begin{align*}Z=\bigcup_{u \in U^{\prime}}Z_u.\end{align*} $$

As U is finite, we have that Z is finite. For each $z\in Z$ , we have

$$ \begin{align*}z={\overline{x}_1}^{\alpha_1(z)}{\overline{x}_2}^{\alpha_2(z)}\cdots {\overline{x}_m}^{\alpha_m(z)}\end{align*} $$

for some $\alpha _i(z)\in \mathbb {N}_0$ . Define

$$ \begin{align*}n=\text{max}\{\alpha_i(z) \mid z\in Z, 1\leq i\leq m\}.\end{align*} $$

Now let $\sim $ be the congruence on S generated by the set $\{(x_i^n,x_i^{n+|H|}) \mid 1 \leq i \leq m \}.$ Each of the finitely many generators of $S/{\sim}$ is periodic and $S/{\sim}$ is commutative, so $S/{\sim}$ is finite. We now show that $[h]_{\sim }=\{h\}$ .

Let $t \sim h$ . We need to show that $t=h$ . Clearly it is sufficient to assume that t is obtained from h by a single application of a pair from the generating set of $\sim $ . So, let $h=sx_i^p$ and $t=sx_i^q$ where $1 \leq i \leq m$ , $s \in S$ and $\{p,q\} = \{n,n+|H|\}$ .

If $(p,q)=(n,n+|H|)$ then $t=hx_i^{|H|}$ . Since $x_i \in \operatorname {\mathrm {Stab}}(H)$ , $[x_i]_{\sigma }$ is an element of the Schützenberger group $\Gamma (H)$ . As $|\Gamma (H)|=|H|$ , it follows that $[x_i^{|H|}]_{\sigma }=[x_i]^{|H|}_{\sigma }=[1]_{\sigma }$ . Hence, $t=hx_i^{|H|}=h$ .

Now we consider the case when $(p,\ q)=(n+|H|, n)$ . As $h \notin \operatorname {\mathrm {Stab}}(H)$ , we have $s\in A_h\!\setminus \!\{0\}$ . Any decomposition of s into generators must contain at least one element from Y. Therefore, we have that $s=us'$ , where $u \in U'$ and $s' \in \operatorname {\mathrm {Stab}}(H)$ . Fix some

$$ \begin{align*}w={\overline{x}_1}^{\beta_1}{\overline{x}_2}^{\beta_2}\cdots {\overline{x}_m}^{\beta_m} \in \operatorname{\mathrm{FC}}_{\overline{X}}\end{align*} $$

such that $\phi (w)=s'$ . As $h=us'x_i^{n+|H|} \in H$ , we have that

$$ \begin{align*}w{\overline{x}_i}^{n+|H|}={\overline{x}_1}^{\beta_1}{\overline{x}_2}^{\beta_2}\cdots {\overline{x}_i}^{\beta_i+n+|H|} \cdots {\overline{x}_m}^{\beta_m} \in I_u.\end{align*} $$

Then there exist $z \in Z$ and $w'={\overline {x}_1}^{\gamma _1} \cdots {\overline {x}_m}^{\gamma _m} \in \operatorname {\mathrm {FC}}_{\overline {x}}$ such that

$$ \begin{align*}zw'=w{\overline{x}_i}^{n +|H|}.\end{align*} $$

For $1 \leq j \leq m$ , we have that

$$ \begin{align*} \alpha_j(z)+\gamma_j=\begin{cases} \beta_{j} &\text{if } j \neq i, \\ \beta_i+n+|H| &\text{if } j=i. \end{cases} \end{align*} $$

As $\alpha _i(z) \leq n$ by definition, it must be the case that $\gamma _i \geq |H|$ . Then

$$ \begin{align*}w{\overline{x}_i}^n=z{\overline{x}_1}^{\gamma_1}\cdots {\overline{x}_i}^{\delta} \cdots {\overline{x}_m}^{\gamma_m}\end{align*} $$

where $\delta =\gamma _i-|H| \geq 0$ . Hence $w{\overline {x}_i}^n \in I_u$ and so $sx_i^n=t \in H$ . As $h=tx_i^{|H|}$ , an argument similar to that above proves that $h=t$ , as required.

Case 2. Now we assume that H is a group. By Proposition 2.10, $A_h$ is either a group or the ideal extension of H by a nilpotent semigroup. Hence, $A_h \subseteq \operatorname {\mathrm {Stab}}(H)$ . If H is the minimum $\mathcal {H}$ -class of S, then $S^1=\text {Stab}(H).$ If H is not minimal, we may assume that it is the minimum nonzero $\mathcal {H}$ -class of S (by taking the Rees quotient by I), in which case we have $S^1\!\setminus \!\{0\}=\text {Stab}(H)$ by Corollary 4.5 and the fact that $A_h \subseteq \operatorname {\mathrm {Stab}}(H)$ .

In either case, let X be a finite generating set for $\operatorname {\mathrm {Stab}}(H)$ . As in Case 1, let $\overline {X}$ be a set in bijection with X, let $\operatorname {\mathrm {FC}}_{\overline {X}}$ denote the free commutative monoid on $\overline {X}$ , and let $\phi : \operatorname {\mathrm {FC}}_{\overline {X}} \to \operatorname {\mathrm {Stab}}(H)$ be the unique extension to a homomorphism of a bijection $\overline {X} \to X$ . Define

$$ \begin{align*}J=\{w \in \operatorname{\mathrm{FC}}_{\overline{X}} \mid \phi(w) \in H\}.\end{align*} $$

Then J is an ideal of $\operatorname {\mathrm {FC}}_{\overline {X}}$ . Let Z be a finite generating set for J (as an ideal), and let n be the maximal exponent of a generator appearing in any word $z \in Z$ .

Let $\sim $ be the congruence on S with generating set $\{(x^n,x^{n+|H|}) \mid x \in X\}.$ An argument essentially the same as that for Case 1 shows that $S/{\sim}$ is finite and $[h]_\sim =\{h\}$ , completing the proof of this direction and of the theorem.

5 Beyond finitely generated commutative semigroups

Given that the three properties of complete separability, strong subsemigroup separability and weak subsemigroup separability coincide for finitely generated commutative semigroups, the following questions naturally arise.

• • For commutative semigroups in general (not necessarily finitely generated), do the properties of complete separability and strong subsemigroup separability coincide?

• • For commutative semigroups in general, do the properties of strong subsemigroup separability and weak subsemigroup separability coincide?

• • For finitely generated semigroups in general (not necessarily commutative), do the properties of complete separability and strong subsemigroup separability coincide?

• • For finitely generated semigroups in general, do the properties of strong subsemigroup separability and weak subsemigroup separability coincide?

In this section we answer all of these questions in the negative. We first deal with the commutative case and then the finitely generated case.

5.1 Nonfinitely generated commutative semigroups

We give an example of a commutative semigroup that is weakly subsemigroup separable but not strongly subsemigroup separable. In order to do this, we first establish the following result.

Proposition 5.1. If a residually finite semigroup S has $\mathbb {N}$ as a homomorphic image then it is weakly subsemigroup separable.

Proof. Let $T \leq S$ be finitely generated and let $x \in S \! \setminus \! T$ . By assumption there exists a homomorphism $f: S \to \mathbb {N}$ . Let $n=f(x)$ . The set $I=\{m \mid m> n\} \subseteq \mathbb {N}$ is an ideal of $\mathbb {N}$ . Let $g: \mathbb {N} \to \mathbb {N}/I$ be the canonical homomorphism.

Since T is finitely generated and $f(st)>f(s)$ for any $s,t \in S$ , it follows that the set

$$ \begin{align*}Y=\{t \in T \mid f(t)=n\}\end{align*} $$

is finite. Since S is residually finite, there exist a finite semigroup P and homomorphism $h: S \to P$ such that $h(x) \notin h(Y)$ . Then $(g \circ f) \times h: S \to \mathbb {N}/I \times P$ separates x from T.

Example 5.2. Consider $S=\mathbb {N} \times \mathbb {Z}$ . Now, S is residually finite since it is the direct product of two residually finite semigroups. As the projection map onto the first factor gives a homomorphic image which is $\mathbb {N}$ , we conclude that S is weakly subsemigroup separable by Proposition 5.1.

We now show that S is not strongly subsemigroup separable. Consider $\mathbb {N} \times \mathbb {N} \leq S$ and the element $(2,0) \notin \mathbb {N} \times \mathbb {N}$ . Let $\sim $ be a finite index congruence on S. Then there exist $i, j \in \mathbb {Z}$ with $i<j$ such that $(1, i) \sim (1, j)$ . Then

$$ \begin{align*}(2, 0)=(1, i)(1, -i) \sim (1,j)(1, -i)=(2, j-i) \in \mathbb{N} \times \mathbb{N}.\end{align*} $$

Hence, S is a commutative semigroup which is weakly subsemigroup separable but not strongly subsemigroup separable.

Remark 5.3. The semigroup $\mathbb {N} \times \mathbb {Z}$ is also an example of a weakly subsemigroup separable semigroup that is a direct product of two semigroups one of which is not weakly subsemigroup separable. This is in contrast to the situation for residual finiteness: the direct product of two semigroups is residually finite if and only if both factors are residually finite [Reference Gray and Ruškuc11, Theorem 2].

We are left to find an example of a strongly subsemigroup separable commutative semigroup that is not completely separable. Our example is a group.

Example 5.4. Let $C_2$ denote the cyclic group of order 2. Let $G=C_2^{\mathbb {N}}$ be the Cartesian product of countably many copies of $C_2$ . By Lemma 2.4, G is not completely separable. But from Theorem 2.8, an abelian group is strongly subsemigroup separable if and only if it is torsion and, for each prime p, the primary p-component is bounded in the exponent. As every nonidentity element in G has order 2, G certainly satisfies these conditions. Hence G is strongly subsemigroup separable.

5.2 Finitely generated semigroups

In the previous section we showed that, in the statement of Theorem 4.3, being commutative is not on its own a sufficient condition. In this section we show that being finitely generated is also not on its own a sufficient condition. That is, we provide two examples of finitely generated semigroups, one of which is weakly subsemigroup separable but not strongly subsemigroup separable, and the other strongly subsemigroup separable but not completely separable.

First we give an example of a finitely generated semigroup that is weakly subsemigroup separable but not strongly subsemigroup separable. We do this by introducing a construction of semigroups and establishing necessary and sufficient conditions for this construction to be weakly subsemigroup separable and finitely generated. For the construction and proof, we will use the following notation. For a subset $Z \subseteq G$ of an abelian group G, let $X_Z=\{x_z \mid z \in Z\}$ be a a copy of $Z$ disjoint from G.

Construction 5.5. Let T be a semigroup, and let G be an abelian group such that there exists a surjective homomorphism $\phi : T \to G$ . Let $N=X_G \cup \{0\}$ be a null semigroup disjoint from T. Let $\mathcal {S}(T,G,\phi )=T \cup N$ , with multiplication inherited from T and N, and for $t \in T$ and $x_g \in X_G$ define the following multiplication:

$$ \begin{align*} \begin{split} & x_g\cdot t=x_{g \phi(t)}, \\ & t \cdot x_g=x_{g (\phi(t))^{-1}}, \\ & t \cdot 0=0 \cdot t =0. \\ \end{split} \end{align*} $$

An exhaustive check confirms this multiplication is associative; and therefore $\mathcal {S}(T,G,\phi )$ is a semigroup.

Remark 5.6. In Construction 5.5, the set $X_G$ forms a nongroup $\mathcal {H}$ -class and the Schützenberger group of this $\mathcal {H}$ -class is isomorphic to G.

To give necessary and sufficient conditions for $\mathcal {S}(T,G, \phi )$ to be weakly subsemigroup separable we use the following lemma.

Lemma 5.7. Let G be a weakly subgroup separable group. Let H be a finitely generated subgroup of G, let

$$ \begin{align*}U=\bigcup_{i=1}^n{Hg_i}\end{align*} $$

be a finite union of cosets of H, and let $x \in G \! \setminus \! U$ . Then x can be separated from U.

Proof. Let $i \in \{1,\ldots ,n\}$ . First we show that x can be separated from $Hg_i$ . As $x \notin Hg_i$ , we have $xg_i^{-1} \notin H$ . As G is weakly subgroup separable, there exist a finite group $K_i$ and homomorphism $\phi _i: G \to K_i$ such that $\phi _i(xg_i^{-1}) \notin \phi _i(H)$ . It follows that $\phi _i(x) \notin \phi _i(Hg_i)$ . Then

$$ \begin{align*}\phi_1 \times \phi_2 \times \cdots \times \phi_n : G \to K_1 \times K_2 \times \cdots \times K_n\end{align*} $$

is a homomorphism into a finite group that separates x from U.

Proposition 5.8. Let T be a semigroup, let G be an abelian group such that there exists a surjective homomorphism $\phi : T \to G$ , and let $S=\mathcal {S}(T,G,\phi )$ . Then S is weakly subsemigroup separable if and only if T is weakly subsemigroup separable and G is weakly subgroup separable.

Proof. ( $\Rightarrow $ ) First assume that S is weakly subsemigroup separable. Since T is a subsemigroup of S, it must be weakly subsemigroup separable by Proposition 2.3. Since G is abelian and isomorphic to a Schützenberger group of S, it follows from Corollary 3.9 that G is weakly subgroup separable.

( $\Leftarrow $ ) Now assume that T is weakly subsemigroup separable and G is weakly subgroup separable. Let $Y \subseteq S$ be a finite set, $U=\langle Y \rangle \leq S$ and $v \in S \! \setminus \! U$ . Let $N\subseteq S$ be as in Construction 5.5. Let $Y_1=Y \cap T$ and $Y_2=Y \cap N$ . We consider three cases.

Case 1. Assume that $v \in T$ . Note that $T \cap U=\langle Y_1 \rangle $ . As T is weakly subsemigroup separable and $v \notin \langle Y_1 \rangle $ , there exist a finite semigroup P and homomorphism $f: T \to P$ such that $f(v) \notin f(T \! \setminus \! \langle Y_1 \rangle )$ . Define $\overline {f}: S \to P^0$ by

$$ \begin{align*} s \mapsto \begin{cases} f(s) & \text{if } s \in T, \\ 0 &\text{otherwise.} \end{cases} \end{align*} $$

Then $\overline {f}$ is a homomorphism and $\overline {f}(v) \notin \overline {f} (U)$ .

Case 2. Now assume that $v\in N$ and $Y_2 = \emptyset $ . Then $U \subseteq T$ . Let $\sim $ be the congruence on S with classes T and N. Then $[v]_{\sim } \neq [u]_{\sim }$ for all $u \in U$ .

Case 3. Finally, assume that $v \in N $ and $Y_2 \neq \emptyset $ . Note that $0\in Y_2^2\subseteq N$ , and hence $v\neq 0$ . Let $v=x_g$ and $Y_2=\{x_{g_1},\ldots ,x_{g_n}\}$ . Let $H \leq G$ be the subgroup generated by the set $\phi (Y_1)$ . Then

$$ \begin{align*}U \cap N =X_Z \cup \{0\},\end{align*} $$

where $Z=\bigcup _{i=1}^n{Hg_i}.$ As $v \notin U$ , it follows that $g \notin \bigcup _{i=1}^n{Hg_i}$ . As G is weakly subgroup separable there exist a finite group K and homomorphism $f: G \to K$ such that $f(g) \notin \bigcup _{i=1}^n{f(Hg_i)}$ by Lemma 5.7. Let $P=\mathcal {S}(K,K,\text {id})=K\cup X_K\cup \{0\}$ . Let $\overline {f}: S \to P$ be given by

$$ \begin{align*} s \mapsto \begin{cases} (f \circ \phi)(s) & \text{if } s \in T, \\ x_{f(g)} & \text{if } s=x_g \text{ for some } g \in G, \\ 0 & \text{if } s=0. \end{cases} \end{align*} $$

Then it is straightforward to check that $\overline {f}$ is a homomorphism with $\overline {f}(v) \notin \overline {f}(U)$ .

The next lemma provides necessary and sufficient conditions for $\mathcal {S}(T,G,\phi )$ to be finitely generated.

Lemma 5.9. Let T be a semigroup, let G be an abelian group such that there exists a surjective homomorphism $\phi : T \to G$ , and let $S=\mathcal {S}(T,G,\phi )$ . Then S is finitely generated if and only if T is finitely generated.

Proof. If S is finitely generated, then, as T is the complement of an ideal, it must also be finitely generated. Conversely, if T is generated by a finite set Y, then it is easy to see that S is generated by $Y\cup \{x_{1_G}\}$ .

We provide an example of a weakly subsemigroup separable semigroup S that has the following properties:

• • S is finitely generated, noncommutative, but not strongly subsemigroup separable;

• • S has a Schützenberger group that is not weakly subsemigroup separable.

Example 5.10. Let $F_2=\{a,b\}^+$ be the free semigroup on $\{a, b\}$ . Let $\phi : F_2 \to \mathbb {Z}$ be given by $a \mapsto 1$ and $b \mapsto -1$ . As $F_2$ is completely separable [Reference Golubov7, Corollary 1] and $\mathbb {Z}$ is weakly subgroup separable, it follows that $\mathcal {S}(F_2,\mathbb {Z},\phi )$ is weakly subsemigroup separable by Lemma 5.8. Since $F_2$ is finitely generated, $\mathcal {S}(F_2,\mathbb {Z},\phi )$ is finitely generated by Lemma 5.9. It is clear that $\mathcal {S}(F_2,\mathbb {Z},\phi )$ is not commutative.

By Remark 5.6 we have that $X_{\mathbb {Z}}$ is an infinite nongroup $\mathcal {H}$ -class. Hence $\mathcal {S}(F_2,\mathbb {Z},\phi )$ is not strongly subsemigroup separable by Proposition 3.2. Also, the Schützenberger group of $X_{\mathbb {Z}}$ is isomorphic to $\mathbb {Z}$ and therefore is not weakly subsemigroup separable. Notice that, due to the way the right and left actions of $F_2$ on $X_{\mathbb {Z}}$ are defined, the $\mathcal {H}$ -class $X_{\mathbb {Z}}$ does not satisfy the condition of Lemma 3.5.

We conclude this section by exhibiting an example of a finitely generated semigroup which is strongly subsemigroup separable but not completely separable.

Example 5.11. Let $F_3=\{a,b,c\}^+$ be the free semigroup on the set $\{a, b,c\}$ . Let $I \leq F_3$ be the ideal generated by the set $\{x^2 \mid x \in F_3\}$ . Let $S=F_3/I$ be the Rees quotient of $F_3$ by I. We can view S as the set of all square-free words over the alphabet $\{a,b,c\}$ with a zero adjoined. Multiplication in S is concatenation, unless concatenation creates a word containing a subword that is a square, in which case the product is zero. Certainly, S is finitely generated by $\{a,b, c\}$ .

First we show that S is not completely separable. It is known that there exists an infinite square-free sequence $w=x_1x_2x_3\cdots $ over $\{a,b,c\}$ ; see [Reference Lothaire23, Ch. 2]. Then every finite prefix of w is a nonzero element of S. Let $w_i=x_1x_2\cdots x_i \in S$ . For $i < j$ , let $v_{i,j}=x_{i+1}x_{i+2}\cdots x_j \in S$ . Let $\sim $ be a finite index congruence class on S. Then there exist $i, j \in \mathbb {N}$ , with $i < j$ , such that $w_i \sim w_j$ . Then

$$ \begin{align*}w_j=w_iv_{i,j}\sim w_jv_{i,j}=w_iv_{i,j}v_{i,j}=0.\end{align*} $$

So we have shown that it is not possible for $0$ to be separated from $S \!\setminus \!\{0\}$ in a finite quotient. Hence, S is not completely separable.

Now let $T \leq S$ . Then $0 \in T$ . For $x \in S\!\setminus \! \{0\}$ let $|x|$ denote the length of x in terms of the generators $\{a,b,c\}$ . Now let $v \notin T$ and set $n = |v|$ . Let

$$ \begin{align*}I=\{x \in S \mid |x|> n\} \cup \{0\}.\end{align*} $$

Then I is an ideal. Clearly the Rees quotient $S/I$ is finite. Furthermore, $[v]_{I}=\{v\}$ . Hence, S is strongly subsemigroup separable.

6 Semigroups with finitely many $\mathcal {H}$ -classes

In Section 3 we asked which of our separability properties are inherited by Schützenberger groups. We showed in Corollaries 3.3 and 3.4 that the properties of complete separability and strong subsemigroup separability are inherited by Schützenberger groups. Although it is not true that every Schützenberger group of a weakly subsemigroup separable semigroup is itself weakly subsemigroup separable, we showed in Corollary 3.6 that weak subsemigroup separability is inherited by Schützenberger groups of commutative semigroups.

One may ask whether the properties are inherited in the opposite direction, that is, if every Schützenberger group of a semigroup S has a separability property must S itself satisfy the same property? This, however, turns out not to be true. Let $\mathscr {P}$ be any of the properties of complete separability, strong subsemigroup separability, weak separability or residual finiteness. A semigroup whose Schützenberger groups all have property $\mathscr {P}$ need not itself have property $\mathscr {P}$ . One example is the bicyclic monoid, given by the monoid presentation $\langle b,c \mid bc=1 \rangle $ . The bicyclic monoid is $\mathcal {H}$ -trivial, meaning that every $\mathcal {H}$ -class is a singleton, so every Schützenberger group is the trivial group and certainly completely separable. However, the bicyclic monoid is not even residually finite [Reference Clifford and Preston3, Corollary 1.12]. In fact this direction fails comprehensively even for commutative semigroups, as the next example shows.

Example 6.1. Let $A=\langle a \rangle \cong \mathbb {N}$ . Let $B=\{b_i \mid i \in \mathbb {N}\}\cup \{0\}$ be the countable null semigroup. Let $S=A \cup B$ with multiplication between A and B as follows:

$$ \begin{align*}a^ib_j=b_ja^i=\begin{cases} b_{j-i} & \text{for } j> i, \\ 0 & \text{otherwise}, \end{cases}\end{align*} $$

$$ \begin{align*}a^i0=0a^i=0.\end{align*} $$

An exhaustive case analysis shows that this evidently commutative multiplication is associative. It is straightforward to check that S is $\mathcal {H}$ -trivial. However, S is not residually finite. Suppose that $\sim $ is a finite index congruence on S. Then there exist $i, j \in \mathbb {N}$ , with $i < j$ , such that $b_i \sim b_j$ . Then

$$ \begin{align*}0=b_ia^{j-1} \sim b_ja^{j-1}=b_1.\end{align*} $$

So we cannot separate $0$ and $b_1$ in a finite quotient; and hence S is not residually finite.

In the remainder of this section we restrict our attention to the class of semigroups that have only finitely many $\mathcal {H}$ -classes, or, equivalently, semigroups that have only finitely many left and right ideals. This is motivated by the following result.

We investigate whether there are analogous results for the properties of complete separability, strong subsemigroup separability and weak subsemigroup separability.

For complete separability, the analogous result holds.

From Corollary 3.3, we know that every Schützenberger group of a strongly subsemigroup separable semigroup is itself strongly subsemigroup separable. However, even when a semigroup has only finitely many $\mathcal {H}$ -classes, every Schützenberger group being strongly subsemigroup separable does not guarantee that the semigroup is strongly subsemigroup separable, as the following example demonstrates.

The final property to consider is that of weak subsemigroup separability. Of all the separability properties considered in this paper, this is the only one that is not necessarily inherited by Schützenberger groups, as demonstrated by Example 5.10. However, when we restrict to a semigroup with only finitely many $\mathcal {H}$ -classes, the following problem remains open.

Indeed, we do not even know if either direction of the above statement holds. In the rest of this section, we restrict our attention to locally finite semigroups with only finitely many $\mathcal {H}$ -classes. By concentrating on this smaller class of semigroups we are be able to invoke Lemma 2.7, which says that a semigroup that is both residually finite and locally finite is weakly subsemigroup separable. This line of investigation allows us to give the following partial answer to Open Problem 6.6.

In particular, we note that a commutative semigroup with finitely many $\mathcal {H}$ -classes is weakly subsemigroup separable if and only if all its Schützenberger groups are weakly subsemigroup separable. To prove Theorem 6.7 we make use of several lemmas.

A semigroup S is called an epigroup if every element of S has a power which lies in a subgroup of S.

At this point on our path to prove Theorem 6.7, we introduce Green’s relation $\mathcal {J}$ on a semigroup S:

$$ \begin{align*}\mathcal{J}=\{(x,y) \mid S^1xS^1=S^1yS^1\}.\end{align*} $$

This is an equivalence relation and $\mathcal {H} \subseteq \mathcal {J}$ . It is also true that any ideal is a union of $\mathcal {J}$ -classes. For more on Green’s $\mathcal {J}$ relation see [Reference Howie16, Ch. 2].

We show that an epigroup with finitely many $\mathcal {J}$ -classes is locally finite if and only if all its maximal subgroups are locally finite. We first consider the case when a semigroup with a zero only has two $\mathcal {J}$ -classes.

A semigroup S with a zero is called 0-simple if $S^2 \neq \{0\}$ and the only ideals of S are $\{0\}$ and S. A 0-simple semigroup is called completely 0-simple if it is both 0-simple and an epigroup. For equivalent definitions of completely 0-simple semigroups see [Reference Howie16, Theorem 3.2.11]. Rees showed that the class of completely 0-simple semigroups coincides with the class of Rees matrix semigroups over zero-groups [Reference Howie16, Theorem 3.2.3]. For a group G, let the zero-group $G^0$ be G with a zero adjoined. Let $I,\Lambda $ be nonempty sets and $P=(p_{\lambda i})$ be a $\Lambda \times I$ matrix with entries from $G^0$ such that no row or column of P consists entirely of zeros. The Rees matrix semigroup over the zero-group $G^{0}$ is $S=M^0[G;I,\Lambda ;P]=(I \times G \times \Lambda ) \cup \{0\}$ with multiplication given as follows:

$$ \begin{align*} \begin{split} (i,a,\lambda)(j,b,\mu)&=\begin{cases} (i,ap_{\lambda j},\mu) &\text{if } p_{\lambda i} \neq 0, \\ 0 &\text{if } p_{\lambda i} =0, \end{cases} \\ (i,a,\lambda)0&=0(i,a,\lambda)=0\cdot0=0. \end{split} \end{align*} $$

In a completely 0-simple semigroup $M^0[G;I, \Lambda; P]$ , the $\mathcal {H}$ -class of the element $(i,a ,\lambda )$ is $\{i\} \times G \times \{\lambda \}$ . This is a maximal subgroup if and only if $p_{\lambda i} \neq 0$ , in which case it is isomorphic to G.

Proof. Suppose that S is locally finite. As S has a subsemigroup isomorphic to G, and local finiteness is inherited by subsemigroups, it follows that G is locally finite.

Now suppose that G is locally finite. Let $T=\langle (i_1,g_1,\lambda _1),\, \ldots , \, (i_n,g_n,\lambda _n) \rangle $ . We show that the intersection of T with a nonzero $\mathcal {H}$ -class is finite. Let K be the subgroup of G generated by the set

$$ \begin{align*}\{g_i \mid 1 \leq i \leq n\} \cup \{p_{\lambda_k i_l} \mid p_{\lambda_k i_l}\neq0, \, 1 \leq k, \, l\leq n\}.\end{align*} $$

As K is finitely generated, it is finite. Let $H=\{i\}\times G \times \{\lambda \}$ be a nonzero $\mathcal {H}$ -class. Then $H \cap T \subseteq \{i\} \times K \times \{\lambda \}$ and hence $H \cap T$ is finite. As T can only intersect finitely many nonzero $\mathcal {H}$ -classes, it follows that T is finite.

Lemmas 6.8 and 6.13 together yield the following result.

We are now in a position to prove Theorem 6.7.

If there is any hope of solving Open Problem 6.6, we must consider cases where an infinite Schützenberger group is weakly subsemigroup separable but not solvable. The authors are aware of only a limited number of such groups. One such example is the Grigorchuk group, which is a finitely generated infinite torsion group that is weakly subgroup separable (and hence weakly subsemigroup separable); see [Reference Grigorchuk and Wilson13]. In particular, the following problem remains open.