2 Completely simple, stable and inverse semigroups

In this section we will show that all DSC semigroups belonging to certain classes are in fact groups. Specifically, we will do this for completely simple, stable and inverse semigroups, in that order.

A nonempty subset I of a semigroup S is said to be an ideal if for all $x\in I$ and all $s\in S$ we have $sx,xs\in I$ . A semigroup is said to be simple if it has no ideals other than itself.

Let S be a semigroup, and let E be the set of idempotents of S. The relation $\leq $ on E defined by $e\leq f {\iff} ef=fe=e$ is a partial order. Any minimal element in this partial order is said to be primitive. A simple semigroup S is said to be completely simple if it has a primitive idempotent. All finite simple semigroups are completely simple.

There is a complete structural description of completely simple semigroups, originally due to Suschkewitsch [Reference Suschkewitsch8]. Let G be a group, let I and J be two index sets, and let $P=(p_{ji})_{j\in J,i\in I}$ be a $J\times I$ matrix with entries from G. The Rees matrix semigroup $\mathcal {M}[G;I,J;P]$ is the set $I\times G\times J$ with multiplication $(i , g , j)(k , h , l) = (i, g p_{jk} h , l)$ . Suschkewitsch’s theorem then asserts that a semigroup S is completely simple if and only if it is isomorphic to some Rees matrix semigroup $\mathcal {M}[G;I,J;P]$ (see [Reference Howie4, Theorem 3.3.1]).

Proof. Let S be a completely simple semigroup. By Suschkewitsch’s theorem, without loss of generality we may assume $S = \mathcal {M}[G;I,J;P]$ . If $|I|> 1$ then pick $i \neq k \in I$ . Now consider the set

$$ \begin{align*} \rho = \{ ((i,g,j),(k,h,j)) \colon g,h\in G,\ j\in J\} \cup \{ (l,g,j),(l,h,j)) \colon l \in I,\ g,h \in G ,\ j \in J\}. \end{align*} $$

It is routine to verify that $\rho $ is a diagonal subsemigroup. For an arbitrary $j\in J$ it is easily seen that $((i,1,j),(k,1,j)) \in \rho $ but $((k,1,j),(i,1,j)) \notin \rho $ . Hence, $\rho $ is not a congruence, contradicting $\mathcal {M}[G;I,J;P]$ being DSC. Therefore, it must then be the case that $|I| = 1$ and, analogously, $|J| = 1$ . It now easily follows that $S\cong G$ , a group.

We know from Theorem 2.1 that a DSC semigroup S must be simple. Whenever we can show that under some additional assumptions S must in fact be completely simple, Theorem 2.2 will force S to be a group. We deploy this strategy for stable and inverse semigroups.

In order to define stability, we need to introduce Green’s equivalences $\mathcal {R}$ , $\mathcal {L}$ and $\mathcal {J}$ on a semigroup S:

$$ \begin{align*} s\mathcal{R} t \Leftrightarrow sS^1=tS^1,\quad s\mathcal{L} t \Leftrightarrow S^1s=S^1t,\quad s\mathcal{J} t \Leftrightarrow S^1sS^1=S^1tS^1 \end{align*} $$

(for a detailed introduction see [Reference Howie4, Section 2.1]). We then say that S is stable if the following implications hold:

$$ \begin{align*} x \mathcal{J} sx \Rightarrow x \mathcal{L} sx \quad\text{and}\quad x \mathcal{J} xs \Rightarrow x \mathcal{R} xs \quad\text{for all } s,x\in S. \end{align*} $$

All finite semigroups are stable [Reference Rhodes and Steinberg7, Theorem A.2.4]. By [Reference Rhodes and Steinberg7, Theorem A.4.15] every stable simple semigroup is completely simple, and so we have the following corollary.

A semigroup S is said to be periodic if for every $s\in S$ there exist distinct $m,n\in \mathbb {N}$ such that $s^m=s^n$ . Every finite semigroup is periodic.

We finish this section with a discussion of inverse semigroups. An element $s\in S$ is said to be regular if $sts=s$ for some $t\in S$ . If in addition $tst=t$ we say that s and t are (semigroup) inverses of each other. It is known that an element is regular if and only if it has an inverse (see [Reference Howie4, page 51]). The semigroup S is regular if every element is regular, and it is inverse if every element has a unique inverse.

We started this paper by introducing the concept of diagonal subalgebra for general algebras and we have looked at the cases when the algebra is a group or semigroup. At this point we have enough theory to answer this question for inverse semigroups as well.

3 Some further infinite non-DSC semigroups

As we have seen in the previous section, there exist no nongroup completely simple, stable or inverse DSC semigroups. So if we want to find a non-DSC semigroup we will have to look a bit harder. We know that any DSC semigroup is simple, and we will explore different constructions leading to examples of simple semigroups.

One such is the Rees matrix semigroup construction, which we have already encountered, but which can be deployed in greater generality. Specifically, instead of starting with a group G, we can start with an arbitrary semigroup S. Keeping the remainder of the definition from Section 2 unchanged, we obtain the Rees matrix semigroup $\mathcal {M}[S;I,J;P]$ . It is easy to check (for example, by using [Reference Howie4, Corollary 3.1.2]), that $\mathcal {M}[S;I,J;P]$ is simple if and only if S is simple. By an analogous proof to that of Theorem 2.2, we can see that if $S^\prime = \mathcal {M}[S;I,J;P]$ is DSC then $|I| = |J| = 1$ . Hence, $S^\prime $ has multiplication $x \cdot y = xay$ for some $a \in S$ . We claim that if $S^\prime $ is DSC then S must also be DSC. Let $\rho $ be a diagonal subsemigroup of S and define $\rho ^\prime \subseteq S^\prime \times S^\prime $ by $(x,y) \in \rho ^\prime {\iff} (x,y) \in \rho $ . As $\rho $ contains the diagonal, so does $\rho ^\prime $ . And if we have $(x,y),(z,t) \in \rho ^\prime $ then $(x,y),(z,t) \in \rho $ , which gives $(xaz,yat) \in \rho $ and hence ${(x\cdot z,y \cdot t) \in \rho ^\prime }$ . Hence, $\rho ^\prime $ is a diagonal subsemigroup and, as $S^\prime $ is DSC, a congruence on $S^\prime $ . It follows that $\rho $ is also a congruence on S, and therefore S must be DSC.

We have just seen that in order for $\mathcal {M}[S;I,J;P]$ to be DSC, S must also be DSC. So for finding more DSC semigroups, taking a Rees matrix semigroup will not be much help as we would have to know whether our original semigroup was DSC to begin with.

Another way to construct simple semigroups is the Bruck–Reilly extension [Reference Howie4, Section 5.6]. Here we take a semigroup S and an endomorphism $\theta $ of S. The Bruck–Reilly extension $ {\textsf {BR}}(S,\theta )$ is the set $\mathbb {N}_0 \times S \times \mathbb {N}_0$ with multiplication

$$ \begin{align*} (m,s,n)(p,t,q) = (m - n + k , (s \theta^{k-n})(t \theta^{k - p}) , q - p + k)\quad \text{where }k = \max(n,p). \end{align*} $$

Under certain conditions the semigroup $ {\textsf {BR}}(S,\theta )$ is simple. The conditions themselves will not concern us, but the reader can consult Proposition 5.6.6 and Exercise 5.25 in [Reference Howie4] for two examples. Unfortunately, again, this construction will not work for us. One can show directly that no Bruck–Reilly extension is DSC, but it is easier to use some of the theory we have built up in the previous section. Additionally, we will need the following result.

Proof. Let $\rho $ be a diagonal subsemigroup of $S/\sigma \times S / \sigma $ . Define $\rho ^\prime \subseteq S \times S$ by

$$ \begin{align*} (x,y) \in \rho ^\prime {\iff} (x\sigma , y\sigma ) \in \rho. \end{align*} $$

For each $x \in S$ , $(x\sigma ,x \sigma ) \in \rho $ so $(x,x) \in \rho ^\prime $ . If $(x,y),(z,t) \in \rho ^\prime $ then $(x \sigma ,y \sigma ), (z \sigma , t\sigma ) \in \rho $ . This implies $(xz \sigma ,yt \sigma ) \in \rho $ and so $(xz,yt) \in \rho ^\prime $ . Hence, $\rho ^\prime $ is a diagonal subsemigroup of $S \times S$ which by assumption is also a congruence. Now

$$ \begin{align*} &(x \sigma,y \sigma) \in \rho \Rightarrow (x,y) \in \rho^\prime \Rightarrow (y,x) \in \rho^\prime \Rightarrow (y \sigma , x \sigma ) \in \rho, \\ & (x \sigma , y \sigma) , (y \sigma , z \sigma) \in \rho \Rightarrow (x,y),(y,z) \in \rho^\prime \Rightarrow (x,z)\in \rho^\prime \Rightarrow (x \sigma,z \sigma) \in \rho. \end{align*} $$

Hence, $\rho $ is a congruence and $S/\sigma $ is DSC.

In particular, the homomorphic image of a DSC semigroup is DSC. This result makes it much easier to check if a semigroup is not DSC.

The final kind of simple semigroups we will look at in this section are the generalised Baer–Levi semigroups (see [Reference Clifford and Preston3, Section 8.1]). Here we take two infinite cardinals p and q with $p \geq q$ . Let X be a set of cardinality p and consider the set

$$ \begin{align*} B(p,q) = \{ f\colon X \to X\: \colon\: f \text{ is injective and } | X \setminus Xf| = q \}. \end{align*} $$

Under the usual composition of functions $B(p,q)$ forms a semigroup. This semigroup is always simple (even right simple) and contains no idempotents [Reference Clifford and Preston3, Theorem 8.2], so potentially makes a good candidate for a nongroup DSC semigroup. Unfortunately, this hope too turns out to be unjustified.

Proof. Let $B = B(p,q)$ . Consider the set

$$ \begin{align*} \rho = \{ (f,g) \in B \times B \colon (X \setminus X f) \cap (X \setminus X g) \neq \emptyset \}. \end{align*} $$

It is easy to check that $\rho $ is a diagonal subsemigroup of $B \times B$ . It is also clear that $\rho $ is symmetric, so we will show that $\rho $ is not transitive. Partition X into two sets A and B with cardinality p. Let $A^\prime $ and $ B^\prime $ be subsets of A and B respectively, both with cardinality q. Let $x \in A^\prime $ . The sets $X \setminus A^\prime , X \setminus (B^\prime \cup {x} )$ and $X \setminus B^\prime $ all have cardinality p. So there are bijections

$$ \begin{align*} f^\prime\colon X \to X \setminus A^\prime ,\quad g^\prime \colon X \to X \setminus (B^\prime \cup {x})\quad \text{and}\quad h^\prime \colon X \to X \setminus B^\prime. \end{align*} $$

Each can be extended to an injection from X to itself; call these functions $f,g$ and h, respectively. Note that $X\setminus Xf = A^\prime $ , $X\setminus Xg = B^\prime \cup \{ x \}$ and $X\setminus Xh = B^\prime $ . So $f,g,h \in B$ and $(f,g),(g,h) \in \rho $ , but $(f,h) \notin \rho $ . Hence, $\rho $ is not transitive.

4 Nongroup DSC semigroups

In Theorem 3.1 we saw that any quotient of a DSC semigroup must be DSC. So if we have a semigroup S with congruence $\sigma $ , for S to be DSC so must $S/\sigma $ . This gives us an extra constraint on being DSC. So we will try looking at congruence-free semigroups. One rather general such construction was introduced by Byleen in [Reference Byleen2]. The construction uses the notions of monoid actions and presentations, which we now briefly review.

Let S be a monoid with identity $1$ and let A be a set. A right action of S on A is a function $A\times S\rightarrow A$ , $(a,s)\mapsto a\triangleright s$ such that $(a\triangleright s)\triangleright t=a\triangleright (st)$ and $a\triangleright 1=a$ for all $a\in A$ and all $s,t\in S$ . The action is said to be faithful if for any two distinct $s,t\in S$ there exists $a\in A$ such that $a\triangleright s\neq a\triangleright t$ . A left action of S on a set B is defined analogously. For more details, see [Reference Howie4, Section 8.1].

Now suppose that X is an alphabet and denote by $X^\ast $ the free monoid on X; it consists of all words over X, including the empty word $\epsilon $ , and the operation is concatenation. A monoid presentation is a pair of the form $\langle X\mid R\rangle $ , where $R\subseteq X^\ast \times X^\ast $ . The monoid defined by this presentation is $S=X^\ast /\rho $ , where $\rho $ is the congruence generated by R. The elements of this semigroup are the $\rho $ -classes $[u]$ , $u\in X^\ast $ . An elementary sequence with respect to $\langle X\mid R\rangle $ is any sequence $w_1,w_2,\ldots , w_n$ ( $n\geq 1$ ) of words from $X^\ast $ such that for every $i=1,\ldots ,n-1$ we have $w_i=w'uw"$ , $w_{i+1}=w'vw"$ for some $w',w"\in X^\ast $ and some $(u,v)\in R$ or $(v,u)\in R$ . For two words $u,v\in X^\ast $ we have $[u]=[v]$ if and only if there exists an elementary sequence starting at u and ending in v. We will abuse notation and write u instead of $[u]$ for a typical element of S, and $u=v$ instead of $(u,v)$ for a typical element of R. For a more detailed basic introduction to presentations see [Reference Howie4, Section 1.6].

Definition 4.1. Let S be a monoid with identity $1$ , and let A and B be sets that are disjoint from each other and from S. Let $\alpha \colon A\times S\rightarrow A, (a,s)\mapsto a\triangleright s$ be a right action, and let $\beta \colon S \times B \rightarrow B, (s,b) \mapsto s \triangleleft b$ be a left action. Let $W = A \cup B \cup S$ and let P be an $A \times B$ matrix with entries in W. Let ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ denote the monoid with monoid presentation

$$ \begin{align*} \langle W \mid ab = p_{a,b} ,\ as = a \triangleright s ,\ sb = s \triangleleft b ,\ st = s \cdot t ,\ 1 = \epsilon \ (a \in A ,\ b \in B ,\ s,t \in S)\rangle. \end{align*} $$

In the above presentation, the relation $st=s \cdot t$ should be interpreted as a word of length $2$ , namely $st$ , being equal to a word of length $1$ , the product of s and t in S. In other words, those relations represent the inclusion of the Cayley table of S in the defining presentation for ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ .

In [Reference Byleen2] it is shown that any element of ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ can be written uniquely in the form $vsu$ where $v \in B^\ast , s \in S$ and $ u \in A^\ast $ . The monoid ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ has identity $1 = \epsilon $ . Calculations are easy in ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ as each relation (other than $1 = \epsilon $ ) replaces a word of length $2$ with a word of length $1$ . In general, this semigroup need not be DSC. We now introduce some additional conditions which will then imply DSC.

We will be interested in $2$ -transitive $A \times B$ matrices with entries in $W = A \cup B \cup S$ . As $|W| \geq |A|,|B|$ , the sets A and B will of necessity be infinite. For an explicit construction when A and B are countably infinite see [Reference Quick and Ruškuc6].

The proof of the following result closely follows Byleen’s proof showing that ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is congruence-free. Our proof will be divided into more cases as we have to work around the parts that use symmetry and transitivity.

Proof. Let $T = {\mathcal {C}^1(S;\alpha ,\beta ;P)}$ . We will show that $\Delta =\{ (t,t)\colon t\in T\}$ and $T \times T$ are the only diagonal subsemigroups of $T \times T$ . To this end we will consider arbitrary distinct $vsu, ytx \in T$ and show that the subsemigroup $\rho $ of $T\times T$ generated by $(vsu,ytx)$ and $\Delta $ is equal to $T \times T$ . Note that if $W \times W \subseteq \rho $ , then for any $w = w_1 \cdots w_n $ and $ {w^\prime = w^\prime _1 \cdots w^\prime _m \in T}$ we have $(w_i,1),(1,w_j^\prime ) \in \rho $ for all $i,j$ , and so

$$ \begin{align*} (w,w^\prime) = (w_1 , 1)\cdots (w_n,1)(1,w_1^\prime)\cdots (1,w_m^\prime) \in \rho. \end{align*} $$

Hence, under this assumption, $\rho = T \times T$ . So it suffices to show that $W \times W \subseteq \rho $ . To do this we will first prove several intermediate claims.

Proof. The result is trivial if $u = \epsilon $ so let $u = a_1 \cdots a_n$ . By $2$ -transitivity of P there are $b_1 , \ldots , b_n \in B$ such that $a_1 b_1 = 1 , a_2 b_2 = b_1 , \ldots , a_n b_n = b_{n-1}$ . If we let $\lambda = (b_n,b_n)$ , then

$$ \begin{align*} (1,1) = (a_1 \cdots a_n b_n ,a_1 \cdots a_n b_n) = (u,u) \lambda .\\[-34pt] \end{align*} $$

Proof. By Claim 2, there exist $\lambda ^\prime \in \Delta $ and $\epsilon \neq p\in A^\ast $ such that $(u,x)\lambda ^\prime = (1,p)$ or $(p,1)$ . We will assume $(u,x)\lambda ^\prime = (1,p)$ ; the case when $(u,x)\lambda ^\prime = (p,1)$ is dual. Let ${p = a_1 \cdots a_n}$ . There exist $b_0,\ldots ,b_n \in B$ with $a_1 b_1 = b_0 , \ldots , a_{n-1} b_{n-1} = b_{n-2} , a_n b_n = b_{n-1}$ and $b_n \neq b_0$ , so $pb_n = b_0$ . Now we can pick $a \in A$ such that $ab_n = w_1$ and ${ab_0 = w_2}$ . If we let $\mu = (a,a)$ and $\lambda = \lambda ^\prime (b_n , b_n)$ , then

$$ \begin{align*} \mu (u,x) \lambda = (a,a)(1,p)(b_n,b_n) = (a b_n , a b_0) = (w_1 , w_2) .\\[-36pt] \end{align*} $$

The next three claims are dual to Claims 1, 2, 3 and we omit their proofs.

Now let $vsu,ytx \in T$ be distinct, and let $w_1,w_2 \in W$ be arbitrary. We will show $(w_1,w_2) \in \rho = \langle (vsu,ytx), \Delta \rangle $ .

If $u = x$ and $v = y$ then it must be the case that $s \neq t$ . By Claims 1 and 4 there exist $\lambda , \mu \in \Delta $ such that $(u,x)\lambda = (1,1) = \mu (v,y)$ . As the right action $\alpha $ is faithful, there is $a \in A$ such that $a_1 = a \triangleright s \neq a \triangleright t = a_2$ . From Claim 3, there exist $\lambda ^\prime , \mu ^\prime \in \Delta $ such that $\mu ^\prime (a_1,a_2) \lambda ^\prime = (w_1,w_2)$ . Thus,

$$ \begin{align*} (w_1,w_2) = \mu^\prime (a,a) \mu(vsu,ytx)\lambda \lambda^\prime \in \rho. \end{align*} $$

Now suppose that $u \neq x$ and $v = y$ (the case when $u = x$ and $v \neq y$ is dual). By Claims 2 and 4 there exist $\lambda , \mu \in \Delta $ and $\epsilon \neq p\in A^\ast $ such that $\mu (v,y) = (1,1)$ and $(u,x) \lambda = (p,1)$ (again the case when $(u,x) \lambda = (1,p)$ is dual). Let a be any element of A and let ${a_1 = a \triangleright s, a_2 = a \triangleright t}$ . As the words p and $1 = \epsilon $ are distinct, the words $a_1 p$ and $a_2$ are also distinct. Hence, by Claim 3, there exist $\lambda ^\prime , \mu ^\prime \in \Delta $ such that $\mu ^\prime (a_1 p ,a_2) \lambda ^\prime = (w_1,w_2)$ . Thus,

$$ \begin{align*} (w_1,w_2) = \mu^\prime (a,a) \mu (vsu,ytx) \lambda \lambda^\prime \in \rho. \end{align*} $$

If instead we had $u \neq x$ and $v \neq y$ then, by Claims 2 and 5, there exist ${\lambda ,\mu \in \Delta }$ and $\epsilon \neq p \in A^\ast $ , $\epsilon \neq q \in B^\ast $ such that

$$ \begin{align*} (u,x) \lambda \in \{ (1,p),(p,1)\}\quad\text{and}\quad \mu (v,y) = \{ (1,q),(q,1)\}. \end{align*} $$

Here we will write $(u,x) \lambda = (p_1,p_2)$ , noting that $| p_1 | \neq | p_2 |$ . We will treat the case when $\mu (v,y) = (1,q)$ ; the other case ( $(v,y) \mu = (q,1)$ ) is dual to this.

Write $q = b_1 \cdots b_n$ , and let a be any element of A. There exist $a_n,\ldots , a_1 \in A$ such that

$$ \begin{align*} a_n b_n = a,\quad a_{n-1} b_{n-1} = a_n ,\quad \ldots ,\quad a_1 b_1 = a_2. \end{align*} $$

So we have $a_1 q = a$ . The words $(a_1 \triangleright s) p_1$ and $(a \triangleright t) p_2$ are distinct (as $p_1$ and $p_2$ have different lengths), so, by Claim 3, there are $\lambda ^\prime $ , $\mu ^\prime \in \Delta $ such that

$$ \begin{align*} \mu^\prime ( (a_1 \triangleright s) p_1 , (a \triangleright t) p_2 ) \lambda^\prime = (w_1,w_2). \end{align*} $$

Hence,

$$ \begin{align*} (w_1,w_2) = \mu^\prime (a_1,a_1) \mu (vsu ,ytx) \lambda \lambda^\prime \in \rho. \end{align*} $$

So in all cases $(w_1,w_2) \in \rho $ and hence $W \times W \subseteq \rho $ .

From here on, when we refer to ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ , we will assume that $\alpha $ and $\beta $ are faithful actions, and that P is $2$ -transitive over $A \cup B \cup S$ .

We will show that there are nongroup DSC semigroups that are bisimple. The Green’s equivalence $\mathcal {D}$ on a semigroup S is defined to be $\mathcal {R} \circ \mathcal {L} = \mathcal {L} \circ \mathcal {R}$ (for a more detailed explanation see [Reference Howie4, Section 2.1]). We say a semigroup S is bisimple if $\mathcal {D} = S \times S$ .

Proof. (i) From Theorem 4.3, we have seen that ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is DSC. To see that it is not a group, take any $a \in A$ and $vsu \in {\mathcal {C}^1(S;\alpha ,\beta ;P)}$ . Then $vsu a \neq 1$ . So, the element a has no group theoretic inverse and hence ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is not a group.

(ii) If S is bisimple, Byleen showed in [Reference Byleen2] that ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is also bisimple. Now suppose S is regular. Let $vsu \in {\mathcal {C}^1(S;\alpha ,\beta ;P)}$ , with $v = b_1 \cdots b_m$ and $u = a_1 \cdots a_n$ . There exist $s^\prime \in S$ , $a_1^\prime , \ldots , a_m^\prime \in A$ and $b_1^\prime , \ldots , b_n^\prime \in B$ such that $s s^\prime s = s, s^\prime s s^\prime = s^\prime $ and

$$ \begin{align*} a_1^\prime b_1 = \cdots = a_m^\prime b_m=a_1 b_1^\prime = \cdots =a_n b_n^\prime =1. \end{align*} $$

If we set

$$ \begin{align*} y = b_n^\prime \cdots b_1^\prime \quad\text{and}\quad x = a_m^\prime \cdots a_1^\prime, \end{align*} $$

then $ys^\prime x$ is an inverse of $vsu$ . Hence, ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is regular. Therefore, if a monoid S is regular and bisimple (such as any group) then ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ is regular and bisimple.

(iii) Consider the subsemigroup $A^\ast $ of ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ . This semigroup is not DSC (as it is not simple). Hence, ${\mathcal {C}^1(S;\alpha ,\beta ;P)}$ has non-DSC subsemigroups.

5 Closing remarks and further questions

Now that we have seen that there exist DSC semigroups that are not groups, one might want to try to understand them better, and perhaps completely classify them. This seems to be out of reach at present but some seemingly easy questions remain. For example, we have seen that not even all (infinite) groups are DSC semigroups. So one may ask whether a description of DSC groups might be possible. Also, we do not know whether DSC semigroups are closed under formation of direct products.

Another direction one may take is to investigate more systematically the degree of interdependence of all four defining properties of a congruence. Specifically, a relation $\rho $ on a semigroup S is a congruence if and only if it is reflexive, symmetric, transitive and compatible. DSC semigroups are precisely those for which reflexivity and compatibility imply symmetry and transitivity. What about other combinations of these properties?