## Abstract

In this paper, completely regular endomorphisms of the join of split graphs are investigated. We give conditions under which all completely regular endomorphisms of the join of two split graphs form a monoid.

Show Summary Details# The join of split graphs whose completely regular endomorphisms form a monoid

#### Open Access

## Abstract

## 1 Introduction and preliminaries

#### Lemma 1.1

#### Lemma 1.2

#### Lemma 1.3

#### Lemma 1.4

## 2 Main results

#### Theorem 2.1

#### Proof

#### Lemma 2.2

#### Proof

#### Lemma 2.3

#### Proof

#### Lemma 2.4

#### Proof

#### Theorem 2.5

#### Proof

#### Corollary 2.6

#### Proof

#### Theorem 2.7

#### Proof

#### Theorem 2.8

#### Proof

## Acknowledgement

## References

## About the article

More options …# Open Mathematics

### formerly Central European Journal of Mathematics

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Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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In this paper, completely regular endomorphisms of the join of split graphs are investigated. We give conditions under which all completely regular endomorphisms of the join of two split graphs form a monoid.

Keywords: Endomorphism; Monoid; Join of split graphs; Completely regular

Endomorphism monoids of graphs are generalizations of automorphism groups of graphs. In recent years, much attention has been paid to endomorphism monoids of graphs and many interesting results concerning graphs and their endomorphism monoids have been obtained (cf. [1–4]). The endomorphism monoids of graphs have valuable applications (cf. [1]) and are related to automata theory (cf. [2, 5]). Let *X* be a graph. Denote by *End*(*X*) the set of all endomorphisms of *X*. It is known that *End*(*X*) forms a monoid with respect to composition of mappings. We call *End*(*X*) the endomorphism monoid of *X*. An element *a* of a semigroup *S* is said to be *regular* if there exists *x* ∈ *S* such that *axa* = *a*. Let *f* ∈ *End*(*X*). Then *f* is called a *regular endomorphism* of *X* if it is a regular element in *End* (*X*). Denote by *rEnd*(*X*) the set of all regular endomorphisms of *X*. For a monoid *S*, the composition of its two regular elements is not regular in general. So it is natural to ask: Under what conditions does the set *rEnd*(*X*) form a monoid for a graph *X*? In [6], Hou, Gu and Shang characterized the regular endomorphisms of the join of split graphs and the conditions under which the regular endomorphisms of the join of split graphs form a monoid were given. The endomorphism monoids of split graphs and the joins of split graphs were studied by several authors (cf. [7–13]).

An element *a* of a semigroup *S* is said to be completely regular if *a* = *axa* and *xa* = *ax* hold for some *x* ∈ *S*. Let *f* ∈ *End*(*X*) for a graph *X*. Then *f* is called a *completely regular endomorphism* of *X* if it is a completely regular element in *End*(*X*). Denote by *cEnd*(*X*) the set of all completely regular endomorphisms of graph *X*. In general, the composition of its two completely regular elements is also not completely regular for a monoid *S*. So it is natural to ask: Under what conditions does the set *cEnd*(*X*) form a monoid for *X*? However, it seems difficult to obtain a general answer to this question. Therefore a natural strategy for work towards answering this question is to find various kinds of conditions for various kinds of graphs. In [14], completely regular endomorphisms of split graphs were characterized and the conditions under which the completely regular endomorphisms of split graphs form a monoid were given. In this paper we give an answer to this question in the range of the joins of split graphs.

The graphs considered in this paper are finite undirected graphs without loops and multiple edges. The vertex set of *X* is denoted by *V*(*X*) and the edge set of *X* is denoted by *E*(*X*). If two vertices *x*_{1} and *x*_{2} are adjacent in *X*, the edge joining them is denoted by {*x*_{1}, *x*_{2}}. For a vertex *v* of *X*, denote by *N _{X}*(

Let *X* and *Y* be two graphs. A mapping *f* from *V*(*X*) to *V*(*Y*) is called a *homomorphism* (from *X* to *Y*) if {*x*_{1}, *x*_{2}} ∈ *E*(*X*) implies that {*f*(*x*_{1}), *f*(*x*_{2})} ∈ *E*(*Y*). A homomorphism *f* from *X* to itself is called an *endomorphism* of *X*. A endomorphism *f* is said to be *half*-*strong* if {*f*(*a*), *f*(*b*)} ∈ *E*(*X*) implies that there exist *x*_{1}, *x*_{2} ∈ *V*(*X*) with *f*(*x*_{1}) = *f*(*a*) and *f*(*x*_{2}) = *f*(*b*) such that {*x*_{1}, *x*_{2}} ∈ *E*(*X*). Denote by *hEnd*(*X*) the set of all half-strong endomorphisms of *X*.

A *retraction* of a graph *X* is an endomorphism *f* from *X* to a subgraph *Y* of *X* such that the restriction *f* |_{Y} to *V*(*Y*) is the identity mapping on *V*(*Y*). It is well known that the idempotents of *End*(*X*) are retractions of *X*. Denote by *Idpt*(*X*) the set of all idempotents of *End*(*X*). Let *f* ∈ *End*(*X*). A subgraph of *X* is called the *endomorphic image* of *X* under *f*, denoted by *I _{f}*, if

We use the standard terminology and notation of semigroup theory as in [2, 15] and of graph theory as in [16]. We list some known results which are used in this paper.

([17]). *Let X be a graph and f* ∈ *End*(*X*). *Then*

*f*∈*hEnd*(*X*)*if and only if I*._{f}is an induced subgraph of X*If f is regular, then f*∈*hEnd*(*X*).

([18]). *Let G be a graph and f* ∈ *End*(*G*). *Then f is completely regular if and only if there exists g* ∈ *I dpt*(*G*) *such that ρ _{g}* =

([6]). *Let X* + *Y be a join of split graphs and f* ∈ *End*(*X* + *Y*). *Then the following statements are equivalent*.

*There exists h*∈*I dpt*(*X*+*Y*)*such that I*=_{h}*I*._{f}*I*+_{f}is an induced subgraph of X*Y and*{*x*,*y*} ∉*E*(*X*+*Y*)*for any x*∈*K*\_{i}*I*∈_{f}and y*S*∩_{i}*I*= 1, 2_{f}(where i*)*.

([6]). *Let X* + *Y be a join of split graphs and f* ∈ *End*(*X* + *Y*). *Then there exists g* ∈ *I dpt*(*X* + *Y*) *such that ρ _{g}* =

Let *X* be a split graph with *V*(*X*) = *K*_{1} ∪ *S*_{1}, where *K*_{1} = {*k*_{1}, *k*_{2}, ⋯, *k _{n}*} is a maximal complete set and

First, we give a characterization of the completely regular endomorphisms for *X* + *Y*.

*Let X* + *Y be a join of split graphs and f* ∈ *End*(*X* + *Y*). *Then f is completely regular if and only if the following conditions hold*:

*I*+_{f}is an induced subgraph of X*Y and*{*x*,*y*} ∉*E*(*X*+*Y*)*for any x*∈*K*\*I*∈_{f}and y*S*∩*I*._{f}*N*(*b*) = ∪_{x∈[b]ρf}*N*(*x*)*for any b*∈*V*(*I*)_{f}*with*[*b*]_{ρf}⊆*S*.*f*(*a*)≠*f*(*b*)*for any a*,*b*∈*V*(*I*)_{f}*with a*≠*b*.

Necessity. Since *f* is completely regular, by Lemma 1.2, there exists *g* ∈ *I dpt*(*X* + *Y*) such that *ρ _{g}* =

Sufficiency. Let *X* + *Y* be a join of split graphs and *f* ∈ *End*(*X* + *Y*) be such that (1), (2) and (3). Note that *f*(*a*) ≠ *f*(*b*) for any *a*, *b* ∈ *V*(*I _{f}*). Then for any

Assume that *K* ⊆ *I _{f}*. Then

Assume that |*K* ∩ *I _{f}*| =

Assume that |*K* ∩ *I _{f}*| =

It is easy to check that *g* ∈ *I dpt*(*X* + *Y*), *ρ _{g}* =

Next, we start to seek the conditions for *X* + *Y* under which *cEnd*(*X* + *Y*) forms a monoid.

*If there exist y _{i}*,

Suppose that there exist *y _{i}*,

*If there exist y _{i}*,

Suppose that there exist *y _{i}*,

Up to now, we have obtained the following necessary conditions for *cEnd*(*X* + *Y*) being a monoid:

*N*(*y*) ⊄_{i}*N*(*y*) for any_{j}*y*,_{i}*y*∈_{j}*S*.|

*N*(*y*)∩_{i}*K*|≠|*N*(*y*)∩_{j}*K*| for any*y*,_{i}*y*∈_{j}*S*with*i*≠*j*.

To show that (*A*) and (*B*) are also sufficient for *cEnd*(*X* + *Y*) being a monoid, we need the following characterization of completely regular endomorphisms of *X* + *Y* satisfying (*A*) and (*B*).

*Let X* + *Y be a join of split graphs satisfying* (*A*) *and* (*B*), *and let f* ∈ *cEnd*(*X* + *Y*). *If there exists y* ∈ *S such that* [*y*]_{ ρf}⊆ *S*, *then* [*y*]_{ ρf} = {*y*}.

Let *f* ∈ *cEnd*(*X* + *Y*). By Theorem 2.1 (3), *f*(*a*)≠ *f*(*b*) for any *a*, *b* ∈ *V*(*I _{f}*). Thus there exists

*Let X* + *Y be a join of split graphs satisfying* (*A*) *and* (*B*), *and let f* ∈ *End*(*X* + *Y*). *Then f* ∈ *cEnd*(*X* + *Y*) *if and only if one of the following conditions hold*:

*For x*∈*K*,*f*(*x*)∈*K*;*for y*∈*S*,*either f*(*y*) =*y*,*or f*(*y*)∈*K*.*f*(*K*)≠*K and I*≅_{f}*K*.*There exist x*_{1}∈*K*,*y*_{1}∈*S with N*(*y*_{1}) =*K*\ {*x*_{1}}*such that f*(*x*_{1}) =*y*_{1}*and f*(*y*_{1}) =*x*_{1};*f*(*K*\ {*x*_{1}}) =*K*\ {*x*_{1}};*for y*∈*S with*{*y*,*y*_{1}} ∉*E*(*X*+*Y*),*f*(*y*)∈*K*\ {*x*_{1}};*for y*∈*S with*{*y*,*y*_{1}} ∈*E*(*X*+*Y*),*either f*(*y*) =*y*,*or f*(*y*)∈*K*.

Necessity. Let *X* + *Y* be a join of split graphs satisfying (*A*) and (*B*) and let *f* ∈ *cEnd*(*X* + *Y*). We divide it into two cases to discuss:

Case 1. Assume that *f*(*K*) = *K*. For any *y* ∈ *S*, if *f*(*y*)∉ *K*, then *f*(*y*) ∈ *S*. Since *f*(*K*) = *K*, [*y*]_{ ρf} ⊆ *S*. By Lemma 2.4, [*y*]_{ρf} = {*y*}. Since *f*(*K*) = *K*, |*N*(*y*)∩ *K* | = |*N*(*f*(*y*)) ∩ *K* |. It follows from (*B*) that *f*(*y*) = *y*. Hence *f* is an endomorphism of *X* + *Y* satisfying (1).

Case 2. Assume that *f*(*K*)≠ *K*. Then there exist *x _{i}* ∈

*x*≠_{i}*x*_{1}. Then*x*_{1}∉*I*. Otherwise, since_{f}*f*(*K*\ {*x*}) =_{i}*K*\ {*x*_{1}},*f*^{−1}(*x*_{1})⊆*S*. Let*y*∈*f*^{−1}(*x*_{1}). By Lemma 2.4, [*y*]_{ρf}= {*y*}. It follows from {*x*_{1},*y*_{1}}∉*E*(*X*+*Y*) that {*x*,_{j}*y*}∉*E*(*X*+*Y*). Thus*y*≠*y*_{1}. It follows from (*B*) that |*N*(*y*)∩*K*| ≤*n*+*m*− 2. Thus there exists*k*_{0}∈*K*\ {*x*} such that {_{i}*k*_{0},*y*}∉*E*(*X*+*Y*). Obviously, {*f*(*y*),*f*(*k*_{0})} = {*x*_{1},*f*(*k*_{0})} ∈*E*(*X*+*Y*). But {*y*,*t*}∉*E*(*X*+*Y*) for any*t*∈ [*k*_{0}]_{ ρf}. Hence*f*is not half-strong. Note that*f*∈*hEnd*(*X*+*Y*) if and only if*I*is an induced subgraph of_{f}*X*+*Y*. This contradicts Theorem 2.1 (1). It follows from Theorem 2.1 (1) that*y*∉*I*for any_{f}*y*∈*S*with {*y*,*x*_{1}} ∈*E*(*X*+*Y*). In particular,*y*∉*I*for any_{f}*y*∈*S*with {*y*,*y*_{1}} ∈*E*(*X*+*Y*). Let*y*∈*S*with {*x*,_{i}*y*} ∈*E*(*X*+*Y*). Then {*f*(*x*),_{j}*f*(*y*)} = {*y*_{1},*f*(*y*)} ∈*E*(*X*+*Y*). Hence*f*(*y*)∈*N*(*y*_{1})∩*K*=*K*\ {*x*_{1}}.Let

*y*∈*S*with {*x*,_{i}*y*}∉*E*(*X*+*Y*). Then*N*(*y*)∩*K*⊆*K*\ {*x*}. If_{i}*f*(*y*) ∈*S*, then {*x*_{1},*f*(*y*)} ∉*E*(*X*+*Y*) since*x*_{1}∉*I*. Thus_{f}*N*(*f*(*y*))⊆*N*(*y*_{1}). It follows from (*A*) that*N*(*f*(*y*)) =*N*(*y*_{1}). By (*B*), we have*f*(*y*) =*y*_{1}. If*f*(*y*)∈*K*, then*f*(*y*)≠*x*_{1}. Otherwise, [*y*]_{ ρf}⊆*S*. Since*f*∈*cEnd*(*X*+*Y*),*I*is an induced subgraph of_{f}*X*+*Y*and [*y*]_{ρf}= {*y*}. Note that {*x*,*x*_{1}} ∈*E*(*X*+*Y*) for any*x*∈*K*\ {*x*_{1}}. Then*N*(*y*)∩*K*=*K*\ {*x*}. Thus we have |_{i}*N*(*y*) ∩*K*| = |*N*(*y*_{1})∩*K*| =*n*+*m*− 1. This contradicts (*B*) . Hence*I*is a subgraph of_{f}*X*+*Y*induced by (*K*\ {*x*_{1}}) ∪ {*y*_{1}} and so*I*≅_{f}*K*.*x*=_{i}*x*_{1}. Then*f*(*x*_{1}) =*y*_{1}and*f*(*K*\ {*x*_{1}}) =*K*\ {*x*_{1}}. Since {*y*_{1},*k*} ∈*E*(*X*+*Y*) for any*k*∈*K*\ {*x*_{1}},*f*(*y*_{1}) is adjacent to every vertex of*f*(*K*\ {*x*_{1}}) =*K*\ {*x*_{1}}. Thus*f*(*y*_{1}) ∈ {*x*_{1},*y*_{1}}.If

*f*(*y*_{1}) =*y*_{1}, then*x*_{1}∉*I*by Theorem 2.1 (3). It follows from Theorem 2.1 (1) that_{f}*y*∉*I*for any_{f}*y*∈*S*with {*x*_{1},*y*} ∈*E*(*X*+*Y*). Note that*y*_{1}is the only vertex in*S*, which is not adjacent to*x*_{1}. Then*I*is a subgraph of_{f}*X*+*Y*induced by (*K*\ {*x*_{1}})∪{*y*_{1}} and so*I*≅_{f}*K*.If

*f*(*y*_{1}) =*x*_{1}, then*x*_{1}∈*I*. Let_{f}*y*∈*S*\ {*y*_{1}} be such that {*y*,*y*_{1}}∉*E*(*X*+*Y*). It follows from (*A*) and (*B*) that {*x*_{1},*y*} ∈*E*(*X*+*Y*). Thus {*f*(*x*_{1}),*f*(*y*)} = {*y*_{1},*f*(*y*)} ∈*E*(*X*+*Y*). If*f*(*y*)∈*S*,*f*(*y*) and*y*_{1}lie in the different*S*(where_{i}*i*∈ {1, 2 Note that [*y*]_{ρf}⊆*S*, then [*y*]_{ρf}= {*y*}. Since*f*is half-strong,*f*((*N*(*y*)∩*K*) \ {*x*_{1}}) = (*N*(*f*(*y*))∩*K*) \ {*x*_{1}}. Hence |*N*(*y*)∩*K N*(*f*(*y*))∩*K*|. This contradicts (*B*). Hence*f*(*y*)∈*K*\ {*x*_{1}} for any*y*∈*S*\ {*y*_{1}} with {*y*,*y*_{1}}∉*E*(*X*+*Y*). Let*y*∈*S*be such that {*y*,*y*_{1}} ∈*E*(*X*+*Y*). If*f*(*y*)∉*K*, then*f*(*y*)∈*S*. Since {*x*_{1},*y*} ∈*E*(*X*+*Y*), {*f*(*x*_{1}),*f*(*y*)} = {*y*_{1},*f*(*y*)} ∈*E*(*X*+*Y*). Thus*f*(*y*)≠*y*_{1}. Then [*y*]_{ρf}⊆*S*and so [*y*]_{ρf}= {*y*} by Lemma 2.4. Note that*I*is an induced subgraph of_{f}*X*+*Y*. Then |*N*(*y*)∩*K N*(*f*(*y*))∩*K*|. It follows from (*B*) that*f*(*y*) =*y*. Hence*f*is an endomorphism of*X*+*Y*satisfying (3).

Sufficiency. This follows directly from Theorem 2.1. □

*Let X* + *Y be a join of split graphs satisfying* (*A*) *and* (*B*), *and let f* ∈ *End*(*X* + *Y*). *If I _{f}*≅

This follows directly from Theorem 2.5. □

Now we prove that *cEnd*(*X* + *Y*) forms a monoid for *X* + *Y* satisfying (*A*) and (*B*).

*Let X* + *Y be a join of split graphs satisfying* (*A*) *and* (*B*). *Then cEnd*(*X* + *Y*) *forms a monoid*.

Let *X* + *Y* be a join of split graphs satisfying (*A*) and (*B*). We only need to show that the composition of any two completely regular endomorphisms of *X* + *Y* is also completely regular in every case. Let *f* be an arbitrary completely regular endomorphism of *X* + *Y*. Then by Theorem 2.5, *f* acts in one of the following ways:

For

*x*∈*K*,*f*(*x*)∈*K*; for*y*∈*S*, either*f*(*y*) =*y*, or*f*(*y*)∈*K*.*f*(*K*)≠*K*and*I*≅_{f}*K*.There exist

*x*_{1}∈*K*,*y*_{1}∈*S*with*N*(*y*_{1}) =*K*\ {*x*_{1}} such that*f*(*x*_{1}) =*y*_{1}and*f*(*y*_{1}) =*x*_{1};*f*(*K*\ {*x*_{1}}) =*K*\ {*x*_{1}}; for*y*∈*S*with {*y*,*y*_{1}}∉*E*(*X*+*Y*),*f*(*y*)∈*K*\ {*x*_{1}}; for*y*∈*S*with {*y*,*y*_{1}} ∈*E*(*X*+*Y*), either*f*(*y*) =*y*, or*f*(*y*)∈*K*.

It is straightforward to see that the composition of any such two completely regular endomorphisms is still a completely regular endomorphism of *X* + *Y*. The proof is complete. □

Up to now we have

*Let X* + *Y be a join of split graphs*. *Then cEnd*(*X* + *Y*) *forms a monoid if and only if*

*N*(*y*) ⊄_{i}*N*(*y*)_{j}*for any y*,_{i}*y*∈_{j}*S*,*and*|

*N*(*y*) ∩_{i}*K*| ≠ |*N*(*y*) ∩_{j}*K*|*for any y*,_{i}*y*∈_{j}*S with i*≠*j*.

Necessity follows directly from Lemmas 2.2 and 2.3.

Sufficiency follows directly from Theorem 2.7. □

The authors want to express their gratitude to the referees for their helpful suggestions and comments.

This research was partially supported by the National Natural Science Foundation of China(No.11301151), the Key Project of the Education Department of Henan Province(No.13A110249), the Subsidy Scheme of Young Teachers of Henan Province(No.2014GGJS-057) and the Innovation Team Funding of Henan University of Science and Technology(No.2015XTD010).

- [1]
Kelarev A.V., Ryan J. and Yearwood J., Cayley graphs as classifiers for data mining: The influence of asymmetries, Discrete Math., 2009, 309, 5360-5369. CrossrefWeb of ScienceGoogle Scholar

- [2]
Kelarev A.V., Graph algebras and automata, Marcel Dekker, New York, 2003. Google Scholar

- [3]
Kelarev A.V. and Praeger C.E., On transitive Cayley graphs of groups and semigroups, Euro.J.Combin., 2003, 24, 59-72.CrossrefGoogle Scholar

- [4]
Wilkeit E., Graphs with regular endomorphism monoid, Arch.Math., 1996, 66, 344-352. CrossrefGoogle Scholar

- [5]
Kelarev A.V., Labelled Cayley graphs and minimal automata, Australasian J. Combinatorics, 2004, 30, 95-101. Google Scholar

- [6]
Hou H., Gu R. and Shang Y., The join of split graphs whose regular endomorphisms form a monoid, Communications in Algebra, 2014, 42, 795-802. Web of ScienceCrossrefGoogle Scholar

- [7]
Fan S., Retractions of split graphs and End-orthodox split graphs, Discrete Math., 2002, 257, 161-164. CrossrefGoogle Scholar

- [8]
Hou H., Luo Y. and Gu R., The join of split graphs whose half-strong endomorphisms form a monoid, Acta Mathematica Sinica, English Series, 2010, 26, 1139-1148.CrossrefWeb of ScienceGoogle Scholar

- [9]
Hou H., Luo Y. and Fan X., End-regular and End-orthodox joins of split graphs, Ars Combinatoria, 2012, 105, 305-318.Google Scholar

- [10]
Hou H., Gu R. and Shang Y., The join of split graphs whose quasi-strong endomorphisms form a monoid, Bulletin of the Australian Mathematical Society, 2015, 91, 1-10.CrossrefWeb of ScienceGoogle Scholar

- [11]
Li W. and Chen J., Endomorphism-regularity of split graphs, European J. Combin., 2001, 22, 207-216.CrossrefGoogle Scholar

- [12]
Lu D., Wu T., Endomorphism monoids of generalized split graphs, Ars Combinatoria, 2013, 111, 357-373.Google Scholar

- [13]
Luo Y., Zhang W., Qin Y. and Hou H., Split graphs whose half-strong endomorphisms form a monoid, Science China Mathematics, 2012, 55, 1303-1320. CrossrefWeb of ScienceGoogle Scholar

- [14]
Hou H. and Gu R., Split graphs whose completely regular endomorphisms form a monoid, Ars Combinatoria, 2016, 127, 79-88. Google Scholar

- [15]
Howie J.M., Fundamentals of semigroup theory, Clarendon Press, Oxford, 1995. Google Scholar

- [16]
Godsil C. and Royle G., Algebraic graph theory, Springer-verlag, New York, 2000. Google Scholar

- [17]
Li W., Graphs with regular monoid, Discrete Math., 2003, 265, 105-118. CrossrefGoogle Scholar

- [18]
Li W., Split graphs with completely regular endomorphism monoids, Journal of Mathematics Research and Exposition, 2006, 26, 253-263. Google Scholar

**Received**: 2016-07-06

**Accepted**: 2017-05-08

**Published Online**: 2017-06-22

**Citation Information: **Open Mathematics, Volume 15, Issue 1, Pages 833–839, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0071.

© 2017 Hou *et al*.. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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