aa r X i v : . [ m a t h . R A ] N ov On a class of semigroup graphs ∗ Li Chen † and Tongsuo Wu ‡ Department of Mathematics, Shanghai Jiaotong University, Shanghai 200240, P.R.China
Abstract.
Let G = Γ( S ) be a semigroup graph, i.e., a zero-divisor graph of a semigroup S with zero element0. For any adjacent vertices x, y in G , denote C ( x, y ) = { z ∈ V ( G ) | N ( z ) = { x, y }} . Assume that in G there exist two adjacent vertices x, y , a vertex s ∈ C ( x, y ) and a vertex z such that d ( s, z ) = 3. In thispaper, we study algebraic properties of S with such graphs G = Γ( S ), giving some sub-semigroups andideals of S . We construct some classes of such semigroup graphs and classify all semigroup graphs withthe property in two cases. Key Words: Zero-divisor semigroup; Sub-semigroup; Zero-divisor graph; Graph classification
1. Introduction
Throughout, G is a simple and connected graph. For a vertex x of G , the neighborhoodof x is denoted as N ( x ), which is the set of all vertices adjacent to x . Denote also N ( x ) = N ( x ) ∪ { x } . The cardinality of N ( x ) is denoted by deg ( x ). The vertex x iscalled an end vertex if deg ( x ) = 1, and an isolated vertex if deg ( x ) = 0. Throughout, S is a commutative semigroup with 0. Recall that for a commutative semigroup (or acommutative ring) S with 0, the zero-divisor graph Γ( S ) is an undirected graph whosevertices are the zero-divisors of S \ { } , and with two vertices a, b adjacent in case ab = 0([3],[2],[1],[7],[5],[12]). If G ∼ = Γ( S ) for some semigroup S with zero element 0, then G iscalled a semigroup graph .Some fundamental properties and possible algebraic structure of S and graphic struc-tures of Γ( S ) were established in [2, 1, 7] among others. For example, it was proved thatΓ( S ) is always connected, and the diameter of Γ( S ) is less than or equal to 3. If Γ( S )contains a cycle, then its core, i.e., the union of the cycles in Γ( S ), is a union of squaresand triangles, and any vertex not in the core is an end vertex which is connected to thecore by a single edge. In [7, Theorem 4], the authors provided a descending chain of ideals I k of S , where I k consists of all elements of S with vertex degree greater than or equalto k in Γ( S ). In [13], the authors continued the study on the sub-semigroup structure ∗ This research is supported by the National Natural Science Foundation of China (Grant No.10671122). † chenli [email protected] ‡ [email protected] S ) contains no cycle if andonly if Γ( S ) is either a star graph or a two-star graph. By [11, Theorem 2.10], the core K ( G ) contains no triangle if and only if Γ( S ) is a bipartite graph, if and only if Γ( S )is one of the following graphs: star graphs, two-star graphs, complete bipartite graphs,complete bipartite graphs with a thorn. By [10, Theorem 2.3], the core K ( G ) containsno rectangle if and only if Γ( S ) is one of the following graphs: an isolated vertex, a stargraph, a two-star graph, a triangle with n thorns ( n = 0 , , , S be a commutative semigroup with zero-element 0, and let G = Γ( S ). For anyadjacent vertices a, b in V ( G ), denote C ( a, b ) = { x ∈ V ( G ) | N ( x ) = { a, b }} and let T a denote the set of all end vertices adjacent to a . Consider the following condition assumedon G = Γ( S )( △ ) There exist in G two adjacent vertices a, b , a vertex s ∈ C ( a, b ) and a vertex z such that d ( s, z ) = 3.In this paper, we study algebraic properties of semigroups S and the graphic structures ofΓ( S ) such that the condition ( △ ) holds for Γ( S ). (We can further assume that trianglesand rectangles coexist in the core K [Γ( S )].) In particular, it is proved that S \ [ C ( a, b ) ∪ T a ∪ T b ] is an ideal of S (Theorem 2.4). Under some additional conditions, it is provedthat S \ C ( a, b ) is a sub-semigroup of S and there exists an element c ∈ C ( a, b ) such that[ S \ C ( a, b )] ∪ { c } is also a sub-semigroup of S . We also use Theorem 2.4 to constructsome classes of semigroup graphs which satisfies the condition ( △ ), and give a completeclassification of such semigroup graphs in two cases.We record a known result on finite semigroups to end this part (see, e.g., [9, Corollary5.9 on page 25 ]). We also include a proof for the completeness. Lemma 1.1.
Any finite nonempty semigroup S contains an idempotent element. Proof.
Take any element x from S and consider the sequence x, x , x , · · · . Since S is afinite set, there exist m < n such that x m = x n . Let r = n − m , and take k such that kr ≥ m . Then x kr = x m · x kr − m = ( x r · x m ) x kr − m = x r · x kr = x r ( x r x kr ) = · · · = ( x kr ) . (cid:3)
2. Properties of S Lemma 2.1.
Let S be a commutative semigroup with 0, Γ( S ) its zero-divisor graph. Forany vertex x ∈ Γ( S ) , if there exists a vertex y ∈ Γ( S ) such that d ( x, y ) = 3 , then x = 0 in S . Proof. As d ( x, y ) = 3, there exist vertices a, z ∈ Γ( S ) such that x − a − z − y , xz = 0and ay = 0. If x = 0, then x z = 0 and thus xz ∈ Ann ( x ). Clearly xz ∈ Ann ( y ). Then2 z ∈ Ann ( x ) ∩ Ann ( y ) = { } , a contradiction. (cid:3) Part of the following result is contained in [13, Proposition 2.8].
Proposition 2.2.
Let G = Γ( S ) be a zero-divisor graph of a semigroup S . For a vertex b ∈ V ( G ) , let T b = { x ∈ V ( G ) | xb = 0 , x = b, deg ( x ) = 1 } . (1) If b = 0 , then T b ∪ { } is a sub-semigroup of S . (2) If b is not an end vertex and T b = ∅ , then { , b } is an ideal of S . Proof. (1) We only need consider as T b = ∅ . If G contains no cycle, then G is either a two-star graph or a star graph by [8, Theorem 1.3] . If G is a star graph, then T b = S \ { b, } .For all x, y ∈ T b , we must have xy = b , since otherwise 0 = xyb = b = 0, a contradiction.This shows that T b ∪ { } is a sub-semigroup of S when G is a star graph. If G is a two-stargraph or a graph with cycles, then B = ∅ where B = { x ∈ V ( G ) | deg ( x ) ≥ , xb = 0 } .For all x ∈ T b , we have x ∈ Ann ( b ) ∪ { } = { } ∪ T b ∪ B If x ∈ B , denote x = v . Then there exists z ∈ S \ { b } such that zv = 0. Since x z = vz = 0, we have xz ∈ Ann ( x ) = { x, b } . If xz = x , then v = x = x z = vz = 0,a contradiction. If xz = b , then 0 = xzb = b = 0, another contradiction. So we musthave x ∈ T b ∪ { } . If | T b | ≥
2, then exists a vertex y ∈ T b such that x = y . If xy ∈ B ,denote xy = v . Then there exists z ∈ S \ { b } such that zv = 0. As xyz = 0, we have xz ∈ Ann ( y ) = { y, b } and yz ∈ Ann ( x ) = { x, b } , and thus xz = y and yz = x . Then x = xyz = vz = 0. On the other hand, xy = x z = 0, a contradiction. So xy ∈ T b ∪ { } ,and hence T b ∪ { } S .(2) Since T b = ∅ , there exists x ∈ T b such that by ∈ Ann ( x ) = { , b, x } for all y ∈ S .By assumption, b is not an end vertex and thus there exists z ∈ S \{ x } such that bz = 0.Then by = x since otherwise, by = x and it implies 0 = bzy = zx = 0, a contradiction.This completes the proof. (cid:3) Remark 2.3.
In Proposition 2.2(1), the conclusion can not hold if b = 0 . For a vertex v of a graph G , if v is not an end vertex and there is no end vertexadjacent to v , then v is said to be an internal vertex . We know prove the main result ofthis section. Theorem 2.4.
Let G = Γ( S ) be a semigroup graph satisfying condition ( △ ) . Denote L = { z ∈ S | d ( s, z ) = 3 } . Then { , a, b } is an ideal of S , S \ [ C ( a, b ) ∪ T a ∪ T b ] is an idealof S and L is a sub-semigroup of S . Furthermore, (1) If both a and b are internal vertices, then S \ C ( a, b ) is an ideal of S . (2) If a is an internal vertex, while b is not an internal vertex and b = 0 , then S \ C ( a, b ) is a sub-semigroup of S . Proof.
Fix some s ∈ C ( a, b ) and let B = { x | x ∈ S, x / ∈ C ( a, b ) , d ( s, x ) = 2 } , L =3 y | y ∈ S, d ( s, y ) = 3 } . By assumption L = ∅ , C ( a, b ) = ∅ , and T a ∪ T b ⊂ B . Noticethat there is no end vertex in B \ ( T a ∪ T b ). By [2, Theorem 2.3] or by [7, Theorem1(2)], S = { , a, b } ∪ C ( a, b ) ∪ B ∪ L and it is a disjoint union of four nonempty subsets.By Lemma 2.1. we have c = 0 , ∀ c ∈ C ( a, b ), and hence Ann ( c ) = { , a, b } . Clearly, a ∈ Ann ( c ), b ∈ { , a, b } and { , a, b } ( B ∪ L ) ⊆ Ann ( c ) = { , a, b } . This shows that { , a, b } is an ideal of S .For any y in L , there exists a vertex x ∈ B such that xy = 0. Then yS ⊆ Ann ( x )while C ( a, b ) ∩ Ann ( x ) = ∅ . Hence LS ∩ C ( a, b ) = ∅ . Furthermore, for any s ∈ S , ys ∈ { , a, b } ∪ L ∪ B . If ys ∈ { , a, b } ∪ B , then it is clear that ys T a ∪ T b whether sy = x or not. Thus LS ∩ ( T a ∪ T b ) = ∅ , and hence( C ( a, b ) ∪ T a ∪ T b ) ∩ LS = ∅ . For any vertex x in B \ ( T a ∪ T b ), x C ( a, b ) and it has degree greater than one. Hencefor any x ∈ B \ ( T a ∪ T b ) and any x ∈ S , there exists a vertex u ∈ B ∪ L such that x u = 0.Then x x ∈ Ann ( u ) and it implies x x / ∈ C ( a, b ). Thus [( B \ ( T a ∪ T b )) S ] ∩ C ( a, b ) = ∅ .Finally, by [7, Theorem 4], the core of G together with 0 forms an ideal of S . Thus thesearguments show that S \ [ C ( a, b ) ∪ T a ∪ T b ] is an ideal of S .Now take any u, v ∈ L and consider uv . Clearly uv = 0 (see also remark(1) proceedingTheorem 3.3). Also uv C ( a, b ). If uv L , then we can assume ( uv ) a = 0. Since { , a, b } is an ideal of S , we would have either ua = 0 or ub = 0. This shows uv ∈ L and hence L is a sub-semigroup of S . Notice 0 L .(1) If both a and b are internal vertices, then S \ C ( a, b ) = S \ [ C ( a, b ) ∪ T a ∪ T b ]. Inthis case, S \ C ( a, b ) is clearly an ideal of S .(2) Now assume that b is not an internal vertex, and b = 0. Again let T b be the setof end vertices adjacent to b . By the above discussion, we already have ([ { , a, b } ∪ L ∪ ( B \ T b )] S ) ∩ C ( a, b ) = ∅ . Since b = 0, we have T b T b ∪ { } by Theorem 2.2(1). Thesefacts show that S \ C ( a, b ) is a sub-semigroup of S , and it completes the proof. (cid:3) Remarks 2.5.(1)
In Theorem 2.4, if there is no z ∈ V ( G ) such that d ( s, z ) = 3, thenthe theorem may not hold. An example is contained in Example 3.1. Remarks 2.5.(2)
Theorem 2.4 can be easily extended for a graph which satisfies thecondition ( K p ): in the definition ( △ ), replace a − b by an induced complete subgraph H with p − p . The condition ( K p ) is rather natural for thering graphs, see our subsequent work [6]. Theorem 2.6.
Let G = Γ( S ) be a semigroup graph satisfying condition ( △ ) . If further a is an internal vertex, while b is not an internal vertex and b = 0 , then both { , a } and { , b } are ideals of S . Proof.
By Theorem 2.2(2), we already have { , b } E S .4et B = { x | x / ∈ C ( a, b ) , d ( c, x ) = 2 } , L = { y | d ( c, y ) = 3 } . For any d ∈ T b , we have ad ∈ Ann ( c ) = { a, b, } . Clearly, ad = 0. We conclude ad = a . In fact, if ad = a , then ad = b and hence ad = bd = 0. By Theorem 2.2(1), we get d ∈ T b ∪ { } , and we have d = 0 by Lemma 2.1. Thus d ∈ T b , and thus ad = 0, a contradiction. The contradictionshows ad = a .For any x ∈ B \ T b , we have ax ∈ Ann ( c ) = { a, b, } . If ax = b , then b = ax = adx = bd = 0, a contradiction. Thus we must have aB ⊆ { , a } .In a similar way, we prove aL ⊆ { , a } . This completes the proof. (cid:3) In the following we proceed to prove that under some additional conditions, thereexists an element c in C ( a, b ) such that [ S \ C ( a, b )] ∪ { c } is a sub-semigroup of S . Forthis purpose, we need the following technical lemma. Lemma 2.7.
Let G = Γ( S ) be a semigroup graph satisfying condition ( △ ) . Assumefurther that one of the following conditions is satisfied: (1) Both a and b are internal vertices. (2) a is an internal vertex, b = 0 and T b = { d } .Then there exists an element c ∈ C ( a, b ) such that [ S \ C ( a, b )] ∪ { c } S , if and only ifthere exists an element c ∈ C ( a, b ) such that c ∈ [ S \ C ( a, b )] ∪ { c } . Proof. ⇒ Clear. In fact, we further have c = 0 by Lemma 2.1. ⇐ Assume that there exists an element c ∈ C ( a, b ) such that c ∈ [ S \ C ( a, b )] ∪ { c } .Then c ∈ { a, b, c } ∪ X , where X is the set of the vertices which are adjacent to a, b andat the same time belong to B .(1) If both a and b are internal vertices, then T a ∪ T b = ∅ . In this case, repeat theproof of Theorem 2.4 and obtain ( c B ∪ c L ) ∩ C ( a, b ) = ∅ . Hence c = c implies[ S \ C ( a, b )] ∪ { c } S .(2) Now assume that a is an internal vertex, while b is not an internal vertex, b = 0and T b = { d } . By Theorem 2.4(2), S \ C ( a, b ) S . Since S \ [ C ( a, b ) ∪ T b ] is an ideal of S , we already have c [ S \ ( C ( a, b ) ∪ T b )] ⊆ S \ ( C ( a, b ).If c T b ⊆ { c } ∪ [ S \ C ( a, b )], then [ S \ C ( a, b )] ∪ { c } is a sub-semigroup of S since c ∈ { c } ∪ [ S \ C ( a, b )]. In the following we assume that c d ∈ C ( a, b ) \ { c } , and denote c d = c . Then d = d since c d = cd = 0 and T b ∪ { } S . Then cd = c . Since c ∈ [ S \ C ( a, b )] ∪ { c } , we have the following four possible subcases.(2.1) c = a . In this case, we have cd = c and c c = c d = ad = a . Then c = c cd = ad = a ∈ Ann ( c ) = { a, b } since { , a, b } S . Thus [ S \ C ( a, b )] ∪ { c } S .(2.2) c = b . In this case, cd = c and cc = c d = bd = 0, a contradiction. Thus thiscase can not occur.(2.3) c = c . In this case, cd = c and cc = c d = c d = c . Then c = cc d = cd = c ,and hence [ S \ C ( a, b )] ∪ { c } S .(2.4) c = x ∈ X . In this case, cd = c and cc = c d = xd C ( a, b ) by Theorem 2.4.5hen c = c cd = xd = xd C ( a, b ), and hence [ S \ C ( a, b )] ∪ { c } S .This completes the proof. (cid:3) Now we are ready to prove
Proposition 2.8.
Let G = Γ( S ) be a semigroup graph satisfying condition ( △ ) . Assumethat | C ( a, b ) | is finite. If one of the following conditions is satisfied, then there exists anelement c ∈ C ( a, b ) such that [ S \ C ( a, b )] ∪ { c } is a sub-semigroup of S : (1) Both a and b are internal vertices. (2) a is an internal vertex, b = 0 and T b = { d } . Proof. If C ( a, b ) is a sub-semigroup of S , then by Lemma 1.1. there is an element c ∈ C ( a, b ) such that c = c . By Lemma 2.7, there exists an element c ∈ C ( a, b )such that [ S \ C ( a, b )] ∪ { c } S . In the following we assume that C ( a, b ) is not a sub-semigroup of S . Then there exist c i , c j ∈ C ( a, b ) such that c i c j C ( a, b ), and this implies c i c j ∈ { a, b } ∪ X where X is the set of the vertices which are adjacent to a, b and at thesame time belong to B .Assume c i c j = a . If i = j , then the result follows from Lemma 2.7. If i = j , then c i c j = a implies c i c j = ac i = 0, and thus c i ∈ Ann ( c j ), i.e. c i ∈ { a, b } . Then we useLemma 2.7 again to obtain the result. When c i c j = b , a similar discussion lead to theresult.Finally, assume c i c j = x ∈ X . In this case, it is only necessary to consider the i = j case. Since x ∈ X , there is an element z ∈ B ∪ L such that xz = 0. Then c i c j z = xz = 0, and hence c j z ∈ Ann ( c i ) = { , a, b } . Thus c j z = ac j = 0 or c j z = bc j = 0,i.e. c j ∈ Ann ( z ). This means c j ∈ { a, b, X } ∩ Ann ( z ). By Lemma 2.7, there exists anelement c ∈ C ( a, b ) such that [ S \ C ( a, b )] ∪ { c } is a sub-semigroup of S . This completesthe proof. (cid:3)
3. Some Examples and complete classifications of the graphs in two cases
In this section, we use Theorem 2.4 to study the correspondence of zero-divisor semigroupsand several classes of graphs satisfying the four necessary conditions of [7, Theorem 1] aswell as the general assumption of Theorem 2.4.
Example 3.1.
Consider the graph G in Fig.3, where both U and V consist of end vertices.We claim that each graph in Fig.3 is a semigroup graph. y y m Fig.3. bdx x x n VU a
In fact, first notice that d ( y i , V ) = 3 , C ( a, b ) = { y , ..., y m } , and C ( a, d ) = { x , ..., x n } .By Theorem 2.4, if G has a corresponding semigroup S = V ( G ) ∪ { } , then the subset S \ ( { y , · · · , y m } ∪ U ) must be an ideal of S . If further a = 0, then S \ { y , · · · , y m } is a sub-semigroup of S . Also by [13, Theorem 2.1], S \ ( { y , · · · , y m } ∪ U ∪ V ) is asub-semigroup of S \ ( { y , · · · , y m } ∪ U ), and thus a sub-semigroup of S .For m = 2 , n = 2, U = { u } and V = { v, v } , it is not very hard to construct asemigroup T such that Γ( T ) = G − { y , y } following the way mentioned above. Thenafter a rather complicated calculation, we succeed in adding two vertices y , y to thistable such that Γ( S ) = G . The multiplication on S is listed in Table 3 and the detailedverification for the associativity is omitted here:Table 3 · a d b x x y y u v va a a ad d d d d b b b b b bx b x x b b x x x x b x x b b x x x y d b b d d y b by d b b d d y b bu d b x x y y u x x v a b x x b b x v vv a b x x b b x v v Notice that S \ { x , x } is not a sub-semigroup of S since uv = x . Notice also that S \ ( U ∪ { x , x , · · · , x n } ) is a sub-semigroup of S .We remark that the construction in Table 3 can be routinely extended for all n ≥ , m ≥ , | U | ≥ | V | ≥
0, where each of m, n, | U | , | V | could be a finite or an infinite7ardinal number. In other words, each graph in Fig.3 has a corresponding semigroup forany finite or infinite n ≥ , m ≥ , | U | ≥ | V | ≥ Remark 3.2.
Consider the graph G in Fig.3 and assume that n ≥ , m ≥ , | U | ≥ , | V | ≥ . (1) If we add an end vertex w which is adjacent to b , then the resulting graph G hasno corresponding zero-divisor semigroup, even if U = ∅ . (2) If we add a vertex w such that N ( w ) = { b, d } , then the resulting graph H has nocorresponding zero-divisor semigroup, even if U = ∅ . Proof. (1) Assume v ∈ V . We only need consider the case when U = ∅ . Suppose that G isthe zero-divisor graph of a semigroup S with V [Γ( S )] = V ( G ). By Proposition 2.2(2), wehave bx = bv = b and dy = dw = d . Clearly, aw, av ∈ Ann ( x ) ∩ Ann ( y ) = { a, } , andthus aw = a and av = a . As awv = av = a , we have wv ∈ [ Ann ( b ) ∩ Ann ( d )] \ Ann ( a ).That means wv = a and a = 0. We have y wv = y a = 0 and x wv = x a = 0.Thus y w = x w = d and y v = x v = b by Lemma 2.1. Consider x y v . We have b = x b = x ( y v ) = y ( x v ) = y b = 0, a contradiction. The contradiction shows that G has no corresponding semigroup.(2) Assume v ∈ V . We only need consider the case U = ∅ . Suppose that H is thezero-divisor graph of a semigroup S . We have bS ⊆ Ann ( y ) ∩ Ann ( w ) ⊆ { , b } . Clearly, wv ∈ [ Ann ( d ) ∩ Ann ( b )] \ Ann ( a ) ⊆ { a, w } since wva = wa = a . If wv = a , then we have wvx = ax = 0 and wvy = ay = 0, which means wx , wy ∈ Ann ( v ) ⊆ { , v, d } . As wx , wy ∈ Ann ( a ) \ { } , we have wx = wy = d . Then 0 = dx = wy x = y d = d , acontradiction. Now assume wv = w and consider wx . wx ∈ Ann ( a ) ∩ Ann ( d ) ∩ Ann ( b ) ⊆{ , a, b, d } . We claim wx = d since otherwise, d = wx = wv · x = wx · v = dv = 0, acontradiction. In a similar way we prove wy ∈ { a, b } . Moreover, wy · x = y · wx = 0whether wx = a or wx = b . Thus wy = a . As x w = y w = x y w = 0, we have x y , x , y ∈ Ann ( a ) ∩ Ann ( w ) ⊆ { b, d, } , but x y = 0. Now consider x y . Weconclude x y = d since otherwise, x y = b and it implies b = bx = x y x = x y = b , acontradiction. Finally, x y = d implies d = dy = y x y = x y ∈ { , b } , a contradiction.This completes the proof. (cid:3) Now come back to the structure of semigroup graphs G satisfying the main assumptionin Theorem 2.4. We use notations used in its proof. The vertex set of the graph wasdecomposed into four mutually disjoint nonempty parts, i.e., V ( G ) = { a, b } ∪ C ( a, b ) ∪ B ∪ L , where after taking a c in C ( a, b ) B = { v ∈ V ( G ) | v / ∈ C ( a, b ) , d ( c, v ) = 2 } , L = { v ∈ V ( G ) | d ( c, v ) = 3 } . (For example, for the graph G in Fig.3, C ( a, b ) = { y j } , B = U ∪ { d } ∪ { x i } , L = V .In particular, L consists of end vertices.) By [7, Theorem 1(4)], for each pair x, y ofnonadjacent vertices of G , there is a vertex z with N ( x ) ∪ N ( y ) ⊆ N ( z ). Then we havethe following observations: 81) No two vertices in L are adjacent in G . Thus a vertex of L is either an end vertexor is adjacent to at least two vertices in B . In particular, the subgraph induced on L is acompletely discrete graph.(2) A vertex in B is adjacent to either a or b . If a vertex k in B is adjacent to a vertex l in L , then k is adjacent to both a and b . Thus B consists of four parts: end vertices in T a that are adjacent to a , end vertices in T b that are adjacent to b , vertices in B that areadjacent to both a and b , and vertices in B that are adjacent to one of a, b and at thesame time adjacent to another vertex in B . By Example 3.1, the structure of the inducedsubgraph on B ∪ B seems to be complicated. In the following, we will give a completeclassification of the semigroup graphs G with | B ∪ B | ≤ | B ∪ B | = 1. Theorem 3.3.
Let G be a connected, simple graph with diameter 3. Assume thatthere exist two adjacent vertices a, b in V ( G ) , and assume that there exist a vertex c ∈ C ( a, b ) and a vertex w ∈ V ( G ) such that d ( c , w ) = 3 . Assume further that | B \ ( T a ∪ T b ) | = 1 . Then G is a semigroup graph if and only if the following condi-tions hold: (1) 1 ≤ | C ( a, b ) | ≤ ∞ , ≤ | W | ≤ ∞ and W consists of end vertices, where W = { s ∈ V ( G ) | d ( c , s ) = 3 } . (2) either T a = ∅ or T b = ∅ . (see Fig.4 with V = ∅ .) Proof. As | B \ ( T a ∪ T b ) | = 1, B \ ( T a ∪ T b ) = B . By the previous observations, we needonly prove the following two facts.(1) If | T a | ≥ T b = ∅ , then G is a subgraph of Fig.3 with C ( a, d ) = ∅ . (seealso Fig.4 with V = ∅ .) We claim that G is a semigroup graph. In fact, if U = ∅ ,delete the three rows and the three columns involving x , x and u in Table 3 to obtainan associative multiplication on S = S \ ( U ∪ { x , x , · · · , x n } ). Clearly, Γ( S ) = G for | C ( a, b ) | = 2 = | V | , | C ( a, d ) | = 0 = | U | in Fig.3. Also, the table can be extended for anyfinite or infinite | C ( a, b ) | ≥ | V | ≥ | U | = 0. If | U | >
0, then we work outa corresponding associative multiplication table listed in Table 4, for C ( a, b ) = { y , y } , U = { u , u } , V = { v , v } . Table 4 · a b d y y u u v v a a ab a a a b bd d d d d d y d d d d d a ay d d d d d a au a d d d y y b bu a d d d y y b bv a b a a b b v v v a b a a b b v v | C ( a, b ) | ≥ | U | ≥ | V | ≥
1. This completes the proof.(2) If both | T a | > | T b | >
0, then we conclude that G is not a semigroup graph. Fig.4.
U VWda bc c c n In fact, in this case, G is a graph in Fig.4, where | W | ≥ , | U | ≥ , | V | ≥
1. Assume u ∈ U , v ∈ V , w ∈ W and c ∈ C ( a, b ). We now proceed to prove that such a graph doesnot have a corresponding semigroup.Suppose that G is the zero-divisor graph of a semigroup S with V [Γ( S )] = V ( G ). ByProposition 2.2(2), we have du = dv = dc = d . Then uc d = ud = d , which implies uc ∈ [ Ann ( a ) ∩ Ann ( b )] \ Ann ( d ) ⊆ { c i , d } .Assume uc = d . Then uc w = dw = 0, and thus c w = a by Lemma 2.1. As c wv = av = a , we have wv ∈ [ Ann ( d ) ∩ Ann ( b )] \ Ann ( c ) ⊆ { d } , thus wv = d . Then a = wvc = dc = d , a contradiction.So uc = c i , and therefore wuc = wc i = 0. We have wu ∈ [ Ann ( a ) ∩ Ann ( d )] \ Ann ( c ) ⊆ { d } , and thus wu = d . Then b = bw = buw = bd = 0, a contradiction. This completes theproof. (cid:3) A natural question arising from Example 3.1 is if L only consists of end vertices. Thefollowing example shows this is not the case. Example 3.4.
Consider the graph G in Fig.5, where C ( a, b ) = { c , c , · · · , c m } , L = { y , y , · · · , y n } , B = { x , x } ∪ V ( m ≥ , n ≥ , | V | ≥ ) and V consists of end verticesadjacent to b . Notice that each of m, n and | V | could be finite or infinite. We concludethat each graph in Fig.5 has a corresponding zero-divisor semigroup. c m a bc Fig.5. Vx x y y y n Proof.
We need only work out a corresponding associative multiplication table for | V | = m = n = 2. We use Theorem 2.4 and list the associative multiplication in Table 5.Clearly, the table can be extended for all finite or infinite m, n ≥
1, and | V | ≥ · a b c c v v x x y y a a a a a ab b bc x x x x x x b bc x x x x x x b bv a x x v v x x a av a x x v v x x a ax x x x x x x x x x x x x x y a b b b a a y y y a b b b a a y y This completes the proof. (cid:3)
We have three remarks to Example 3.4.(1)
Let n ≥ , m ≥ . If we add to G in Fig.5 an end vertex u such that au = 0 , thenthe resulting graph G has no corresponding zero-divisor semigroup. Proof. (1) Suppose to the contrary that G is the zero-divisor graph of a semigroup P with V [Γ( P )] = V ( G ). By Proposition 2.2(2), we have a ∈ { , a } and b ∈ { , b } . First,we have v y ∈ Ann ( b ) ∩ Ann ( x ) ∩ Ann ( x ) = { a, b, } and similarly, uy , c y ∈ { a, b } .Then v y = a and a = a since av y = ay = a . On the other hand, auy = 0 and itimplies uy = b . Similarly, we have c y = b . Consider c uy . We have b = ub = u ( c y ) = c ( uy ) = c b = 0, a contradiction. This completes the proof. (cid:3) (2) Let n ≥ , m ≥ . If we add to G in Fig.5 an end vertex y such that yx = 0 ,then the resulting graph G has no corresponding zero-divisor semigroup, whether or not b = ∅ . Proof. (2) Assume { y , y } ⊆ L , where y is an end vertex adjacent to x . Suppose tothe contrary that G is the zero-divisor graph of a semigroup P with V [Γ( P )] = V ( G ).First, x y ∈ Ann ( a ) ∩ Ann ( x ) ∩ Ann ( y ) = { x , } . Thus x y = x , and hence x = 0.By Proposition 2.2(2), we have c x = x and therefore, c x = x . Thus c ∈ { c i , x | i } .We have c y = 0 since c y ∈ Ann ( a ) ∩ Ann ( x ) ∩ Ann ( x ) = { a, b, } . Since c = x , c y ∈ { a, b, x } and c y = x y = x , it follows that c y = x . Finally, c yy = x y = 0and by Lemma 2.1, we have c y = x , contradicting c y ∈ { a, b } . This completes theproof. (cid:3) (3) Let n ≥ , m ≥ and assume V = ∅ in Fig.5. If further we add to G anedge connecting x and x , then the resulting graph G has no corresponding zero-divisorsemigroup. Proof.
Suppose to the contrary that G is the zero-divisor graph of a semigroup P with V [Γ( P )] = V ( G ). By Lemma 2.1, we have c x ∈ Ann ( y ) = { x , x , } andsimilarly, c x ∈ { x , x } . Then we have c x = 0 and c x = 0, which means c ∈ [ Ann ( a ) ∩ Ann ( b )] \ [ Ann ( x ) ∪ Ann ( x )] = { c i | i = 1 , , · · · , m } since c = 0 by Lemma2.1. Similarly, we have y ∈ { y i | i = 1 , , · · · , n } . Clearly, we have c y ∈ Ann ( a ) ∩ Ann ( x ) ⊆ { a, b, x , x , } . Then as c y = c i y = 0 for some i ∈ { , , · · · , m } , we have c y ∈ { x , x } . Finally, 0 = ( c y ) y = c y = c y i = 0 (for some i ∈ { , , · · · , n } ), acontradiction. This completes the proof. (cid:3) Combining the above results, we now classify all semigroup graphs satisfying the mainassumption of Theorem 2.4 with | B ∪ B | = 2: Theorem 3.5.
Let G be a connected, simple graph with diameter 3. Assume that thereexist two adjacent vertices a, b in V ( G ) , and assume that there exist a vertex y ∈ V ( G ) and a vertex c ∈ C ( a, b ) such that d ( c , y ) = 3 . Assume further | B \ ( T a ∪ T b ) | = 2 . (1) If B = B \ ( T a ∪ T b ) , then G is a semigroup graph if and only if G is a graph inFig.5, where ≤ m ≤ ∞ , ≤ n ≤ ∞ and ≤ | V | ≤ ∞ . (2) If | B | = 1 , then G is a semigroup graph if and only G is a graph in Fig.3, where n = 1 , ≤ m ≤ ∞ , ≤ | V | ≤ ∞ , ≤ | U | ≤ ∞ . Proof. (1) By Example 3.4, each graph in Fig.5 is a semigroup graph. Clearly, B = B \ ( T a ∪ T b ) and it consists of two vertices. Conversely, the result follows from [13,Theorem 2.1] and the three remarks after Example 3.4.(2) If | B | = 1, then assume B \ ( T a ∪ T b ) = { x , x } , where a − x − b . In this case, x − x in G . If x − a in G , then there is no end vertex adjacent to x . In this subcase, G is a semigroup graph if and only if T b = ∅ by Example 3.1 and Remark 3.2(1), the caseof | C ( a, d ) | = 1. The other subcase is x − b in G , and it is the same with the above12ubcase. This completes the proof. (cid:3) It is natural to ask the following question : Can one give a complete classification ofsemigroup graphs G = Γ( S ) with | B ∪ B = n | for any n ≥
3? At present, it seems tobe a rather difficult question.Add two end vertices to two vertices of the complete graph K n to obtain a new graph,and denote the new graph as K n + 2. By [14, Theorem 2.1], K n + 2 has a unique zero-divisor semigroup S such that Γ( S ) ∼ = K n + 2 for each n ≥
4. Having Theorem 2.4 inmind, it is natural to consider graphs obtained by adding some caps to K n + 2. Example 3.6.
Consider the graph G in Fig.6. The subgraph G induced on the vertexsubset S ∗ = { a, b, x , x , y , y } is the graph K + 2 , i.e., K together with two end vertices y , y . Then G has a unique corresponding zero-divisor semigroup S = S ∗ ∪ { } by [14,Theorem 2.1]. We can work out the corresponding associative multiplication table, andlist it in Table 6: a bx x y y Fig.6. c c c n cd Table 6 · a b x x y y a a a ab x x x x x x y a x x y ay a x x a y (1) If we add to G a vertex c such that N ( c ) = { a, b } , then the resulting graph H has no corresponding zero-divisor semigroup. If we add to G a vertex d such that N ( d ) = { a, x } , then the resulting graph H has no corresponding zero-divisor semigroup. (3) If we add to G vertices c i ( i ∈ I ) such that N ( c i ) = { x , x } , then the resultinggraph H has corresponding zero-divisor semigroups, where I could be any finite or infiniteindex set. In each of the above three cases, we say that a cap is added to the subgraph K + 2. Proof. (1) Suppose that H is the zero-divisor graph of a semigroup S with V [Γ( S )] = V ( H ). Then by Theorem 2.4, S is an ideal of S = S ∪ { c } . Thus we only need check theassociative multiplication of S based on the table of S already given in Table 6. First, wehave cx = x by Proposition 2.2(2). Consider y bc . Clearly, 0 = 0 y = ( cb ) y = c ( by ) = cx = x , a contradiction. This completes the proof.(2) Suppose that H is the zero-divisor graph of a semigroup S = S ∪ { d } with V [Γ( S )] = V ( H ). If x = 0, then by Theorem 2.4(2), S is a sub-semigroup of S . ThenΓ( S ) = K + 2, and it implies x = 0 by Table 6, a contradiction. In the following weassume x = 0.By Lemma 2.1, we have d = 0, and thus ay , ay ∈ Ann ( d ) = { a, x , } . Clearly ay = 0 and we can have ay = a . (Otherwise, ay = x and we have 0 = x y = ay y =( ay ) y = 0, a contradiction.) Then ay y = 0, and thus y y ∈ [ Ann ( x ) ∩ Ann ( x )] \ Ann ( a ). It means y y = a and a = 0. Clearly by y = 0, and thus by = x , by = x by Lemma 2.1. Similarly, cy y = ca = 0 and thus cy = x , cy = x . Finally, consider bcy . We have 0 = bx = b ( cy ) = c ( by ) = cx = x , a contradiction. This completes theproof.(3) Suppose that H is the subgraph of G in Fig.6 induced on the vertex set S ∗ ∪{ c i | i ∈ I } . Assume that H is the zero-divisor graph of a semigroup P with V [Γ( P )] = V ( H ).Clearly, it dose not satisfy the condition of Theorem 2.4. For | I | = 2, we work out anassociative multiplication table and list it in Table 7:Table 7 · a b x x y y c c a a a a a ab b b b b bx x x x x x y a b x y c c c y a b x c y c c c a b c c c c c a b c c c c The table can be easily extended for any finite or infinite index set I . (cid:3) We remark that in Example 3.6, replace K by K n for any n ≥
5, the results still14old. There exists no difficulty to generalize the proofs to the general cases. Thus wehave proved the following general result.
Theorem 3.7.
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