On Group-like Properties of Left Groups

Abstract

A left group is a semigroup that is the direct product of a left zero semigroup and a group. In this paper, we investigate some group-like properties of left groups. In particular, we characterise monogenic left group monoids and prove some fundamental results on left group morphisms. We also characterise Green’s equivalences on left groups and, analogously to groups, establish a correspondence between congruences and normal sub-left groups. Finally, we characterise DSC left groups.

1. Introduction and Preliminaries

A non-empty set S together with an associative binary operation is called a semigroup. An element a of a semigroup S is called a left zero element if $ab=a$ for all b in S. A semigroup consisting entirely of left zero elements is called a left zero semigroup. A right zero semigroup is defined analogously. A semigroup S is called $\mathit{left}$ $simple$ if $S=Sa$ for all $a\in S.$ If for all $a,b,c$ in $S,ac=bc$ implies $a=b$, we say that S is a right-cancellative semigroup. Dually, one can define a right simple semigroup and a left-cancellative semigroup.

A semigroup S is called a left group if it is left simple and right cancellative. Every group is a left group, but the converse is not true (see [1]). In Exercises 2.6 (5(b)) of [1], the Bear–Levi semigroup is not a left group. Following Clifford and Preston [2] (Theorem 1.27), a semigroup S is a left group if, and only if, $S=L\times G$, where L is a left zero semigroup and G is a group. If S is a left group, then the following are true: the set $E\left(S\right)$ of idempotents of S is non-empty and is a left zero subsemigroup of S; every idempotent is a left identity of S; and $eS$ is a subgroup of S for every $e\in E\left(S\right)$ (see Exercise 2.6 (6) of [1]). Analogously, a semigroup S is called a right group if it is right simple and left cancellative. Alternatively, by [2] (Theorem 1.27), S is a right group if, and only if, $S=R\times G$, where R is a right zero semigroup and G is a group. The analogue of the above-mentioned facts about left groups also holds for right groups. For more details on left (right) groups, the reader is referred to [2] (Section 1.11, pp. 37–40) and [1] (Exercises 2.6 (5–6)). In this paper, we shall only consider the study of left groups. The analogous results for right groups can be proven in a similar way and shall not be a part of this paper.

Left groups are peculiar in the sense that, on one hand, they behave like groups, but on the other hand, they fail in some of the basic algebraic properties which hold in a group. For example, a non-trivial left group has no identity element, and no element has a two-sided inverse. Left groups play a foundational role in semigroup theory, particularly within the study of regular semigroups and their structural properties. They provide a structural foundation for decomposing more complex semigroups. Specifically, a semigroup is a left group if, and only if, it is left simple and contains at least one idempotent element. The idempotent theory for left regular bands of groups has been developed by researchers. This structure generalises group algebras of finite groups and left regular band algebras, showing how concepts from left groups are used in the representation theory of more complex algebraic structures. The study of left groups has led to the generalisation of semigroup regularities. For instance, concepts such as left regular elements and left magnifying elements are important in studying regular semigroups, which have applications in other areas of mathematics and theoretical computer science.

Semigroups have applications in automata theory, coding theory and formal languages, where concepts such as left and right ideals and Green’s relations are used to understand the structure and behaviour of automata. In short, left groups serve as fundamental building blocks in the structural theory of semigroups, providing a deep understanding of regularity and the decomposition of more complex semigroups into simpler, group-like components.

This motivated the first and third author [3] to consider studying left groups, where weaker notions such as L-identity and L-inverse have been introduced. They also characterised all morphisms between any two left groups, and a characterisation of congruences on a left group is also provided, besides investigating various properties of the monoid of endomorphisms of a left group $S.$ In [4], these authors further investigated the structure and endomorphisms of a strong semilattice of left groups.

This paper is a sequel to [3,4] and aims to investigate the group-like properties of left groups. Cyclic groups play a central role in group theory, and the analogous notion of a monogenic semigroup exists for semigroups. In Section 2, we explore monogenic left group monoids and provide some necessary and sufficient conditions for a left group monoid to be a monogenic monoid. In Section 3, we exhibit group-like properties for left group homomorphisms. We extend the notion of the kernel of group homomorphism to a left group homomorphism, which we call the L-kernel of a left group homomorphism, and prove that some of the basic properties of the kernel hold for the L-kernel.

Green’s equivalences are very important relations defined on a semigroup. Exploring them provides deep insight into the structure of semigroups. This motivates us to explore these equivalences of left groups. In Section 4, we characterise Green’s equivalences on left groups. It is a well-known result that for a group G, the notion of a congruence relation is equivalent to that of a normal subgroup. This motivates us to see the analogy for a left group. In Section 4, we introduce the concept of a normal sub-left group of a left group and prove a similar equivalence between a congruence and a normal sub-left group. Finally, in Section 5, we study the relation between a diagonal subsemigroup and a congruence on a left group.

In summary, since this paper explores the basic structure of a left group, it will open up new dimensions for further research on this important algebraic object. In the future, we intend to apply our expertise on automorphism groups of semigroups (see [5]) and epimorphisms, dominions and amalgams (see [6,7]) to left groups and their strong semilattices in order to produce some exciting and fruitful research.

If a semigroup S does not contain the identity element, then $S^1=S\cup \left\{1\right\}$ denotes the monoid obtained by adjoining the identity element 1 to S, where the extended operation on $S^1$ is defined as $1s=s=s1$ for all $s\in S$ and $11=1$. Note that the element 1 does not have the left zero property, so a left zero semigroup L cannot contain 1. Thus, a left group $S=L\times G$ cannot be a monoid. This prompts us to consider the monoid $L^1$ obtained by adjoining the element 1 to the left zero semigroup L and the monoid $S^1=L^1\times G$ with the identity element $(1,e)$, where e is the identity of the group G. We call $L^1$ a left zero monoid and $S^1$ a left group monoid. In this paper, the terms ‘left group’ and ‘left group monoid’ shall be distinct and shall be mentioned explicitly.

Suppose S is a left group. Then, the element $(1,e)$ is not in S. In this case, for each $a=(l,g)\in S$, there exists a unique element $a_e=(l,e)\in S$ such that $a{a}_e=a_{e}a=a$. We call $a_e$ the L-identity for a. Also, there exists a unique element $a^{\perp }=(l,g^{-1})$ for each $a=(l,g)\in S$ such that $a{a}^{\perp }=a^{\perp }a=(l,e)=a_e$, and we call it the L-inverse of a (see [3] for details).

An element a of a semigroup S is said to be an idempotent if $a^2=a$. Note that every element of a left zero semigroup L is idempotent. Let $a=(l,g)$ be an idempotent element of a left group S. Then, $a^2=a$ implies $(l,g^2)=(l,g)$. As $g^2=g$ in a group G if, and only if, $g=e$, it follows that $a=(l,e)=a_e$. Hence, if $E\left(S\right)$ denotes the set of idempotents of the left group S, then $E\left(S\right)=\{a_e:a\in S\}$. Before we move on to the next sections, we want to end this section by presenting some simple but nontrivial examples to clarify the definitions introduced so far.

Example 1.

Let L be any element left zero semigroup and let $G={\mathbb{Z}}_2=\{0,1\}$ be the cyclic group of order 2. Consider the left group $S=L\times G$. Then,

$$S=\left\{\right(l,0),(l,1):l\in L\}.$$

The set $E\left(S\right)$ of L-identities (idempotents) is $E\left(S\right)=\left\{\right(l,0):l\in L\}$. This semigroup possesses very nice algebraic properties.

(i) 

For each $l\in L$, ${(l,1)}_e=(l,0)$, i.e., $(l,0)$ is the L-identity of $(l,1)$.

(ii) 

For each $x\in S$, $x^{\perp }=x$, i.e., every element of S is a self L-inverse.

(iii) 

For any $x,y\in S\setminus E\left(S\right)$, we have $xy=x_e$ and $yx=y_e$, i.e., the product of any two non-idempotent elements is an idempotent element and is the L-identity of the element appearing on the left in the product.

Example 2.

Let L be any left zero semigroup and $G=K_4=\{1,a,b,c\}$ be the Klein 4-group, the non-cyclic abelian group of order 4. Consider the left group $S=L\times G$. Then,

$$S=\left\{\right(l,1),(l,a),(l,b),(l,c):l\in L\}.$$

Some important properties of S are as follows:

(i) 

The set of L-identities of S is $E\left(S\right)=\left\{\right(l,1):l\in L\}$.

(ii) 

For each $x\in S$, $x^{\perp }=x$, every element of S is a self L-inverse.

(iii) 

For each fixed $l\in L$, the subset $S_l=\{(l,1),(l,g):g\in G\}$ of S is a group and $S_l\cong G$.

2. Monogenic Left Group Monoids

Cyclic groups are important objects in the theory of groups and equally important is the structure of a monogenic semigroup in semigroup theory. Since a left group is a direct product of a left zero semigroup and a group, it is important to investigate how the cyclic structure in the second component affects the monogenic part of the first component and vice versa. Also worth investigating are cases in which a left group is monogenic. In this section, we answer these questions.

Let S be a semigroup and $a\in S$. Then, $<a>=\{a^n:n\in \mathbb{N}\}$ is called the monogenic subsemigroup of S generated by a. If S is a monoid with identity 1 and $a\in S$, then the monogenic submonoid of S generated by a will contain 1 and thus it is $<a{>}^{*}=\{a^n:n\in {\mathbb{N}}^0\}$. The semigroup S is said to be monogenic if $S=<a>$ for some $a\in S$. A monogenic monoid can be defined similarly. We begin by proving a very basic result on monogenic left group monoids.

Proposition 1.

Let $S^1=L^1\times G$ be a left group monoid. S is a monogenic monoid if, and only if, $L^1={\left\{l\right\}}^1=\left\{l\right\}\cup \left\{1\right\}$ and G is a monogenic monoid.

Proof. 

Suppose that $S^1$ is a monogenic monoid. Then, $S^1=<a{>}^{*}$ for some $a=(l,g)\in S$. For any $x\in S$, there exists $k\in {\mathbb{N}}^0$ such that $x=a^k$. If $x=(1,e)$, then by taking $k=0$, we have $a^0={(l,g)}^0=(1,e)$. Suppose now that $x=(l^{\prime },g^{\prime })\ne (1,e)$. Then, $k\in \mathbb{N}$ with $(l^{\prime },g^{\prime })={(l,g)}^k=(l^k,g^k)=(l,g^k)$. This means $l^{\prime }=l$ and $g^{\prime }=g^k$. Hence, $L^1={\left\{l\right\}}^1$ and $G=<g{>}^{*}$ is a monogenic monoid. Conversely assume that these two conditions are satisfied. Take any $x=(l^{\prime },g^{\prime })\in S^1$. If $x=(1,e)$, then for any $a=(l,g)\in S$, $x=a^0$. On the other hand, if $x\ne (1,e)$, then $l^{\prime }=l$ and $g^{\prime }=g^r$ for some $r\in \mathbb{N}$, where $G=<g>$. Therefore, $x=(l^{\prime },g^{\prime })=(l,g^r)={(l,g)}^r$, where $r\in {\mathbb{N}}^0$. Hence, $S^1$ is a monogenic monoid with identity $1=(1,e)$, as required. □

From the proof of the proposition above, one sees that a left zero monoid $L^1$ is monogenic if, and only if, $L^1={\left\{l\right\}}^1$. Therefore, we have:

Corollary 1.

A left group monoid $S^1=L^1\times G$ is a monogonic monoid if, and only if, $L^1$ and G are monogenic monoids.

Next, we prove that the terms ‘cyclic’ and ‘monogenic’ are equivalent in the case of finite groups. We also illustrate with an example that the same is not true for all infinite groups.

Proposition 2.

Let G be a finite group. G is a monogenic monoid if, and only if, it is a cyclic group.

Proof. 

Let $\left|G\right|=m$. Suppose that $G=<a>$ is a cyclic group. Then, m is the least positive integer such that $a^m=1=a^0$ and $a^0,a^1,a^2,\dots ,a^{m-1}$ are all distinct. Since all elements of G are obtained as non-negative powers of a, it follows that $G=<a>=\{1,a,a^2,\dots ,a^{m-1}\}$ is a monogenic monoid. Conversely, assume that $G=<a>$ is a monogenic monoid of order m. Since G is a group, it must be a cyclic group of order m. □

Remark 1.

The above proposition is not valid if we take G as an infinite group. Take $G=(\mathbb{Z},+)$. Then, G is an infinite cyclic group generated by both 1 and $-1$, respectively. However, it is not a monogenic monoid since for any negative integer k, we have $k\ne m.1$ for any positive integer m. A similar argument applies when 1 is replaced by $-1$ and k is a positive integer. Also, as every infinite cyclic group is isomorphic to $(\mathbb{Z},+)$, the above proposition is not valid for all infinite groups G.

Recall that a finite cyclic group of order n is isomorphic to $Z_n$, the group of integers modulo n. This enables us to deduce a very important result regarding left groups.

Corollary 2.

A finite left group monoid $S^1=L^1\times G$ is a monogenic monoid if, and only if, $L={\left\{l\right\}}^1$ and G is $Z_n$ for some positive integer n.

The connection between monogenic semigroups and homomorphisms is fundamental to understanding the structure of these semigroups. For monogenic semigroups, this relationship allows for a complete classification of their structure and reveals a direct link to a special class of quotient semigroups and cyclic groups. Let F be the free monogenic semigroup, which is isomorphic to $({\mathbb{Z}}^{+},+)$. This semigroup consists of all formal powers of a single generator, x, with no relations imposed. For any monogenic semigroup $S=\langle a\rangle$, a canonical surjective homomorphism $f:F\to S$ can be defined by mapping the generator of F to the generator of S: $f\left(x^n\right)=a^n$. The properties of this homomorphism f directly reflect the structure of S. If S is infinite, f is an isomorphism, mapping F bijectively to S. If S is finite with index m and period r, the homomorphism f is many-to-one. The equality $a^k=a^l$ in S means that the corresponding powers $x^k$ and $x^l$ in F map to the same element, $a^k$, in S. The kernel of a finite monogenic semigroup is the minimal ideal, which is a cyclic group.

This cyclic group is one of the congruence classes of a canonical homomorphism, illustrating how the homomorphism’s kernel reveals key structural components of the monogenic semigroup. In the next section, we explore the group-like properties of left group homomorphisms.

3. Homomorphisms of Left Groups

Homomorphisms play a central and unifying role in algebra. They are structure-preserving maps between algebraic structures that respect the operations defined on those structures. Some of the well-known important properties of a group homorphism f between the groups G and H are that it preserves the identity, preserves inverses and preserves integer powers. Also its kernel, $kerf$, is a normal subgroup of G and its image, and $Imf$ is a subgroup of H. The homomorphism f is injective if, and only if, $kerf$ is trivial. The preimage of a subgroup is a subgroup, and the image of a cyclic group is a cyclic group. Not all of these properties are necessarily satisfied by a semigroup homomorphism. As left groups are a special kind of semigroup, we shall try to investigate which of these properties can be generalised to a left group homomorphism.

If S and T are any two semigroups (groups), by $Hom(S,T)$, we mean the set of all semigroup (group) homomorphisms from S into T. The next lemma is an important result from [3], which characterises the set $Hom(S,T)$, where S and T are left groups.

Lemma 1

([3] Theorem 4.1). Let $L_1\times G$ and $L_2\times H$ be two left groups. Take $t\in Hom(G,H)$ and s∈$Hom(L_1,L_2)$. Define $f:L_1\times G\to L_2\times H$ by

$$(l,x)f=\left(\right(l)s,(x\left)t\right)$$

for every $(l,x)\in$$L_1\times G$. Then, f is a homomorphism, and conversely, every homomorphism from $L_1\times G$ into $L_2\times H$ can be so constructed. Moreover, f is bijective if, and only if, s and t are bijective.

We can now adopt the notation for a homomorphism f between the left groups $S=L_1\times G$ and $T=L_2\times H$ as $f=(s,t)$, where s and t are given in Lemma 1.

In the next proposition, we show that some of the crucial properties of a group homomorphism listed at the beginning of this section can be extended to a left group homomorphism.

Proposition 3.

Let $S=L_1\times G$ and $T=L_2\times H$ be two left groups and $f=(s,t)$∈$Hom(S,T)$. Then, the following statements are true.

(i) 

$\left(a^{\perp }\right)f={\left[\left(a\right)f\right]}^{\perp }$$\forall a\in S$, i.e., a left group homomorphism preserves the L-inverse of an element.

(ii) 

$\left(E\right(S\left)\right)f\subseteq E\left(T\right)$, i.e., a left group homomorphism carries the set idempotents of S into the set of idempotents of T. The equality holds if s is onto.

(iii) 

$\left(a_e\right)f$ is the L-identity of $\left(a\right)f$$\forall a\in S$, i.e., f preserves the L-identity of an element.

(iv) 

For each $n\in {\mathbb{N}}^0$ and each $(l,g)\in S,\left({(l,g)}^n\right)f={\left((l,g)f\right)}^n.$

Proof. 

Let $S=L_1\times G$ and $T=L_2\times H$ be two left groups and $f=(s,t)\in Hom(S,T)$.

(i)

Let $a=(l,g)$ be any element of $L_1\times G$. Then,

$\left(a\right)f=(l,g)f=(\left(l\right)s,\left(g\right)t)=(l^{\prime },k)$, where $l^{\prime }=\left(l\right)s$ and $k=\left(g\right)t$.

Now,

$$\begin{array}{cc}\hfill \left(a^{\perp }\right)f& =(l,g^{-1})f\hfill \\ & =(\left(l\right)s,\left(g^{-1}\right)t)\hfill \\ & =(l^{\prime },{\left[\left(g\right)t\right]}^{-1})(ast\in Hom(G,H)).\hfill \\ & =(l^{\prime },k^{-1})\hfill \\ & ={\left[\left(a\right)f\right]}^{\perp }.\hfill \end{array}$$

(ii)

Let $(l,e)\in E\left(S\right)$. Then, $(l,e)f=(\left(l\right)s,\left(e\right)t)=(l^{\prime },e^{\prime })\in E\left(T\right)$ for some $l^{\prime }\in L_2$ and $e^{\prime }$ is the identity of H; thus, $\left(E\right(S\left)\right)f\subseteq E\left(T\right)$.

Take any $(l^{\prime },e^{\prime })\in E\left(T\right),$ as s is onto $l^{\prime }=\left(l\right)s$ for some $l\in L_1$ and t being in $Hom(G,H)$ implies $e^{\prime }=\left(e\right)t$. Therefore, $(l^{\prime },e^{\prime })=(\left(l\right)s,\left(e\right)t)=(l,e)f$. This implies $E\left(T\right)\subseteq \left(E\right(S\left)\right)f$, and so we have the desired equality.

(iii)

Take any $a\in S$. As already noted, $a{a}_e=a=a_{e}a=a$. Therefore, $\left(af\right)\left(a_{e}f\right)=\left(a{a}_e\right)f=\left(a\right)f$. Similarly, $\left(a_{e}f\right)\left(af\right)=\left(a\right)f$. Thus, $\left(a_e\right)f$ is the L-identity of $\left(a\right)f$.

(iv)

This is straightforward.

□

Part (ii) of Proposition 3 motivates us to investigate cases in which elements other than idempotent elements of S are mapped to idempotent elements of T. For this, let $S=L_1\times G$ and $T=L_2\times H$ be two left groups and $f\in Hom(S,T)$. The L-kernel of f, denoted by $L-Ker\left(f\right)$, is the set of elements $a\in S$ such that $\left(a\right)f\in E\left(T\right)$, i.e.,

$$L-Ker\left(f\right)=\{(l,g)\in S:(l,g)f=(l^{\prime },e^{\prime })\mathrm{where}{l}^{\prime }\in L^{\prime }\mathrm{and}{e}^{\prime }\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{of}H\}.$$

Equivalently,

$$L-ker\left(f\right)=\{(l,g)\in S:\left(l\right)s=l^{\prime }\mathrm{and}g\in Ker\left(t\right)\}.$$

Clearly, $E\left(S\right)\subseteq L-ker\left(f\right)$, so it is non-empty. We say that $L-ker\left(f\right)$ is trivial if $L-ker\left(f\right)=E\left(S\right)$. Also, we note that if the left group S is trivial, then $L-ker\left(f\right)$ reduces to the kernel of a group homomorphism.

Following [3], a non-empty subset U of a left group S is said to be the sub-left group of S if U itself is a left group under the operation of S. The next two results from [3] give the criteria for a subsemigroup of a left group to be a sub-left group.

Lemma 2

([3] Lemma 2.6). Let $S=L_1\times G$ be a left group and T be a subset of S. Then, T is a sub-left group of S if, and only if, $T=L_2\times H$, where $L_2$ is a subsemigroup of $L_1$ and H is a subgroup of G.

Lemma 3

([3] Proposition 2.8). A subset T of a left group $S=L_1\times G$ is a sub-left group of S if, and only if, $a{b}^{\perp }\in T$ for all $a,b\in T$.

A sub-left group T of a left group S is said to be a normal sub-left group of S if for each $a\in S$ and $b\in T$, we have $ab{a}^{\perp }\in T$.

The next proposition gives a characterisation of a normal sub-left group of a left group.

Proposition 4.

Let $S=L\times G$ be a left group and $T=L^{\prime }\times H$ be a sub-left group of S. T is a normal sub-left group of S if, and only if, $L^{\prime }=L$ and H is a normal subgroup of G.

Proof. 

Suppose that T is a normal sub-left group of G. Then, for each $a=(l,g)\in S$ and $b=(k,h)\in T$, we have $ab{a}^{\perp }\in T$. This means that $(l,gh{g}^{-1})\in L^{\prime }\times H$. Therefore, $l\in L^{\prime }$ and $gh{g}^{-1}\in H$. Thus, $L^{\prime }=L$ and H is a normal subgroup of G. Conversely, suppose that $L^{\prime }=L$ and H is normal in G. Take any $a=(l,g)\in S$ and $b=(k,h)\in T$. Now, $ab{a}^{\perp }=(l,gh{g}^{-1})$, and from the given conditions, we have $l\in L^{\prime }$, $gh{g}^{-1}\in H$. Thus, $ab{a}^{\perp }\in T$, proving that T is a normal sub-left group of G. □

The next two examples illustrate how to construct sub-left groups and normal sub-left groups of well-known left groups.

Example 3.

Consider the left group $S=L\times {\mathbb{Z}}_2$. The only subgroups of ${\mathbb{Z}}_2$ are ${\mathbb{Z}}_2$ and the trivial subgroup $\left\{0\right\}$, which are clearly normal in ${\mathbb{Z}}_2$. Thus, the sub-left groups of S are of the form $L^{\prime }\times {\mathbb{Z}}_2$ and $L^{\prime }\times \left\{0\right\}$, where $L^{\prime }$ is a subsemigroup of L. There are only two normal sub-left groups, namely, S itself and $L\times \left\{0\right\}$.

Example 4.

Consider the left group $S=L\times K_4$. The subgroups of $K_4$ are $\left\{1\right\}$, $\{1,a\}$, $\{1,b\}$, $\{1,c\}$ and $K_4$ itself. All the subgroups of $K_4$ are normal in $K_4$. Any sub-left group of S will be of any of the following types:

(i) 

$L^{\prime }\times K_4$;

(ii) 

$L^{\prime }\times \left\{1\right\}$;

(iii) 

$L^{\prime }\times \{1,a\}$;

(iv) 

$L^{\prime }\times \{1,b\}$;

(v) 

$L^{\prime }\times \{1,c\}$;

where $L^{\prime }$ is a subsemigroup of L. On the other hand, by taking $L^{\prime }=L$ in these cases, we obtain all the 5 normal sub-left groups of S.

The next theorem is the main result of this section, where we generalise some of the properties being preserved by a group homomorphism to a left group homomorphism.

Theorem 1.

Let $S=L_1\times G$ and $T=L_2\times H$ be two left groups, R be any sub-left group of S and $f\in Hom(S,T)$. The following statements are true.

(i) 

$\left(R\right)f$ is a sub-left group of T.

(ii) 

$L-ker\left(f\right)$ is trivial if, and only if, t is one-to-one.

(iii) 

$L-ker\left(f\right)$ is a normal sub-left group of S.

(iv) 

$(L-ker(f\left)\right)f\subseteq E\left(T\right)$. Moreover $(L-ker(f\left)\right)f=E\left(T\right)$ if s is onto, and in this case, the former is a normal sub-left group of T.

Proof. 

(i)

By Lemma 2, $R=L_1^{\prime }\times K$, where $L_1^{\prime }$ is a subsemigroup of $L_1$ and K is a subgroup of G. Let $x=(l_2,h),y=(l_2^{\prime },h^{\prime })\in \left(R\right)f$. Then, there exist $l_1,l_1^{\prime }\in L_1^{\prime }$, $g,g^{\prime }\in K$ such that $(l_2,h)=(l_1,g)f$ and $(l_2^{\prime },h^{\prime })=(l_1^{\prime },g^{\prime })f$. Now, $x{y}^{\perp }=(l_1,g)f{\left((l_1^{\prime },g^{\prime })f\right)}^{\perp }=(l_1,g)f{(l_1^{\prime },g^{\prime })}^{\perp }f$ (by Proposition 3 (ii)). Since $f\in Hom(S,T)$, we have $(l_1,g)f{(l_1^{\prime },g^{\prime })}^{\perp }f=\left((l_1,g){(l_1^{\prime },g^{\prime })}^{\perp }\right)f$. Since R is a sub-left group of S, by Lemma 3, we have $(l_1,g){(l_1^{\prime },g^{\prime })}^{\perp }\in R$. Thus, $x{y}^{\perp }\in \left(R\right)f$ as required.

(ii)

We have to show that $L-ker\left(f\right)\subseteq E\left(S\right)$. Take any $(l,g)\in L-ker\left(f\right)$. Then, $(l,g)f=(l^{\prime },e^{\prime })$, so $(\left(l\right)s,\left(g\right)t)=(l^{\prime },e^{\prime })$, where $l^{\prime }=\left(l\right)s$ and $e^{\prime }$ is the identity of H. Now, $\left(g\right)t=e^{\prime }$ if, and only if, $g\in ker\left(t\right)$. As t is one-to-one, $ker\left(t\right)=\left\{e\right\}$, and therefore, $g=e$. Thus, $(l,g)\in E\left(S\right)$, as required.

(iii)

Note that $L-ker\left(f\right)=L_1\times ker\left(t\right)$. Since $ker\left(t\right)$ is a normal subgroup of G, using Lemma 2, we have that $L-ker\left(f\right)$ is a normal sub-left group of S, as required.

(iv)

Clearly, $(L-ker(f\left)\right)f\subseteq E\left(T\right)$. For the last statement, as $E\left(S\right)\subseteq L-ker\left(f\right)$, this implies that $\left(E\right(S\left)\right)f\subseteq (L-ker(f\left)\right)f.$ as s is onto according to Proposition 3 (ii) $\left(E\right(S\left)\right)f=E\left(T\right)$. Therefore, $E\left(T\right)\subseteq (L-ker(f\left)\right)f$, showing the reverse inclusion. As s is onto, $\left(kert\right)t$ is a normal subgroup of H. Now, $(L-ker\left(f\right))f=(L_1\times ker\left(t\right))f=\left(L_1\right)s\times \left(ker\left(t\right)\right)t=L_2\times H^{\prime }$, where $L_2=\left(L_1\right)s$ is a subsemigroup of $L_2$ and $H^{\prime }=\left(ker\left(t\right)\right)t$ is a normal subgroup of H. Thus, $(L-ker(f\left)\right)f$ is a normal subgroup of T, as required.

□

In the next example, we exhibit a non-trivial left group homomorphism of an infinite left group where the L-kernel is normal but not trivial.

Example 5.

Let L be a non-trivial left zero semigroup and $G=\mathbb{Z}$ be the group of integers. Suppose $H={\displaystyle \frac{\mathbb{Z}}{n\mathbb{Z}}}$ with $n\ge 2$. Consider the left groups $S=L\times G$ and $T=L\times H$. Let $f:S⟶T$ be a homomorphism given by $f(l,k)=(l,k(modn\left)\right)$. Then, $L-ker\left(f\right)=L\times ker\left(t\right)=L\times n\mathbb{Z}$. As $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$ by Proposition 4 $L-ker\left(f\right)$ is a normal sub-left group of S and is clearly non-trivial.

In a semigroup, the relationship between homomorphisms and Green’s equivalences is complex, providing insight into the semigroup’s internal structure. Homomorphisms do not necessarily preserve the fine structure of Green’s classes, but they do map principal ideals to ideals. That is, if $f:S\to T$ is a homomorphism, then $f\left(S^{1}a\right)=f{\left(S\right)}^{1}f\left(a\right)$. The Green’s classes in the image of a homomorphism, $f\left(S\right)$, are often smaller than their corresponding Green’s classes in S. For example, a single $\mathcal{L}$ class in S may be mapped to multiple $\mathcal{L}$ classes in T by a homomorphism. Specialised semigroups called ‘transformation semigroups’ are defined by maps that explicitly preserve a given equivalence relation. These are a good example of how homomorphisms can be constructed to uphold a specific Green’s relation. In regular semigroups, homomorphisms can have a well-defined relationship with Green’s relations. For instance, in an inverse semigroup, which is a special type of regular semigroup, each $\mathcal{L}$ class and $\mathcal{R}$ class contains a unique idempotent. Studies on specific semigroup classes, such as completely regular semigroups, have shown that homomorphisms can facilitate the decomposition of these semigroups based on Green’s relations.

Like any other algebraic structure, homomorphisms are well behaved with congruences in semigroups. For any homomorphism $f:S\to T$, the kernel of f is a semigroup congruence on S. This congruence, denoted by $kerf$, relates elements that map to the same image in T, i.e., $(a,b)\in kerf$ if, and only if, $f\left(a\right)=f\left(b\right)$. A homomorphism $f:S\to T$ also induces an isomorphism between the quotient semigroup $S/kerf$ and the image $f\left(S\right)$. This means the structure of the image can be understood by examining the congruence classes of the kernel. In the next section, we further explore these relations in the case of left groups.

4. Green’s Equivalences and Congruences on Left Groups

Green’s equivalences (or Green’s relations) are fundamental tools in semigroup theory. They partition a semigroup S into classes that reveal its internal structure, especially regarding ideals, principal ideals and factorisation properties. For a group G, Green’s equivalences simplify drastically because every element is invertible. This forces every group into a single Green’s class and thus there is no finer partition because invertibility collapses the distinctions.

Congruences are central to semigroup theory because they provide the primary mechanism for decomposing and analysing semigroups. In semigroup theory, a congruence is the analogue of a normal subgroup in group theory. It is a way to define a quotient structure that remains a semigroup and enables the factorisation of a semigroup into simpler components, analogous to factor groups.

In this section, we characterise Green’s equivalences in the case of left groups. We also generalise some well-known results regarding congruences from groups to left groups.

Let $S=L\times G$ be a left group and let

$$\rho =\{((l_1,g_1),(l_2,g_2)):l_1,l_2\in L,g_1,g_2\in G\}$$

be a binary relation on S. Then, $\rho$ gives rise to binary relations $\sigma$ and $\tau$ on L and G, respectively, given by $((l,g),(l^{\prime },g^{\prime }))\in \rho$ if, and only if, ($l,l^{\prime })\in \sigma$ and $(g,g^{\prime })\in \tau$. Conversely, every binary relation $\sigma$ on L and $\tau$ on G gives a binary relation $\rho$ on S. Therefore, $\rho$ can be viewed as the pair $(\sigma ,\tau )$.

Let S be a semigroup and $a,b\in S$, $S^{1}a(a{S}^1,S^{1}a{S}^1)$ denotes, respectively, the principal left (right, two-sided) ideal of S generated by a. The five Green’s equivalences are defined in terms of principal ideals as follows:

$$\begin{array}{c}\hfill a\mathcal{L}b\mathrm{if}, \mathrm{and} \mathrm{only} \mathrm{if},S^{1}a=S^{1}b\\ \hfill a\mathcal{R}b\mathrm{if}, \mathrm{and} \mathrm{only} \mathrm{if},a{S}^1=b{S}^1\\ \hfill a\mathcal{J}b\mathrm{if}, \mathrm{and} \mathrm{only} \mathrm{if},S^{1}a{S}^1=S^{1}b{S}^1\\ \hfill a\mathcal{H}b\mathrm{if}, \mathrm{and} \mathrm{only} \mathrm{if},a\mathcal{L}b\mathrm{and}a\mathcal{R}b\\ \hfill \mathcal{D}=\mathcal{L}\circ \mathcal{R}=\mathcal{R}\circ \mathcal{L}=\mathcal{L}\vee \mathcal{R}\end{array}$$

One can easily check that $\mathcal{H}=\mathcal{L}\cap \mathcal{R}$, $\mathcal{L}\subseteq \mathcal{J}$ and $\mathcal{R}\subseteq \mathcal{J}$ since $\mathcal{D}$ is the join of $\mathcal{L}$ and $\mathcal{R}$, $\mathcal{D}\subseteq \mathcal{J}$. If the semigroup S is a group G, then all reduce to the universal relation on G, i.e., $\mathcal{L}=\mathcal{R}=\mathcal{D}=\mathcal{J}=G\times G$.

A relation $\rho$ on a semigroup is considered left compatible if for all $a,b,t\in S$, we have $(a,b)\in S$ implies $(ta,tb)\in S$ and $rightcompatible$ if $\forall a,b,t\in S,(a,b)\in S$ implies $(at,bt)\in S.$ It is called $compatible$ if it is both left compatible and right compatible. A compatible equivalence relation on S is called a $congruence$ on S.

Remark 2.

It is easy to verify that, if $\rho =(\sigma ,\tau )$ is an equivalence on a left group $S=L\times G$, then ρ is a congruence on S if, and only if, σ is a congruence on L and τ is a congruence on G.

Further details on Green’s equivalences and congruences can be found in any standard text on semigroup theory. In particular, the reader is referred to Howie [1] and Clifford and Preston [2].

Next, we first characterise Green’s equivalences on left groups. Unlike groups, it turns out that not all the Green’s relations on a left group are universal. However, like in a group G, each of the Green’s equivalence is proven as a congruence on a left group S. First, we characterise Green’s equivalences on a left zero semigroup L. This, together with the results on groups, provides a characterisation of left groups.

For any semigroup S, $I^S$ denotes the identity relation on S.

Proposition 5.

Let L be a left zero semigroup. Then, the following statements are true.

(i) 

$\mathcal{R}=\mathcal{H}=I^L$.

(ii) 

$\mathcal{L}=\mathcal{J}=\mathcal{D}=L\times L.$

Proof. 

Let L be a left zero semigroup.

(i)

Take any $a,b\in L$ such that $a\mathcal{R}b$. Then, there exists $u,v\in L^1$ such that $au=b,bv=a.$ By the left zero property of L, $a=b.$ This implies $\mathcal{R}=I^L$. Since $\mathcal{H}=\mathcal{L}\cap \mathcal{R}=\mathcal{L}\cap I^L=I^L$; therefore, $\mathcal{R}=\mathcal{H}=I^L.$

(ii)

For any $(a,b)\in L\times L$ by the left zero property of L, we have $ab=a$ and $ba=b$. This implies $a\mathcal{L}b$, and thus $L\times L=\mathcal{L}$. Now, $L\times L=\mathcal{L}\subseteq \mathcal{J}\subseteq L\times L=\mathcal{L},$ so it follows that $\mathcal{L}=\mathcal{J}=L\times L$. Finally, as $\mathcal{D}=\mathcal{L}\circ \mathcal{R}=\mathcal{R}\circ \mathcal{L}$; therefore, $\mathcal{D}=\mathcal{L}\circ I^L=I^L\circ \mathcal{L}=\mathcal{L}.$ Hence, $\mathcal{L}=\mathcal{J}=\mathcal{D}=L\times L$, as required.

□

Note that the identity relation $I^S$ and the universal relation $S\times S$ are easily seen as compatible on any semigroup S, so the next corollary is immediate.

Corollary 3.

Let S be a left zero semigroup. Then, all the Green’s equivalences are congruences on S.

If $S=L\times G$ is a left group and $\rho \in \{\mathcal{L},\mathcal{R},\mathcal{H},\mathcal{J},\mathcal{D}\}$ on S, then we denote it by ${\rho }^S$, and so ${\rho }^S=({\rho }^L,{\rho }^G)$, where ${\rho }^L$ and ${\rho }^G$ denote the corresponding relations on L and G, respectively. We are now in a position to characterise Green’s equivalences on the left group S.

Theorem 2.

Let $S=L\times G$ be a left group. Then, the following statements are true.

(i) 

${\mathcal{L}}^S={\mathcal{J}}^S={\mathcal{D}}^S=S\times S.$

(ii) 

${\mathcal{R}}^S={\mathcal{H}}^S=(I^L,G\times G)=I^L\times (G\times G)$

Proof. 

Let $S=L\times G$ be a left group and ${\rho }^S=({\rho }^L,{\rho }^G)\in \{\mathcal{L},\mathcal{R},\mathcal{H},\mathcal{J},\mathcal{D}\}$.

(i)

Let $\rho =\mathcal{L}$. Then, ${\mathcal{L}}^S=({\mathcal{L}}^L,{\mathcal{L}}^G)$. By Proposition 5 ${\mathcal{L}}^L=L\times L$, and as already mentioned above (see [1]), ${\mathcal{L}}^G=G\times G$. Therefore, ${\mathcal{L}}^S=(L\times L,G\times G)=(L\times G)\times (L\times G)=S\times S.$ Since ${\mathcal{D}}^G={\mathcal{J}}^G=G\times G$, we can apply a similar argument to prove that ${\mathcal{J}}^S={\mathcal{D}}^S=S\times S$.

(ii)

By Proposition 5 ${\mathcal{H}}^L={\mathcal{R}}^L=I^L$, and as mentioned above, ${\mathcal{H}}^G={\mathcal{R}}^G=G\times G$. Now, ${\mathcal{R}}^S=({\mathcal{R}}^L,{\mathcal{R}}^G)=(I^L,G\times G)=I^L\times (G\times G)$. Since, ${\mathcal{H}}^G=G\times G$, we can apply a similar argument to prove that ${\mathcal{H}}^S=I^L\times (G\times G)$.

□

The equalities established in the above theorem may not hold for non-left group semigroups. By Exercise 2.6 (1) of [1], if C is a cancellative semigroup without identity, then $\mathcal{L}=\mathcal{R}=\mathcal{D}=I^C$, but there is a cancellative semigroup S without identity such that $\mathcal{J}=S\times S$. Again, by Exercise 2.6 (2) of [1], in the bicyclic semigroup, $\mathbb{B}={\mathbb{N}}^0\times {\mathbb{N}}^0$, $\mathcal{L}\subset \mathcal{D}=\mathcal{J}=\mathbb{B}\times \mathbb{B}$. Thus, it would be interesting for future research to explore the non-left group semigroup classes in which the equalities established in the above theorem are true.

Corollary 4.

Let $S=L\times G$ be a left group and ${\rho }^S\in \{{\mathcal{L}}^S,{\mathcal{R}}^S,{\mathcal{H}}^S,{\mathcal{J}}^S,{\mathcal{D}}^S\}$. Then, ${\rho }^S$ is a congruence on S.

Proof. 

We complete the proof in the form of two cases. Suppose first that ${\rho }^S\in \{{\mathcal{L}}^S,{\mathcal{J}}^S,{\mathcal{D}}^S\}$. By part (i) of Theorem 2, ${\rho }^S=S\times S$ so trivially is a congruence on S. Suppose next that ${\rho }^S={\mathcal{R}}^S$ and take any $a,b,s\in S$ such that $(a,b)\in {\rho }^S$. Let $a=(l,g),b=(l^{\prime },g^{\prime })$ and $s=(l^{″},g^{″})$. Now, $(a,b)\in {\rho }^S$ implies $(l,l^{\prime })\in {\rho }^L$ and $(g,g^{\prime })\in {\rho }^G$. By part (ii) of Theorem 2, ${\rho }^L=I^L$ and ${\rho }^G=G\times G$ are congruences on L and G, respectively. Thus, ${\rho }^S={\mathcal{R}}^S=({\mathcal{R}}^L,{\mathcal{R}}^G)$ is a congruence on G. By the similar argument ${\mathcal{H}}^S$ is a congruence on S. □

In group theory, every congruence on a group G corresponds to a unique normal subgroup and vice versa. This fundamental correspondence allows us to translate problems on congruences into normal subgroups, which are more familiar and easy to handle. It is natural to expect a similar kind of correspondence in the case of left groups. We are able to provide such a correspondence on left groups. We first prove a basic lemma, which is required to prove the main result.

Lemma 4.

Let $S=L\times G$ be a left group. Then, for any $a=(l,g),b=(l^{\prime },g^{\prime })\in S$, ${\left(ab\right)}^{\perp }=b^{\perp }{a}^{\perp }$.

Proof. 

Let $a=(l,g),b=(k,h)\in S$. Then,

$$\begin{array}{cc}\hfill & \left(ab\right)b^{\perp }{a}^{\perp }\left(ab\right)=(l,gh{h}^{-1}{g}^{-1}gh)=(l,gh)=(l,g)(k,h)=ab.\hfill \end{array}$$

Similarly, $b^{\perp }{a}^{\perp }\left(ab\right)b^{\perp }{a}^{\perp }=b^{\perp }{a}^{\perp }$. Therefore, we get ${\left(ab\right)}^{\perp }=b^{\perp }{a}^{\perp }$, as required. □

Now, we prove the main result of this section.

Theorem 3.

Let $S=L\times G$ be a left group. The following statements hold.

(i) 

For any normal sub-left group T of S,

$${\rho }_T=\{(x,y)\in S\times S:x{y}^{\perp }\in T\}$$

is a congruence on S.

(ii) 

Let $\rho =(\sigma ,\tau )$ be a congruence on S. For each idempotent $(l,e)$ in S, the set $T=(l,e)\rho$ is a normal sub-left group of S if, and only if, $l\sigma =L$, and so $\rho ={\rho }_T$.

(iii) 

If $T,U$ are normal sub-left groups of S, then ${\rho }_T\cap {\rho }_U={\rho }_{T\cap U}$ and ${\rho }_T\circ {\rho }_U={\rho }_{TU}.$

Proof. 

(i)

Let T be a normal sub-left group of a left group S. By Proposition 4, $T=L\times H$, where H is a normal subgroup of G. First, we show ${\rho }_T$ is an equivalence on S.

• Reflexive: Take any $a=(l,g)\in S$. As T is a normal sub-left group of S, it follows that $a{a}^{\perp }=(l,g{g}^{-1})=(l,e)\in T$. This implies $(a,a)\in {\rho }_T$.
• Symmetric: Take any $a=(l,g),b=(k,h)\in S$. Let $(a,b)\in {\rho }_T$. Then, $a{b}^{\perp }=(l,g{h}^{-1})\in T$. Again, as S is a normal sub-left group of S, we have $b{a}^{\perp }=(k,h{g}^{-1})=(k,{\left(g{h}^{-1}\right)}^{-1})\in T$. Thus, $(b,a)\in {\rho }_T$.
• Transitive: Let $(a,b)$ and $(b,c)\in {\rho }_T$. Then, $a{b}^{\perp }$ and $b{c}^{\perp }\in T$. Now $b{b}^{\perp }=b_e$ and $a{b}_e=a$, so T being a sub-left group implies $a{b}^{\perp }b{c}^{\perp }=a{c}^{\perp }\in T$. Thus, $(a,c)\in {\rho }_T$.
  Now, it remains to be shown that ${\rho }_T$ is left and right compatible. Take any $a=(l,g)$, $b=(k,h)$ and $s=(m,n)\in S$. Suppose that $(a,b)\in {\rho }_T$ and so $a{b}^{\perp }\in T$. By Lemma 4, $sa{\left(sb\right)}^{\perp }=sa{b}^{\perp }{s}^{\perp }$, and as $a{b}^{\perp }\in T$ and T is a normal sub-left group, it follows that $sa{\left(sb\right)}^{\perp }=s\left(a{b}^{\perp }\right)s^{\perp }\in T$. Thus, $(sa,sb)\in {\rho }_T$, proving the left compatibility of ${\rho }_T$. Again, using Lemma 4, we get $as{\left(bs\right)}^{\perp }=as{s}^{\perp }{b}^{\perp }=a{s}_e{b}^{\perp }=a{b}^{\perp }\in T$. Thus, $(as,bs)\in {\rho }_T$, proving the right compatibility. Hence, ${\rho }_T$ is a congruence on S.

(ii)

Take any congruence $\rho =(\sigma ,\tau )$ on $S=L\times G$ and let $T=(l,e)\rho$, where $(l,e)$ is an idempotent in S. Now, as $(l,e)\rho =l\sigma \times e\tau$, we have to show that $l\sigma \times e\tau$ is a normal sub-left group, which, by Proposition 4, is equivalent to showing that $l\sigma =L$ and $e\tau$ is a normal subgroup of G. By Proposition 1.8.2 of [1] $e\tau$ is a normal subgroup of G. Therefore, the statement holds if, and only if, $l\sigma =L$.

Next, we have to prove that $\rho =(\sigma ,\tau )={\rho }_T={\rho }_{(l,e)}=({\sigma }_l,{\tau }_e)$, and this happens if, and only if, $\sigma ={\sigma }_l$ and $\tau ={\tau }_e$. Since $l\sigma =l$, it follows that ${\sigma }_l=L\times L$, so $\sigma \subseteq {\sigma }_l$. On the other hand, if $(l_1,l_2)\in {\sigma }_l$, then $(l_1,l)\in \sigma$ and $(l,l_2)\in \sigma$, and because $\sigma$ is transitive, it follows that $(l_1,l_2)\in \sigma$. This proves the reverse inclusion, and thus $\sigma ={\sigma }_l$. For the remaining case, $(g,h)\in \tau$ if, and only if, $(g{h}^{-1},h{h}^{-1})\in \tau$ if, and only if, $(g{h}^{-1},e)\in \tau$ if, and only if, $g{h}^{-1}\in e\tau$ if, and only if, $(g,h)\in {\tau }_e$. Thus, $\tau ={\tau }_e$ as required.

(iii)

Finally, if T and U are normal sub-left groups of S, then

$$\begin{array}{cc}\hfill & (a,b)\in {\rho }_T\cap {\rho }_U\hfill \\ \hfill \iff & a{b}^{\perp }\in T\mathrm{and}a{b}^{\perp }\in U\hfill \\ \hfill \iff & a{b}^{\perp }\in T\cap U\hfill \\ \hfill \iff & (a,b)\in {\rho }_{T\cap U}.\hfill \end{array}$$

Next, take any $(a,b)\in {\rho }_T\circ {\rho }_U$ there exists some $c\in S$ such that $(a,c)\in {\rho }_T$ and $(c,b)\in {\rho }_U$. This implies $a{c}^{\perp }\in T$ and $c{b}^{\perp }\in U$, and so $a{b}^{\perp }=a{c}^{\perp }c{b}^{\perp }\in TU$. Thus, $(a,b)\in {\rho }_{TU}$, proving that ${\rho }_T\circ {\rho }_U\subseteq {\rho }_{TU}$. For the reverse inclusion, assume that $(a,b)\in {\rho }_{TU}$, which implies $a{b}^{\perp }\in TU$. Thus, $a{b}^{\perp }=tu$ for some $t\in T$ and $u\in U$. Take $c=ub$. Then, $a{c}^{\perp }=a{\left(ub\right)}^{\perp }=a{b}^{\perp }{u}^{\perp }=tu{u}^{\perp }=t\in T$, and so $(a,c)\in {\rho }_T$. Also, $c{b}^{\perp }=ub{b}^{\perp }=u\in U$, proving that $(c,b)\in {\rho }_U$. Thus, $(a,b)\in {\rho }_T\circ {\rho }_U$, as required.

Hence, ${\rho }_T\cap {\rho }_U={\rho }_{T\cap U}$ and ${\rho }_T\circ {\rho }_U={\rho }_{TU}$. This completes the proof of the theorem. □

This correspondence between normal sub-left groups and congruences cannot be extended to more general classes of semigroups because one cannot exploit the existence of L-inverses, L-identities and idempotents in a general semigroup. For example, if S is a commutative semigroup, then there exists a congruence $\eta$ on S such that the quotient semigroup $S/\eta$ is a semilattice. Since a semilattice is not a left group, this congruence cannot correspond to a normal sub-left group.

In a semigroup S, a congruence is a special type of equivalence relation that can be characterised as a diagonal subsemigroup of the direct product $S\times S$ with additional properties. The connection reveals how a congruence can be understood as an algebraic substructure of the larger semigroup product. The connection is established by showing that a congruence on S is precisely a diagonal subsemigroup of $S\times S$ that is also a reflexive, symmetric and transitive relation. For a group G, the connection becomes simpler. In this case, every diagonal subgroup of $G\times G$ is automatically a congruence on G. The equivalence between diagonal subsemigroups and congruences does not hold for general semigroups, showing a key difference from groups. This difference motivated the definition of a DSC semigroup—a semigroup where every diagonal subsemigroup is necessarily a congruence. In the next section, we explore the DSC property in the case of left groups.

5. Diagonal Subsemigroups and Left Groups

Let S be a semigroup and $\rho \subseteq S\times S$. Then, $\rho$ is said to be a diagonal subsemigroup of $S\times S$ if $\rho$ is a subsemigroup of $S\times S$ containing the identity relation $I^S$ on S. It is natural to then determine for which semigroups S a congruence $\rho$ on S corresponds to a diagonal subsemigroup of $S\times S$ and vice versa. This study has been undertaken in [8], and this motivated us to investigate this interesting question for a left group S. The next lemma is immediate, but for completeness, we include its proof.

Lemma 5.

If ρ is a congruence on a semigroup S, then it is a diagonal subsemigroup of $S\times S$.

Proof. 

Let $\rho$ be a congruence on S. Since $\rho$ is reflexive, it follows that $I^S\subseteq \rho$. Take any $(a,b)$ and $(c,d)$ in $\rho$. Since $\rho$ is compatible, it follows that $(a,b)(c,d)=(ac,bd)\in \rho$. This proves that $\rho$ is a diagonal subsemigroup of $S\times S$. □

It is important to determine when the converse of the above lemma holds. In particular, is the converse true if $S=L\times G$ is a left group? In this section, we shall answer this question.

Lemma 6.

Let $S=L\times G$ be a left group and $\rho =(\sigma ,\tau )\subseteq S\times S$. Then, ρ is a diagonal subsemigroup of $S\times S$ if, and only if, σ is a diagonal subsemigroup of $L\times L$ and τ is a diagonal subsemigroup of $G\times G$.

Proof. 

Observe that $\sigma \subseteq L\times L$ and $\tau \subseteq G\times G$. Also, ${\rho }^2\subseteq \rho$ if, and only if, ${\sigma }^2\subseteq \sigma$ and ${\tau }^2\subseteq \tau$. Thus, $\rho$ is a subsemigroup of $S\times S$ if, and only if, $\sigma$ is a subsemigroup of $L\times L$ and $\tau$ is a subsemigroup of $G\times G$. Finally, as $I^S=(I^L,I^G)$, it follows that $I^S\subseteq \rho$ if, and only if, $I^L\subseteq \sigma$ and $I^G\subseteq \tau$. Hence, we conclude that $\rho$ is a diagonal subsemigroup of $S\times S$ if, and only if, $\sigma$ is a diagonal subsemigroup of $L\times L$ and $\tau$ is a diagonal subsemigroup of $G\times G$. □

The next result shows that for a group G, the notions of a diagonal subgroup and a congruence are equivalent.

Proposition 6.

Let G be a group and τ be a diagonal subgroup of $G\times G$. Then, τ is a congruence on G.

Proof. 

As $\tau$ is a subgroup of $G\times G$, the compatibility follows automatically. So, we only need to show that $\tau$ is an equivalence on G.

(i)

Reflexive: Since $I^S\subseteq \tau$, clearly $\tau$ is reflexive.

(ii)

Symmetric: Let $(g,h)\in \tau$. As $(h^{-1},h^{-1})\in \tau$ and $\tau$ is a subgroup of $G\times G$, it follows that $(g,h)(h^{-1},h^{-1})=(g{h}^{-1},e)\in \tau$. Again, as $(g^{-1},g^{-1})\in \tau$, by the same argument, we have $(g^{-1},g^{-1})(g{h}^{-1},e)=(h^{-1},g^{-1})\in \tau$. Since $\tau$ is a subgroup, $(h,g)={(h^{-1},g^{-1})}^{-1}\in \tau$, as required.

(iii)

Transitive: Let $(g,h)$ and $(h,k)\in \tau$. As $(h^{-1},h^{-1})\in \tau$ and $\tau$ is a subgroup of $G\times G$, we get $(g,k)=(g,h)(h^{-1},h^{-1})(h,k)\in \tau$. This shows that $\tau$ is transitive.

Hence, $\tau$ is a congruence on G. □

By Example 1.3 of [8], for a 2-element left zero semigroup L, it is not true in general that a diagonal subsemigroup of $L\times L$ is a congruence on L. The next example illustrates that the same holds for a 3-element left zero semigroup.

Example 6.

Consider the left zero semigroup $L=\{l,m,n\}$ and take

$$\sigma =\left\{\right(l,l),(l,m),(m,m),(l,n),(n,n\left)\right\}.$$

Clearly, σ is a diagonal subsemigroup of $L\times L$, but it is not a congruence on L, as it is not symmetric.

Following [8], a semigroup S is said to be DSC if every diagonal subsemigroup of $S\times S$ is a congruence on S. Thus, by the above examples, any 2-element and 3-element left zero semigroup is not DSC. The next proposition allows us to extend it to any left zero semigroup.

Proposition 7.

A 3-element left zero semigroup can be embedded into any left zero semigroup T such that $\left|T\right|\ge 3$.

Proof. 

Let $S=\{a_1,a_2,a_3\}$ be a 3-element left zero semigroup and let $T=\{b_1,b_2,b_3,\dots \}$ be any left zero semigroup, where $\left|T\right|\ge 3$. Since $\left|T\right|\ge 3$, we can select three distinct elements from T to be the images of the elements of S. Let us define a map $\varphi :S\to T$ as follows: $\varphi \left(a_1\right)=b_1,\varphi \left(a_2\right)=b_2$ and $\varphi \left(a_3\right)=b_3$. The map $\varphi$ is clearly injective, as it maps distinct elements of S to distinct elements of T. For any $x,y\in S$, we have $\varphi \left(xy\right)=\varphi \left(x\right)$ and $\varphi \left(x\right)\varphi \left(y\right)=\varphi \left(x\right)$. Since $\varphi \left(xy\right)=\varphi \left(x\right)\varphi \left(y\right)$, the map is a homomorphism and thus an embedding. □

The next proposition is important in deciding the DSC property for the case of left groups.

Proposition 8.

Let $S=L\times G$ be a left group. Then, S is DSC if, and only if, both L and G are DSC.

Proof. 

By Remark 2, an equivalence $\rho =(\sigma ,\tau )$ is a congruence on S if, and only if, $\sigma$ is a congruence on L and $\tau$ is a congruence on G. By Lemma 6, $\rho =(\sigma ,\tau )$ is a diagonal subsemigroup of $S\times S$ if, and only if, $\sigma$ is a diagonal subsemigroup of $L\times L$ and $\tau$ is a diagonal subsemigroup of $G\times G$. This completes the proof of the lemma. □

Thus, through Propositions 7 and 8, we conclude the following important results regarding the DSC property of left zero semigroups and left groups.

Corollary 5.

A left zero semigroup L is DSC if, and only if, it is trivial.

By Corollary 2.5 of [8], a finite semigroup S is DSC if, and only if, it is a group. In particular, it implies that every finite group G is DSC. The next example illustrates that an infinite group is not necessarily DSC.

Example 7

([8] Example 1.5). Let $\mathbb{Z}$ denote the infinite cyclic group. Then, the relation $\tau =\left\{\right(x,y):x\le y\}$ is a diagonal subsemigroup of $\mathbb{Z}\times \mathbb{Z}$ but not a congruence on $\mathbb{Z}$.

Corollary 6.

If $S=L\times G$ is a non-trivial left group (finite or infinite), then S is not necessarily DSC.

In fact, we have a stronger result.

Corollary 7.

A left group $S=L\times G$ is DSC if, and only if, $\left|L\right|=1$ and G is a finite group.

Proof. 

Suppose that $S=L\times G$ is a DSC semigroup. Then, by Lemma 6, both L and G must be DSC. If $\left|L\right|\ge 2$, then by Example 6 and [8], Example 1.3 L cannot be DSC. Thus, $\left|L\right|$ must be 1. Also by Example 7 and Corollary 2.5 of [8], G must be a finite group. Conversely, if these two conditions are satisfied, then $S=\left\{l\right\}\times G$ is a finite semigroup isomorphic to G. As G being a finite group is DSC, it follows that S is DSC, as required. □

Next, we illustrate the above corollary with the help of an example.

Example 8.

Consider the left group S of Example 1 with $L=\{l_1,l_2\}$. Let $\sigma =\{(l_1,l_1),(l_1,l_2),$$(l_2,l_2)\}$. Then, σ is a diagonal subsemigroup of $L\times L$ but not a congruence on L. Take any diagonal subsemigroup τ of $G\times G$. Then, τ is a congruence on G. By Lemma 8, S cannot be DSC.

We end this section by observing whether or not the DSC property is preserved under the common construction of left groups, such as direct products or strong semilattices of left groups. By Corollary 7, a left group is DSC if, and only if, it is isomorphic to a finite group G. As any finite direct product of finite groups is again a finite group, it must be DSC. Thus, we get an important result.

Corollary 8.

Let $S_1,S_2,\dots ,S_n$ be DSC left groups and let $S=S_1\times S_2\times \cdots \times S_n$ be their finite direct product. Then, S is DSC.

By Theorem 4.2.1 of [1], a semigroup S is a strong semilattice of groups if, and only if, S is a Clifford semigroup; that is, if, and only if, S is regular and the idempotents of S are central. A finite strong semilattice of finite groups is not necessarily a finite group and so cannot be DSC. The simplest example is the two-element semigroup $S=\{0,1\}$ with usual multiplication of integers. It is clearly a regular semigroup with two central idempotents 0 and 1. Thus, S is a finite non-group Clifford semigroup, as 0 has no inverse in S. Hence, we conclude that a strong semilattice of DSC left groups may not be a DSC semigroup.

6. Conclusions

In this paper, we explored the group-like properties of left groups. We considered the study of monogenic left group monoids and proved that a left group monoid is monogenic if, and only if, the left zero monoid component, and the group component are monogenic. We extended properties of group homorphisms to that of left group homomorphisms. In particular, we introduced the notion of the L-kernel, which extends the kernel of a group homomorphism and forms a normal sub-left group. We also characterised Green’s equivalences and congruences on left groups. The major contribution of this characterisation is the establishment of a correspondence between congruences and normal sub-left groups, analogous to a well-known correspondence between congruences and normal subgroups in groups. Finally, we explored the DSC property for left groups and proved that a non-trivial group fails to satisfy the DSC property. Basic but non-trivial examples have been provided to the reader to explain the new concepts and to justify the analogous results. Detailed counter examples have been provided for certain results that do not hold. The impacts of the results on their applications have been highlighted in the introduction. Overall, the paper has achieved its goal of exploring group-like properties for left groups. However, this paper has opened a new direction for future research. Specifically, one may explore the deep structure of left groups, their endomorphisms and their automorphisms by taking some well-known groups. One could also explore these properties for rectangular groups, strong semilattices of left groups and strong semilattices of rectangular groups (see [9,10,11,12,13,14,15]).