# Isomorphism Theorems for Groupoids and Some Applications

Authors: Jesús Ávila; Víctor Marín; Héctor Pinedo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

Using an algebraic point of view we present an introduction to the groupoid theory; that is, we give fundamental properties of groupoids as uniqueness of inverses and properties of the identities and study subgroupoids, wide subgroupoids, and normal subgroupoids. We also present the isomorphism theorems for groupoids and their applications and obtain the corresponding version of the Zassenhaus Lemma and the Jordan-Hölder theorem for groupoids. Finally, inspired by the Ehresmann-Schein-Nambooripad theorem we improve a result of R. Exel concerning a one-to-one correspondence between partial actions of groups and actions of inverse semigroups.


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## Body

## 1. Introduction

The concept of groupoid from an algebraic point of view appeared for the first time in [1]. From this setting, a (Brandt) groupoid can be seen as a generalization of a group, that is, a set with partial multiplication on it that could contain many identities.Brandt groupoids were generalized by Ehresmann in [2], where the author added further structures such as topological and differentiable structures. Other equivalent definitions of groupoids and their properties are given in [3], where a groupoid is defined as a small category where each morphism is invertible.In Definition 1.1 of [4], the author follows the definition given by Ehresmann and presents the notion of groupoid as a particular case of universal algebra, and he defines strong homomorphism for groupoids and proves the correspondence theorem in this context. The Cayley theorem for groupoids is also presented in Theorem 3.1 of [5].Recently, some applications of groupoids to the study of partial actions are presented in different branches, for instance, in [6] the author constructs a Birget-Rhodes expansion GBR associated with an ordered groupoid G and shows that it classifies partial actions of G on sets, in the topological context in [7] is treated the globalization problem, connections between partial actions of groups and groupoids are given in [8, 9]. Also, ring theoretic and cohomological results of global and partial actions of groupoids on algebras are obtained in [10–16]. Galois theoretic results for groupoid actions are obtained in [12, 17–19]. Finally, the globalization problem for partial groupoid actions has been considered in [7, 20, 21].In [19], Paques and Tamusiunas give some structural definitions in the context of groupoid such as abelian groupoid, subgroupoid, and normal subgroupoid and showed necessary and sufficient conditions for a subgroupoid to be normal. Furthermore, they built quotient groupoids.Due to the applications of the groupoids to partial actions and their usefulness, we will give an elementary introduction to the theory of groupoids from an axiomatic definition following Lawson [22].Our principal goal in this work is to continue the algebraic development of a groupoid theory. The paper is organized as follows. After the introduction, in Section2, we present groupoids from an axiomatic point and show some properties of them. In Section 3 we recall the notions of some substructures of groupoids, such as subgroupoid, wide subgroupoid, and normal subgroupoid. In Section 4, we prove the correspondence and isomorphism theorems for groupoids. In the final section we show an application of section four, we prove the Zassenhaus Lemma and Hölder theorem for groupoids, and we improve Theorem 4.2 of [23] using the Ehresmann-Schein-Nambooripad theorem.It is important to note that the notion of the groupoid can be presented from categories, algebraic structures, and universal algebra. In the last setting, the isomorphism theorems are valid, but the idea is to do an algebraic presentation and verify which assumptions are necessary. So it is possible to reach a wider audience.


## 2. Groupoids

Now, we give two definitions of groupoids from an algebraic point of view.Definition 1.
(see [22], p. 78). Let G be a set equipped with a partial binary operation on G×G which is denoted by concatenation. If g,h∈G and the product gh is defined, we write ∃gh. An element e∈G is called an identity if(1)∃egimplieseg=g,∃g′eimpliesg′e=g′.
The set of identities ofG is denoted by G0. Then G is said to be a groupoid if the following axioms hold:(i)
∃ghl, if and only if ∃ghl and ghl=ghl(ii)
∃ghl, if and only if ∃gh and ∃hl(iii)
For eachg∈G, there are unique identities dg and rg such that ∃gdg and ∃rgg(iv)
For eachg∈G, there is an element g−1∈G such that ∃g−1g, ∃gg−1, dg=g−1g, and rg=gg−1
The following definition of groupoid is presented in Definition 1.1 of [24].Definition 2.
A groupoid is a set endowed with a product map(2)G2∃g,h⟼gh∈G,where the set G2⊆G×G is called the set of composible pairs and an inverse map G∋g⟼g−1∈G such that for all g,h,l∈G the following relations are satisfied.(G1)g−1−1=g(G2) Ifg,h,h,l∈G2, then gh,l,g,hl∈G2 and ghl=ghl(G3)g−1,g∈G2 and if g,h∈G2, then g−1gh=h(G4)g,g−1∈G2 and if l,g∈G2, then lgg−1=l
We shall check that Definitions1 and 2 are equivalent. First, we need a couple of lemmas.Lemma 1 (see [25], Lemma 1.1.4).
Suppose thatG is a groupoid in the sense of Definition 1. Let g,h∈G. Then ∃gh, if and only if dg=rh.Proof.
Letg,h∈G such that ∃gh. By (iv) of Definition 1, we have that ∃g−1,∃h−1, dg=g−1g, and rh=hh−1. Since ∃gh, then ∃g−1ghh−1. That is, ∃dgrh. Now, since dg and rh are identities, then dg=dgrh=rh. Conversely, if dg=rh, then ∃grh, and since grh=ghh−1 we have ∃ghh−1. Whence by (ii) of Definition 1, we have that ∃gh.Lemma 2.
Suppose thatG is a groupoid in the sense of Definition 1. Then, the element g−1 in (iv) is unique and g−1−1=g.Proof.
For eachg∈G, assume that there exists y,z∈G such that ∃yg, ∃gy, ∃zg, ∃gz, yg=dg=zg, and gy=rg=gz. Notice that yg=zg which implies that ygg−1=zgg−1, which is defined by (ii) of Definition 1, and then by associativity, yrg=zrg. Thus, ydy=zdz, so y=z. It is analogous for gy=gz. In particular, the inverse is unique.
Finally, the equalityg−1−1=g follows from the uniqueness of the inverse of g−1.
We give the following.Proposition 1.
LetG be a set. Then, it is a groupoid in the sense of Definition 1, if and only if it is a groupoid in the sense of Definition 2.Proof.
LetG2=g,h∈G×G∃gh. By using (iv) of Definition 1, we define G∋g⟼g−1∈G. Then, by Lemma 2 this map is well defined. We shall check G1–G4 of Definition 2:(G1) It is the second assumption in Lemma2.(G2) Ifg,h,h,l∈G2, then ∃gh and ∃hl. By (i) and (ii), ∃ghl and ∃ghl, that means gh,l,g,hl∈G2 and ghl=ghl.(G3) By item (iv), we get thatg−1,g∈G2. Let h∈G with g,h∈G2. By Lemma 1, we get that dg=rh, and by using (iii), we obtain g−1gh=h.(G4) This is proved analogously to the previous item.
Conversely, suppose thatG is a set. We define a partial binary operation on G by ∃gh, if and only if g,h∈G2 and gh=mg,h. We shall check that properties (i)–(iv) in Definition 1 hold:(i)
Letg,h,l∈G such that ∃ghl. Then g,hl,h,l∈G2 and by G4, l,l−1∈G2 and h=hll−1. Thus, g,hl,hl,l−1∈G2 and by G2, ghl,l−1∈G2 and ghll−1=ghll−1=gh. In particular, g,h∈G2. We conclude that g,h,h,l∈G2 and by using G2, we get that ghl=ghl. Conversely, suppose that ∃ghl. Then, g,h,gh,l∈G2 and by G3, we have that g−1,g∈G2 and g−1gh=h. Thus, g−1,gh,gh,l∈G2 and by G2, h,l∈G2. Finally, since g,h∈G2 we obtain, again by G2, that g,hl∈G2. Hence, ∃ghl.(ii)
This is shown analogously to the previous items.(iii)
(iv) Ifg∈G, then g−1,g,g,g−1∈G2. Thus, we set rg=gg−1 and dg=g−1g. Hence, by G2, G3, and G4, ∃gdg,∃dgg, and the equalities gdg=g=rgg hold.Remark 1.
The interested reader can find another two equivalent definitions of groupoids in [26, 27].
From now on in this workG denotes a groupoid.
For the sake of completeness, we give the proof of some known consequences of Definition1.Proposition 2 (see [25], Lemma 1.1.4).
For eachg,h,k,l∈G we have:(i)
If∃gh, then dgh=dh and rgh=rg.(ii)
∃gh, if and only if ∃h−1g−1, and in this case gh−1=h−1g−1.Proof.
(i)
For the first equality, we prove thatdh satisfies the axiom (iii) from Definition 1. Indeed, assume that ∃gh. Then, ∃ghdh, dg=rh, and(3)ghdh=ghdh=gh.In a similar way, it is possible to show thatrgh=rg.(ii)
We have that∃gh, if and only if dg=rh. Notice that for any l∈G we have that(4)dl=l−1l=l−1l−1−1=rl−1.
Then,dh−1=rg−1. That is, ∃h−1g−1. Furthermore,(5)ghh−1g−1=ghh−1g−1=grhg−1=gdgg−1=gg−1=rg=rgh,h−1g−1gh=h−1g−1gh=h−1dgh=h−1rhh=h−1h=dh=dgh.
Therefore, by the uniqueness of the inverse element, we get thatgh−1=h−1g−1.
The following statements also follow from the definition of groupoid.Proposition 3.
Letg∈G. Then, the following statements hold:(6)idg=rg−1,iiddg=dg,iiirrg=rg,ivdrg=rg,vrdg=dg.Proof.
(i)
This is (4)(ii)
ddg=dg−1g=dg, where the last equality follows from (i) of Proposition 2(iii)
rrg=rgg−1=rg, where the last equality also follows from (i) of Proposition 2
Items (iv) and (v) are proved analogously.Remark 2.
LetG be a groupoid. In ([12], p. 3660), Bagio and Paques called an element e∈G an identity if e=dg, for some g∈G.Proposition 4.
LetG be a groupoid. An element e of G is an identity in the sense of Bagio and Paques, if and only if it satisfies (1).Proof.
Suppose thate=dh is an identity in the sense of Bagio and Paques, for some h∈G. By (i) of Proposition 3, e=dh=rh−1. Now, let g,g′∈G such that ∃eg and ∃g′e. By Lemma 1 and (ii)–(v) of Proposition 3, we have that dg′=re=e=de=rg, then eg=rgg=g and g′e=g′dg′=e. Therefore, e satisfies (1).
Conversely, suppose thate∈G satisfies (1). By (iii) of Definition 1, we get ∃ede and e=ede. Thus e=de, and it follows that e is an identity in the sense of Bagio and Paques.Remark 3.
It follows from the proof of Proposition4 that de=e=re, ∃ee, and ee=e=e−1 for any e∈G0. Moreover, note that the elements of G0 are the unique idempotents of G. In fact, if ∃g2 and g2=g, then ggg−1=gg−1 and so grg=gg−1=rg. Since dg=rg, it follows that g=gdg=grg=rg.Proposition 5.
Lete∈G0. Then, the set Ge=g∈Gdg=rg=e is a group.Proof.
By Remark3, we have that de=re=e. Thus e∈Ge. If g,h∈Ge, then dg=e=rh, and so ∃gh thanks to Lemma 1. Now, (i) of Proposition 2 implies that dgh=dh=e and rgh=rg=e. Hence, gh∈Ge. If g∈Ge, then by Lemma 1, ∃ge and ∃eg and we have that ge=gdg=g and eg=rgg=g. Therefore, e is the identity element of Ge. Finally, let g∈Ge. By Proposition 3, dg−1=rg=e and rg−1=dg=e. Hence, g−1∈Ge, gg−1=g−1g=e, and we conclude that Ge is a group.Definition 3.
The groupGe is called the isotropy group associated with e. The isotropy subgroupoid (see Definition 5) or the group bundle associated to G is defined by the disjoint union IsoG=∪e∈G0Ge.Remark 4.
A concept of abelian groupoid was presented in ([19], p. 111) as follows: a groupoid G is abelian if dg=rg for each g∈G; and gh=hg, for all g,h∈G with dg=rh.
We have the following.Proposition 6.
A groupoidG is abelian in the sense of Paques and Tamusiunas, if and only if G=IsoG and Ge is abelian for all e∈G0.
In the light of Proposition6, we prefer to use the following definition of the abelian groupoid.Definition 4.
(see [28], Definition 1.1).  A groupoid G is called abelian if all its isotropy groups are abelian.
Note that ifG is abelian in the sense of Paques and Tamusiunas, then it is abelian in the sense of Definition 4. Now, consider the groupoid G=g,g−1,dg,rg with dg≠rg. Then, we have that Gdg=dg and Grg=rg. That is, G is an abelian groupoid in the sense of Definition 4, but it is not a union of abelian groups.


## 3. Normal Subgroupoids, the Quotient Groupoid and Homomorphisms

In this section, we present a theory of substructures in a groupoid. We follow the definition of subgroupoid given in [19].Definition 5.
LetG be a groupoid and H a nonempty subset of G. H is said to be a subgroupoid of G if it satisfies: for all g,h∈H,(i)
g−1∈H(ii)
If∃gh, then gh∈H
IfH is a subgroupoid of G, then it is called wide if H0=G0.Remark 5.
It is clear that ifH is a subgroupoid of G, then it is a groupoid with the product (2), restricted to H2=H×H∩G2.Example 1.
LetG be a groupoid.(1)
Takea∈G such that da=ra. The set Ca=g∈Gdaga=ag is a subgroupoid of G. Indeed, first of all note that by assumption a∈Ca. If x,y∈Ca, then x,y∈Gda, xa=ax, and ya=ay. Since Gda is a group, then ∃xya and(7)xya=xya=xay=xay=axy=axy.That is,xy∈Ca. If x∈Ca, then x−1∈Gda. Hence, we have that ∃x−1a and x−1a=ax−1 since ax=xa. Observe that this example generalizes the concept of centralizer in groups.(2)
Suppose thatG is abelian and n>1. Then the set Hn=ana∈IsoG is a subgroupoid of G. If x,y∈Hn, then x=an,y=bn for some a,b∈IsoG. If ∃xy then ∃anbn, and this implies that ∃ab and thus a,b∈Ge for some e∈G0. Then, ab=ba, and so xy=abn∈H. Now, if x∈H, then x=an for some a∈G. Thus, x−1=an−1=a−1n∈Hn. Finally, note that for g∈G, dg=dgn∈Hn. Hence, G0⊆H, and we conclude that H is wide.(3)
Suppose thatG is abelian. Then, the set TorG=g∈IsoGgn∈G0for somen∈ℕ is a wide subgroupoid of G. First, it is clear that G0⊆TorG. If g,h∈TorG, then gn=e,hk=f for some n,k∈ℕ and some e,f∈G0. Thus, we obtain that dg=dgn=e=rgn=rg and dh=dhk=f=rhk=rh. If ∃gh, then dg=rh and thus e=f and gh=hg, since G is an abelian groupoid. Then ghnk=gnkhnk=gnkhkn=dg, that is, gh∈TorG. Now, since gn=e we have g−1n=gn−1=e, and hence g−1∈TorG. We conclude that TorG is a wide subgroupoid of G. Note that if we take a fixed n∈ℕ and define the set Dn=g∈Ggn∈G0, then Dn is a wide subgroupoid of G and Dn⊆TorG. That is, Dn is a subgroupoid of TorG. Observe that this example generalizes the concept of torsion subgroup in abelian groups.Proposition 7.
LetG be a groupoid and H, K subgroupoids of G. Then,(i)
IfHK is non-empty, then HK is a subgroupoid of G, if and only if HK=KH(ii)
IfH and K are wide and HK is a subgroupoid, then HK is wideProof.
The proof of (i) is similar to the group case. To prove (ii), it is enough to observe thatG0=H0=K0 and if e∈G0, then e=ee∈HK.
Now, we present the notion of anormal subgroupoid and prove several properties of them, which generalize well-known results in group theory. We follow the definition given in [19].Definition 6.
LetG be a groupoid. The subgroupoid H of G is said to be normal, denoted by H⊲G, if g−1Hg≠∅ and g−1Hg⊆H, for all g∈G, where g−1Hg=g−1hgh∈H∩Grg.Remark 6.
By the proof of Lemma 3.1 of [19], one has that g−1Hg≠∅, if and only if H is wide. Also the assertion g−1Hg⊆H is equivalent to g−1Hrgg=Hdg, for all g∈G.
Several examples of normal groupoids are presented in [19], p. 110-111.
Given a wide subgroupoidH of G, in [19], Paques and Tamusiunas define a relation on G as follows: for every g,l∈G,(8)g≡Hl⟺∃l−1gandl−1g∈H.
Furthermore, they prove that this relation is a congruence, which is an equivalence relation that is compatible with products. The equivalence class of≡H containing g is the set gH=ghh∈H∧rh=dg. This set is called the left coset of H in G containing g. Then, we have the following.Proposition 8.
(see [19], Lemma 3.12). Let H be a normal subgroupoid of G and let G/H be the set of all left cosets of H in G. Then, G/H is a groupoid such that ∃gHlH, if and only if ∃gl and the partial binary operation is given by gHlH=glH.The groupoidG/H in Proposition 8 is called the quotient groupoid of G by H.Now, we present the notion of groupoid homomorphism and prove several properties of them, which generalize well-known results in homomorphisms of groups.Definition 7.
LetG and G′ be groupoids. A map ϕ:G⟶G′ is called groupoid homomorphism if for all x,y∈G, ∃xy implies that ∃ϕxϕy, and in this case ϕxy=ϕxϕy.
Notice thatj:G⟶G/H defined by g⟼gH for all g∈G, is a surjective groupoid homomorphism.Definition 8.
Letϕ:G⟶G′ be a homomorphism of groupoids. We define the following sets:(i)
ForH⊆G, write ϕH=ϕh∈G′h∈H, the direct image of H. In particular, the set ϕG is called the image of ϕ.(ii)
Kerϕ=g∈Gϕg∈G0′, the kernel of ϕ.(iii)
Letℋ′⊆G′, ϕ−1ℋ′=g∈Gϕg∈ℋ′, the inverse image of H′ by ϕ.(iv)
ϕ is called a monomorphism if it is injective, an epimorphism if it is surjective, and an isomorphism if it is bijective.Remark 7.
IfG is abelian and H is a subgroupoid of G, then it is not difficult to show that H is abelian. Moreover, if G′ is another groupoid, such that there is a groupoid epimorphism ϕ:G⟶G′, then G′ is also abelian.Proposition 9.
Letϕ:G⟶G′ be a groupoid homomorphism. Then,(i)
For eacha∈G, ϕda=dϕa, ϕra=rϕa and ϕa−1=ϕa−1.(ii)
IfH′ is a subgroupoid of G′, then ϕ−1H′ is a subgroupoid of G. Moreover, if H′ is wide then ϕ−1H′ is wide, and it contains Kerϕ.(iii)
IfH′⊲G′, then ϕ−1H′⊲G and Kerϕ⊆ϕ−1H′. In particular, Kerϕ⊲G.Proof.
(i)
Leta∈G. Since ∃ada then ∃ϕaϕda and ϕa=ϕada=ϕaϕda. Thus, by the uniqueness of the identities ϕda=dϕa. Analogously, ϕra=rϕa. Finally, since ∃aa−1 and ∃a−1a, then ∃ϕaϕa−1 and ∃ϕa−1ϕa. Moreover,(9)ϕaϕa−1=ϕaa−1=ϕra=rϕa,ϕa−1ϕa=ϕa−1a=ϕda=dϕa,Which implies thatϕa−1=ϕa−1.(ii)
It is not difficult to show thatϕ−1H′ is a subgroupoid of G. Now suppose that H′ is wide. By item (i), we know that ϕG0⊆G0′⊆H′, that is, G0⊆ϕ−1H′. Finally, if x∈Kerϕ, then ϕx∈G0′⊆H′ and hence x∈ϕ−1H′, as desired.(iii)
By item (ii), it is enough to see thatg−1ϕ−1H′g⊆ϕ−1H′ for all g∈G. Indeed, let g−1lg∈g−1ϕ−1H′g with l∈ϕ−1H′ and dl=rl=rg. Then, ∃lg and thus ϕdl=ϕrg. We have(10)dϕg−1ϕl=dϕl=rϕg.
Then,∃ϕg−1ϕlϕg, and since ϕl∈H′ and H′⊲G′ we obtain that(11)ϕg−1lg=ϕg−1ϕlϕg=ϕg−1ϕlϕg∈H′.
Finally, to show thatKerϕ⊲G, it is enough to observe that ϕ−1G0′=Kerϕ and G0′ is normal in G′.


### 3.1. Strong Isomorphism Theorems for Groupoids

In this section, we present a special type of groupoid homomorphism, called thestrong groupoid homomorphism. Using these homomorphisms, we show the correspondence theorem and the isomorphism theorems for groupoids. This notion of strong groupoid homomorphism has been considered before by several authors (see [4], Remark 2.2).Definition 9.
Letϕ:G⟶G′ be a groupoid homomorphism. ϕ is called strong if for all x,y∈G, ∃ϕxϕy implies that ∃xy.Example 2.
LetX be a nonempty set and X2=X×X. Then X2 is a groupoid, where the product is given by: y,zx,y=x,z, for x,y,z∈X. Then, the map f:G∋g⟼dg,rg∈G02 is a strong groupoid homomorphism with kernel IsoG.Proposition 10.
Letϕ:G⟶G′ be a strong groupoid homomorphism. Then,(i)
IfH<G, then ϕH<G′, and KerϕH=ϕ−1ϕH. In particular, Imϕ=ϕG and KerϕHare subgroupoids of G′ and G, respectively.(ii)
IfH⊲G, then ϕH⊲ϕG.(iii)
([29], Proposition 9) ϕ is an injective homomorphism, if and only if Kerϕ=G0.(iv)
(The Correspondence Theorem for Groupoids) There exists a one-to-one correspondence between the setsU=HH<G∧Kerϕ⊆H and B=H′H′<ϕG′. Moreover, this correspondence preserves normal subgroupoids.Proof.
(i)
It is clear thatϕH≠∅. Let s,t∈ϕH and suppose that ∃st. Then, s=ϕx,t=ϕy for some x,y∈H. Since ϕ is strong, we have that ∃xy. Thus, st=ϕxϕy=ϕxy∈ϕH. Now, if y∈ϕH, then ϕx=y for some x∈H, and we have y−1=ϕx−1=ϕx−1∈ϕH.Now, we check the equalityKerϕH=ϕ−1ϕH. If g∈ϕ−1ϕH, then there exists h∈H with ϕg=ϕh. Since ϕ is strong, we get that ∃gh−1 and gh−1∈Kerϕ. Hence, g=gh−1h∈KerϕH. The other inclusion is clear.(ii)-(iii)
These are similar to the group case.(iv)
First, define the functionsα:U⟶B by αH=ϕH for each H∈U, and β:B⟶U by βH′=ϕ−1H′ for each H′∈B. By (i) of Proposition 9 and (ii) of Proposition 10, it has that β∘α=id and α∘β=idB. That is, α is a bijective function. The remaining proof follows from item iii of Proposition 9 and (ii) above.
Now, we use strong homomorphisms to extend to the groupoid context, a well-known result concerning the product of groups.Proposition 11.
LetH and K be subgroupoids of G. If K is normal then(i)
HK is a subgroupoid of G(ii)
IfH is normal, then HK is a normal subgroupoid of G(iii)
IfH is wide, then H∩K is a normal subgroupoid of HProof.
(i)
Consider the groupoid epimorphismϕ:G∋g⟼gK∈G/K. Then, by the definition of G/K the map ϕ is strong and Kerϕ=K. Thus, by (i) of Proposition 10 we get that KH=ϕ−1ϕH is a subgroupoid of G. Hence, the result follows from Proposition 7.(ii)
By the previous item,HK is a subgroupoid of G. Moreover, it is clear that HK is wide. Let g∈G and x∈HK∩Grg. Then, x=hk, with h∈H, k∈K, dh=rk and rh=dk=rg. Thus, ∃g−1hkg and we have that(12)g−1hkg=g−1hrkkg=g−1hrgkg=g−1hgg−1kg∈HK,that is,HK is a normal subgroupoid of G.(iii)
It is clear thatH∩K is a wide subgroupoid of H. Let g∈H and h∈H∩K with rh=dh=rg. Then ∃g−1hg and by assumptions it follows that g−1hg∈H∩K.
Next we present the isomorphism theorems for groupoids.Theorem 1.
(the first isomorphism theorem). Letϕ:G⟶G′ be a surjective strong groupoid homomorphism. Then there exists a strong isomorphism ϕ¯:G/Kerϕ⟶G′ such that ϕ=ϕ¯∘j, where j is the canonical homomorphism of G onto G/Kerϕ.Proof.
LetK=Kerϕ. We define ϕ¯:G/K⟶G′ as ϕ¯xK=ϕx, for each xK∈G/K. First of all, we show that ϕ¯ is a well defined function. Indeed, assume that xK=yK. Then ∃y−1x and y−1x∈K. That is ϕy−1x=dl for some l∈G′, and then ϕy−1ϕx=dl. Since ϕ is surjective, then l=ϕz, for some z∈G. Multiplying the above equation by ϕy, we have that(13)ϕyϕy−1ϕx=ϕydl=ϕydϕz=ϕyϕdz=ϕydz.
Then,dy=dz. So ϕydz=ϕydy=ϕy, whence ϕx=ϕy. Hence, ϕ¯ is well defined.
Now, note thatϕ¯ is a surjective strong homomorphism. Finally, we prove that ϕ¯ is injective. Indeed, assume that ϕ¯xK=ϕ¯yK, that is, ϕx=ϕy. Then, as ϕ is strong we have that ϕy−1x∈G0′. Thus, y−1x∈K and we have that xK=yK.Example 3.
(1)
Consider the identity functioniG of the groupoid G. Then, it is clear that iG is a surjective strong homomorphism and KeriG=G0. Thus, by the first isomorphism theorem, we obtain that G/G0≅G.(2)
Consider the functionθ:IsoG⟶IsoG, defined by θg=dg for all g∈IsoG. For g,h∈IsoG suppose that ∃gh. Then, dg=rh=dh, ∃dgdh, and(14)θgh=dgh=dh=dgdh=θgθh.Now, letg,h∈IsoG such that ∃θgθh. Then, ∃dgdh which implies that dg=dh and since dh=rh we obtain ∃gh. In conclusion, θ is a strong homomorphism, with Kerθ=IsoG and Imθ=G0. Whence, by the first isomorphism theorem, we obtain IsoG/IsoG≅G0.(3)
LetG and G′ be a groupoids. The set G×G′ is a groupoid with the product defined by ∃x,y⋅z,w iff ∃x⋅z∧∃y⋅w, and in this case x,y⋅z,w=x⋅z,y⋅w. Moreover, note that G×G′0=G0×G0′. If H⊲G and K⊲G′, then H×K⊲G×G′ and G×G′/H×K≅G/H×G′/K. Indeed, it is clear that H×K⊲G×G′. For the second affirmation, define ψ:G×G′⟶G/H×G′/K by g,g′⟼gH,g′K, and note that ψ is a strong homomorphism. Moreover,(15)Kerψ=g,g′∈G×G′ψg,g′∈G/H0×G′/K0=g,g′∈G×G′gH,g′K∈G0H×G0′K=g,g′∈G×G′g,g′∈H×K=H×K.
Thus, by the first isomorphism theorem the result follows.Theorem 2.
(the second isomorphism theorem). LetG be a groupoid, ℳ a wide subgroupoid of G, and N a normal subgroupoid of G. Then, ℳ∩N⊲ℳ and(16)ℳℳ∩N≅ℳNN.Proof.
First, note that by (i) of Proposition11, ℳN is a subgroupoid of G. Moreover, since N⊲G we have N⊲ℳN. Also, it is clear that ℳ∩N⊲ℳ.
We considerψ:ℳ⟶ℳN/N given by ψm=mN for all m∈ℳ. Then, it is clear that ψ is a strong homomorphism. Furthermore, if mnN∈ℳN/N, then mnN=mN. Thus, ψ is surjective. Now,(17)ℳNN0=dxNxN∈ℳNN=dxNx∈ℳN=dxNx=mn∧dm=rn=dnNn∈N.
On the other hand,Kerψ=m∈ℳmN=dnNfor  somen∈N=ℳ∩N. Indeed, if t∈ℳ∩N, then tN=dt−1N and thus t∈Kerψ. For the other inclusion, if m∈Kerψ, then m∈ℳ and mN=dnN for some n∈N. Thus, ∃dnm and dnm∈N. That is, dn=rm, and we have dnm=rmm=m∈N. Finally, by Theorem 1, we conclude that ℳ/ℳ∩N≅ℳN/N, as desired.Remark 8.
GivenM and N as in Theorem 2, we saw in the proof of the same theorem that N⊲ℳN, which implies that N0⊆ℳ0. Indeed, let e∈N0. By Proposition 4, there is n∈Nsuch that e=rn. Since n=m′n′ with m′,n′∈M×N∩G2, then e=rm′∈ℳ0. Conversely, the condition N0⊆ℳ0, clearly implies that N⊲ℳN. From this, we conclude that for ℳ and N subgroupoids of G, we have that N⊲ℳN, if and only if N0⊆ℳ0.Theorem 3.
(the third isomorphism theorem). LetG be a groupoid, H⊲G and K⊲G with K⊆H. Then, H/K⊲G/K and(18)G/KH/K≅GH.Proof.
Defineφ:G/K⟶G/H by φgK=gH. First of all, we show that φ is a well-defined function. Indeed, if gK=lK, then ∃l−1g and l−1g∈K. Since K<H, we have l−1g∈H, and hence gH=lH. Now,(19)Kerφ=gKgH∈GH0=gKgH∈G0H=gKg∈H=HK.
Thus,H/K⊲G/K and Theorem 1 implies that(20)G/KH/K≅GH.


## 3.1. Strong Isomorphism Theorems for Groupoids

In this section, we present a special type of groupoid homomorphism, called thestrong groupoid homomorphism. Using these homomorphisms, we show the correspondence theorem and the isomorphism theorems for groupoids. This notion of strong groupoid homomorphism has been considered before by several authors (see [4], Remark 2.2).Definition 9.
Letϕ:G⟶G′ be a groupoid homomorphism. ϕ is called strong if for all x,y∈G, ∃ϕxϕy implies that ∃xy.Example 2.
LetX be a nonempty set and X2=X×X. Then X2 is a groupoid, where the product is given by: y,zx,y=x,z, for x,y,z∈X. Then, the map f:G∋g⟼dg,rg∈G02 is a strong groupoid homomorphism with kernel IsoG.Proposition 10.
Letϕ:G⟶G′ be a strong groupoid homomorphism. Then,(i)
IfH<G, then ϕH<G′, and KerϕH=ϕ−1ϕH. In particular, Imϕ=ϕG and KerϕHare subgroupoids of G′ and G, respectively.(ii)
IfH⊲G, then ϕH⊲ϕG.(iii)
([29], Proposition 9) ϕ is an injective homomorphism, if and only if Kerϕ=G0.(iv)
(The Correspondence Theorem for Groupoids) There exists a one-to-one correspondence between the setsU=HH<G∧Kerϕ⊆H and B=H′H′<ϕG′. Moreover, this correspondence preserves normal subgroupoids.Proof.
(i)
It is clear thatϕH≠∅. Let s,t∈ϕH and suppose that ∃st. Then, s=ϕx,t=ϕy for some x,y∈H. Since ϕ is strong, we have that ∃xy. Thus, st=ϕxϕy=ϕxy∈ϕH. Now, if y∈ϕH, then ϕx=y for some x∈H, and we have y−1=ϕx−1=ϕx−1∈ϕH.Now, we check the equalityKerϕH=ϕ−1ϕH. If g∈ϕ−1ϕH, then there exists h∈H with ϕg=ϕh. Since ϕ is strong, we get that ∃gh−1 and gh−1∈Kerϕ. Hence, g=gh−1h∈KerϕH. The other inclusion is clear.(ii)-(iii)
These are similar to the group case.(iv)
First, define the functionsα:U⟶B by αH=ϕH for each H∈U, and β:B⟶U by βH′=ϕ−1H′ for each H′∈B. By (i) of Proposition 9 and (ii) of Proposition 10, it has that β∘α=id and α∘β=idB. That is, α is a bijective function. The remaining proof follows from item iii of Proposition 9 and (ii) above.
Now, we use strong homomorphisms to extend to the groupoid context, a well-known result concerning the product of groups.Proposition 11.
LetH and K be subgroupoids of G. If K is normal then(i)
HK is a subgroupoid of G(ii)
IfH is normal, then HK is a normal subgroupoid of G(iii)
IfH is wide, then H∩K is a normal subgroupoid of HProof.
(i)
Consider the groupoid epimorphismϕ:G∋g⟼gK∈G/K. Then, by the definition of G/K the map ϕ is strong and Kerϕ=K. Thus, by (i) of Proposition 10 we get that KH=ϕ−1ϕH is a subgroupoid of G. Hence, the result follows from Proposition 7.(ii)
By the previous item,HK is a subgroupoid of G. Moreover, it is clear that HK is wide. Let g∈G and x∈HK∩Grg. Then, x=hk, with h∈H, k∈K, dh=rk and rh=dk=rg. Thus, ∃g−1hkg and we have that(12)g−1hkg=g−1hrkkg=g−1hrgkg=g−1hgg−1kg∈HK,that is,HK is a normal subgroupoid of G.(iii)
It is clear thatH∩K is a wide subgroupoid of H. Let g∈H and h∈H∩K with rh=dh=rg. Then ∃g−1hg and by assumptions it follows that g−1hg∈H∩K.
Next we present the isomorphism theorems for groupoids.Theorem 1.
(the first isomorphism theorem). Letϕ:G⟶G′ be a surjective strong groupoid homomorphism. Then there exists a strong isomorphism ϕ¯:G/Kerϕ⟶G′ such that ϕ=ϕ¯∘j, where j is the canonical homomorphism of G onto G/Kerϕ.Proof.
LetK=Kerϕ. We define ϕ¯:G/K⟶G′ as ϕ¯xK=ϕx, for each xK∈G/K. First of all, we show that ϕ¯ is a well defined function. Indeed, assume that xK=yK. Then ∃y−1x and y−1x∈K. That is ϕy−1x=dl for some l∈G′, and then ϕy−1ϕx=dl. Since ϕ is surjective, then l=ϕz, for some z∈G. Multiplying the above equation by ϕy, we have that(13)ϕyϕy−1ϕx=ϕydl=ϕydϕz=ϕyϕdz=ϕydz.
Then,dy=dz. So ϕydz=ϕydy=ϕy, whence ϕx=ϕy. Hence, ϕ¯ is well defined.
Now, note thatϕ¯ is a surjective strong homomorphism. Finally, we prove that ϕ¯ is injective. Indeed, assume that ϕ¯xK=ϕ¯yK, that is, ϕx=ϕy. Then, as ϕ is strong we have that ϕy−1x∈G0′. Thus, y−1x∈K and we have that xK=yK.Example 3.
(1)
Consider the identity functioniG of the groupoid G. Then, it is clear that iG is a surjective strong homomorphism and KeriG=G0. Thus, by the first isomorphism theorem, we obtain that G/G0≅G.(2)
Consider the functionθ:IsoG⟶IsoG, defined by θg=dg for all g∈IsoG. For g,h∈IsoG suppose that ∃gh. Then, dg=rh=dh, ∃dgdh, and(14)θgh=dgh=dh=dgdh=θgθh.Now, letg,h∈IsoG such that ∃θgθh. Then, ∃dgdh which implies that dg=dh and since dh=rh we obtain ∃gh. In conclusion, θ is a strong homomorphism, with Kerθ=IsoG and Imθ=G0. Whence, by the first isomorphism theorem, we obtain IsoG/IsoG≅G0.(3)
LetG and G′ be a groupoids. The set G×G′ is a groupoid with the product defined by ∃x,y⋅z,w iff ∃x⋅z∧∃y⋅w, and in this case x,y⋅z,w=x⋅z,y⋅w. Moreover, note that G×G′0=G0×G0′. If H⊲G and K⊲G′, then H×K⊲G×G′ and G×G′/H×K≅G/H×G′/K. Indeed, it is clear that H×K⊲G×G′. For the second affirmation, define ψ:G×G′⟶G/H×G′/K by g,g′⟼gH,g′K, and note that ψ is a strong homomorphism. Moreover,(15)Kerψ=g,g′∈G×G′ψg,g′∈G/H0×G′/K0=g,g′∈G×G′gH,g′K∈G0H×G0′K=g,g′∈G×G′g,g′∈H×K=H×K.
Thus, by the first isomorphism theorem the result follows.Theorem 2.
(the second isomorphism theorem). LetG be a groupoid, ℳ a wide subgroupoid of G, and N a normal subgroupoid of G. Then, ℳ∩N⊲ℳ and(16)ℳℳ∩N≅ℳNN.Proof.
First, note that by (i) of Proposition11, ℳN is a subgroupoid of G. Moreover, since N⊲G we have N⊲ℳN. Also, it is clear that ℳ∩N⊲ℳ.
We considerψ:ℳ⟶ℳN/N given by ψm=mN for all m∈ℳ. Then, it is clear that ψ is a strong homomorphism. Furthermore, if mnN∈ℳN/N, then mnN=mN. Thus, ψ is surjective. Now,(17)ℳNN0=dxNxN∈ℳNN=dxNx∈ℳN=dxNx=mn∧dm=rn=dnNn∈N.
On the other hand,Kerψ=m∈ℳmN=dnNfor  somen∈N=ℳ∩N. Indeed, if t∈ℳ∩N, then tN=dt−1N and thus t∈Kerψ. For the other inclusion, if m∈Kerψ, then m∈ℳ and mN=dnN for some n∈N. Thus, ∃dnm and dnm∈N. That is, dn=rm, and we have dnm=rmm=m∈N. Finally, by Theorem 1, we conclude that ℳ/ℳ∩N≅ℳN/N, as desired.Remark 8.
GivenM and N as in Theorem 2, we saw in the proof of the same theorem that N⊲ℳN, which implies that N0⊆ℳ0. Indeed, let e∈N0. By Proposition 4, there is n∈Nsuch that e=rn. Since n=m′n′ with m′,n′∈M×N∩G2, then e=rm′∈ℳ0. Conversely, the condition N0⊆ℳ0, clearly implies that N⊲ℳN. From this, we conclude that for ℳ and N subgroupoids of G, we have that N⊲ℳN, if and only if N0⊆ℳ0.Theorem 3.
(the third isomorphism theorem). LetG be a groupoid, H⊲G and K⊲G with K⊆H. Then, H/K⊲G/K and(18)G/KH/K≅GH.Proof.
Defineφ:G/K⟶G/H by φgK=gH. First of all, we show that φ is a well-defined function. Indeed, if gK=lK, then ∃l−1g and l−1g∈K. Since K<H, we have l−1g∈H, and hence gH=lH. Now,(19)Kerφ=gKgH∈GH0=gKgH∈G0H=gKg∈H=HK.
Thus,H/K⊲G/K and Theorem 1 implies that(20)G/KH/K≅GH.


## 4. Normal and Subnormal Series for Groupoids

In this section, we present some applications of the isomorphism theorems of groupoids to normal and subnormal series. In particular, we show that the Jordan-Hölder Theorem is also fulfilled in the context of groupoids. First, we introduce the following natural definitions.Definition 10.
LetG be a groupoid. Then,(i)
A subnormal series of a groupoidG, is a chain of subgroupoids G0=G>G1>⋯>Gnsuch that Gi+1 is normal in Gi for 0≤i≤n. The factors of the series are the quotient groupoids Gi/Gi+1. The length of the series is the number of strict inclusions. A subnormal series such that Gi is normal in G for all i, is called normal.(ii)
LetS:G=G0>G1>⋯>Gn be a subnormal series. A one-step refinement of this series is any series of the form G=G0>⋯>Gi>N>Gi+1>⋯>Gn or G=G0>⋯>Gn>N, where N is a normal subrgoupoid of Gi and Gi+1 is normal in N (if i<n ). A refinement of a subnormal series S is any subnormal series obtained from S by a finite sequence of one-step refinements. A refinement of S is called to be proper if it is larger than the length of S.(iii)
A subnormal seriesG=G0>G1>⋯>Gn=G0 is a composition series if each factor is simple; that is, its only normal subgroupoids are G and G0, and it is solvable if each factor is abelian.Remark 9.
It follows from (iv) of Proposition10 that if N is a normal subgroupoid of a groupoid G, every normal subgroupoid of G/N is of the form H/N where H is a normal subroupoid of G, which contains N. Thus, if G≠N, then G/N is simple, if and only if N is a maximal element in the set of all the normal subgroupoids ℳ of G, such that ℳ≠G.Proposition 12.
LetG be a groupoid. Then,(i)
IfG is finite, then it has a composition series(ii)
Every refinement of a solvable series is a solvable series(iii)
A subnormal series is a composition series, if and only if it has no proper refinementsProof.
(i)
LetG1 be a maximal normal subgroupoid of G. Then, G/G1 is simple by (iv) of Proposition 10. Let G2 be a maximal normal subgroupoid of G1, and so on. Now, since G is finite, this process must end with Gn=G0. Thus, G=G0>G1>⋯>Gn=G0 is a composition series.(ii)
Here we use Remark7 to observe that if Gi/Gi+1 is abelian and Gi+1⊲H⊲Gi, then H/Gi+1 is abelian since it is a subgroupoid of Gi/Gi+1. Moreover, Gi/H is abelian since it is isomorphic to Gi/Gi+1/H/Gi+1 by Theorem 3.(iii)
It follows from (iv) of Proposition10 and that a subnormal series G=G0>G1>⋯>Gn=G0 has a proper refinement, if and only if there is a subgroupoid H such that for some i,Gi+1⊲H⊲Gi with H proper in Gi and Gi+1 proper in H.Definition 11.
Two subnormal seriesS and T of a groupoid G are equivalent, if there is a one-to-one correspondence between the nontrivial factors of S and the nontrivial factors of T, such that the corresponding factors are isomorphic groupoids.Lemma 3.
IfS is a composition series of a groupoid G, then any refinement of S is equivalent to S.Proof.
LetS:G=G0>G1>⋯>Gn=G0. By Proposition 12 (iii), S has no proper refinement. Thus, the only possible refinements of S are obtained by inserting additional copies of each Gi. Whence, any refinement of S has exactly the same nontrivial factors as S. Therefore, it is equivalent to S.Lemma 4.
(Zassenhaus theorem for groupoids). LetA∗,A,ℬ∗,ℬ be wide subgroupoids of a groupoid G such that(i)
A∗ is normal in A(ii)
ℬ∗ is normal in ℬ
Then,A∗A∩ℬ and ℬ∗A∩ℬ are subgroupoids of G such that(i)
A∗A∩ℬ∗ is a normal subgroupoid of A∗A∩ℬ(ii)
ℬ∗A∗∩ℬ is a normal subgroupoid of ℬ∗A∩ℬ(iii)
A∗A∩ℬ/A∗A∩ℬ∗≅ℬ∗A∩ℬ/ℬ∗A∗∩ℬProof.
(i) Sinceℬ∗ is normal in ℬ, A∩ℬ∗=A∩ℬ∩ℬ∗ is a normal subgroupoid of A∩ℬ thanks to (iii) of Proposition 11; similarly, A∗∩ℬ is normal in A∩ℬ. Then, D=A∗∩ℬA∩ℬ∗ is a normal subgroupoid of A∩ℬ by (ii) of Proposition 11. Also, by this same proposition, we have that A∗A∩ℬ and ℬ∗A∩ℬ are subgroupoids of A and ℬ respectively. Now, we define(21)τ:A∗A∩ℬ∋ac⟼τac=Dc∈A∩ℬD,for all a∈A∗,c∈A∩ℬ. The map τ is well defined since ac=a1c1 with a,a1∈A∗;c,c1∈A∩ℬ, implies that(22)c1c−1=a1−1a∈A∩ℬ∩A∗=A∗∩ℬ⊆D,whence Dc1=Dc. The map τ is clearly a strong epimorphism, and the equality Kerτ=A∗A∩ℬ∗ is shown in an analogous way to the group case.
Thus, Proposition9 (iv) implies that A∗A∩ℬ∗ is normal in A∗A∩ℬ, and by the first isomorphism theorem, we get A∗A∩ℬ/A∗A∩ℬ∗≅A∩ℬ/D.
A symmetric argument shows thatℬ∗ℬ∩A∗ is normal in ℬ∗A∩ℬ and ℬ∗A∩ℬ/ℬ∗A∗∩ℬ≅A∩ℬ/D. Whence (iii) follows.Proposition 13.
(Schreier theorem for groupoids). Any two subnormal (resp. normal) series of a groupoidG have subnormal (resp. normal) refinement, which are equivalent.Proof.
It follows from Lemma4, (ii) of Proposition 11, and Proposition 7 (1).Proposition 14.
(Jordan-Hölder theorem for groupoids). Any two composition series of a groupoidG are equivalent.Proof.
It follows from Proposition13 and Lemma 3.


### 4.1. Some Remarks on the Equivalence between Inductive Groupoids and Inverse Semigroups

Recall that an inverse semigroup, is a semigroupS such that for any s∈S there is a unique s∗∈S such that s∗=s∗ss∗ and s=ss∗s. Now, let X be a set and consider the inverse semigroup(23)GX=f:A⟶BA⊆X,B⊆Xandfis  a  bijection.We recall the following.Definition 12.
LetS be a semigroup. An action of S on X is a semigroup homomorphism ϕ:S⟶GX.
It follows from Theorem 4.2 of [23]) that partial actions of a group G on X are in one-to-one correspondence with actions of EG on X, where EG is the semigroup generated by the symbols gg∈G under the following relations: for g,h∈G,(24)g−1gh=g−1gh,ghh−1=ghh−1,g1=g.
The semigroupEG was introduced in [23], and it is called the Exel semigroup ofG.Remark 10.
Now, we present some facts aboutEG.(1)
The semigroupEG is a monoid with 1EG=1.(2)
([23], Proposition 2.5). For each g∈G let γg=gg−1. Then, γg is an idempotent of EG, each element α∈EG may be uniquely written (up to the order of the si’s) as(25)α=γs1,γs2,…,γsng,for somes1,s2,…,sn,g∈G, with g≠si≠sj≠g,i≠j and si≠1, for i∈1,…,n. From equation (25), it follows that any idempotent in EG has the form γs1γs2…γsn for some (uniquely) s1,s2,…,sn,g∈G.(3)
([23], Theorem 3.4) The set EG is an inverse semigroup. In particular, the idempotents of EG commute (see [22], Theorem 3)).
Given an inverse semigroupS and s,t∈S, one defines the restricted product(26)s⋅texists if and only ifs∗s=tt∗.
It follows from Proposition 3.1.4 and Proposition 4.1.1 thatS,⋅,≤ is an inductive groupoid (see [22], p. 108), where ≤ is the natural partial order defined on S. Then, by using the restricted product in GX, we have that(27)G2X=f,g∈GX×GXimg=domf,and GX is a groupoid with the product given by composition of maps restricted to G2X. Moreover, GX0=idAA⊆X.
With respect to the semigroupEG, we have the following result.Proposition 15.
Letα=γs1γs2…γsng and β=γt1γt2…γtml where s1,s2,…,sn,t1,t2,…,tm,g,l∈G are as in (25) of Remark 10. Then αα∗=β∗β, if and only if s1,s2,…,sn,g=t1,t2,…,tn,l−1.Proof.
We have thatαα∗=γs1,γs2,…,γsngg−1γs1,γs2,…,γsn=γs1,γs2,…,γsn,γg. Then,(28)αα∗=β∗β⟺γs1,γs2,…,γsn,γg=γt1,γt2,…,γtm,γt−1⟺s1,s2,…,sn,g=t1,t2,…,tn,l−1,where the last equivalence follows from (2) of Remark 10.
Using the restricted product to provideEG with a groupoid structure we get by Proposition 15 that,(29)EG2=γs1,…,γsng,γt1,…,γtmls1,…,sn,g=t1,…,tn,l−1,and EG0=γs1γs2…γsnsi≠sj,i≠j,n∈ℕ.
From [7], it follows that a global action β of a groupoid G on X is a family of bijections β=βg:Xg−1⟶Xgg∈G such that(i)
X=∪e∈G0Xe(ii)
βe=idXe, for all e∈G0(iii)
βg∘βh=βgh, for all g,h∈G2
Then according to ([7], Proposition 10), global actions of G on X correspond to groupoid homomorphism G⟶GX. On the other hand, in the case when G is a group we obtain the definition of a partial group action on a set (see [2], Definition 1.2)
IfG is an inductive groupoid, then Proposition 4.1.7 of [22]) implies that G,⊗ is an inverse semigroup, where ⊗ denotes the pseudo product defined on G (see [22], p. 112).
Then, we have the following.Proposition 16.
For every group G and any set X, there is a one-to-one correspondence between(1)
Partial actions ofG on X(2)
Unital semigroup actions ofEG on X(3)
Groupoid homomorphismsEG⟶GX(4)
Groupoid actions ofEG on XProof.
We have already observed that there is a one-to-one correspondence between partial actions ofG on X and semigroup actions of EG on X, and between groupoid homomorphisms EG⟶GX, and global actions of G on X. Moreover, given a semigroup action ϕ:EG⟶GX, then let βα=ϕα and Xα=imϕα. Since ϕ1=idX, one has that the family β=βα:Xα−1⟶Xαα∈EG is a global action of EG on X. Conversely, given a global action β=βα:Xα−1⟶Xαα∈EG of EG on X, let φ:EG∋α⟼βα∈GX. Then, φ is an action of EG on X. Indeed, if α,β∈EG, then by Proposition 4.1.7, we have that αγ=α⊗γ and ϕαγ=ϕα⊗γ=ϕα⊗ϕγ=ϕα∘ϕγ, as desired.


## 4.1. Some Remarks on the Equivalence between Inductive Groupoids and Inverse Semigroups

Recall that an inverse semigroup, is a semigroupS such that for any s∈S there is a unique s∗∈S such that s∗=s∗ss∗ and s=ss∗s. Now, let X be a set and consider the inverse semigroup(23)GX=f:A⟶BA⊆X,B⊆Xandfis  a  bijection.We recall the following.Definition 12.
LetS be a semigroup. An action of S on X is a semigroup homomorphism ϕ:S⟶GX.
It follows from Theorem 4.2 of [23]) that partial actions of a group G on X are in one-to-one correspondence with actions of EG on X, where EG is the semigroup generated by the symbols gg∈G under the following relations: for g,h∈G,(24)g−1gh=g−1gh,ghh−1=ghh−1,g1=g.
The semigroupEG was introduced in [23], and it is called the Exel semigroup ofG.Remark 10.
Now, we present some facts aboutEG.(1)
The semigroupEG is a monoid with 1EG=1.(2)
([23], Proposition 2.5). For each g∈G let γg=gg−1. Then, γg is an idempotent of EG, each element α∈EG may be uniquely written (up to the order of the si’s) as(25)α=γs1,γs2,…,γsng,for somes1,s2,…,sn,g∈G, with g≠si≠sj≠g,i≠j and si≠1, for i∈1,…,n. From equation (25), it follows that any idempotent in EG has the form γs1γs2…γsn for some (uniquely) s1,s2,…,sn,g∈G.(3)
([23], Theorem 3.4) The set EG is an inverse semigroup. In particular, the idempotents of EG commute (see [22], Theorem 3)).
Given an inverse semigroupS and s,t∈S, one defines the restricted product(26)s⋅texists if and only ifs∗s=tt∗.
It follows from Proposition 3.1.4 and Proposition 4.1.1 thatS,⋅,≤ is an inductive groupoid (see [22], p. 108), where ≤ is the natural partial order defined on S. Then, by using the restricted product in GX, we have that(27)G2X=f,g∈GX×GXimg=domf,and GX is a groupoid with the product given by composition of maps restricted to G2X. Moreover, GX0=idAA⊆X.
With respect to the semigroupEG, we have the following result.Proposition 15.
Letα=γs1γs2…γsng and β=γt1γt2…γtml where s1,s2,…,sn,t1,t2,…,tm,g,l∈G are as in (25) of Remark 10. Then αα∗=β∗β, if and only if s1,s2,…,sn,g=t1,t2,…,tn,l−1.Proof.
We have thatαα∗=γs1,γs2,…,γsngg−1γs1,γs2,…,γsn=γs1,γs2,…,γsn,γg. Then,(28)αα∗=β∗β⟺γs1,γs2,…,γsn,γg=γt1,γt2,…,γtm,γt−1⟺s1,s2,…,sn,g=t1,t2,…,tn,l−1,where the last equivalence follows from (2) of Remark 10.
Using the restricted product to provideEG with a groupoid structure we get by Proposition 15 that,(29)EG2=γs1,…,γsng,γt1,…,γtmls1,…,sn,g=t1,…,tn,l−1,and EG0=γs1γs2…γsnsi≠sj,i≠j,n∈ℕ.
From [7], it follows that a global action β of a groupoid G on X is a family of bijections β=βg:Xg−1⟶Xgg∈G such that(i)
X=∪e∈G0Xe(ii)
βe=idXe, for all e∈G0(iii)
βg∘βh=βgh, for all g,h∈G2
Then according to ([7], Proposition 10), global actions of G on X correspond to groupoid homomorphism G⟶GX. On the other hand, in the case when G is a group we obtain the definition of a partial group action on a set (see [2], Definition 1.2)
IfG is an inductive groupoid, then Proposition 4.1.7 of [22]) implies that G,⊗ is an inverse semigroup, where ⊗ denotes the pseudo product defined on G (see [22], p. 112).
Then, we have the following.Proposition 16.
For every group G and any set X, there is a one-to-one correspondence between(1)
Partial actions ofG on X(2)
Unital semigroup actions ofEG on X(3)
Groupoid homomorphismsEG⟶GX(4)
Groupoid actions ofEG on XProof.
We have already observed that there is a one-to-one correspondence between partial actions ofG on X and semigroup actions of EG on X, and between groupoid homomorphisms EG⟶GX, and global actions of G on X. Moreover, given a semigroup action ϕ:EG⟶GX, then let βα=ϕα and Xα=imϕα. Since ϕ1=idX, one has that the family β=βα:Xα−1⟶Xαα∈EG is a global action of EG on X. Conversely, given a global action β=βα:Xα−1⟶Xαα∈EG of EG on X, let φ:EG∋α⟼βα∈GX. Then, φ is an action of EG on X. Indeed, if α,β∈EG, then by Proposition 4.1.7, we have that αγ=α⊗γ and ϕαγ=ϕα⊗γ=ϕα⊗ϕγ=ϕα∘ϕγ, as desired.



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