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# End-regular and End-orthodox generalized lexicographic products of bipartite graphs

Rui Gu
• Rui Gu: Mathematics and Science College, Shanghai Normal University, Shanghai, 200234, China
/ Hailong Hou
• Corresponding author
• Hailong Hou: School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, 471023, China
• Email:
Published Online: 2016-04-06 | DOI: https://doi.org/10.1515/math-2016-0021

## Abstract

A graph X is said to be End-regular (End-orthodox) if its endomorphism monoid End(X) is a regular (orthodox) semigroup. In this paper, we determine the End-regular and the End-orthodox generalized lexicographic products of bipartite graphs.

MSC 2010: 05C25

## 1 Introduction and preliminary concepts

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 (see [1, 2, 5, 1114] and references therein). The aim of this research is to develop further relations between graph theory and algebraic theory of semigroups and to apply the theory of semigroups to graph theory. Petrich and Reilly [18] pointed out that, in the great range of special classes of semigroups, regular semigroups take a central position from the point of view of richness of their structural regularity. So it is natural to ask: which graphs have regular endomorphism monoids? This question was posed by L.Marki [17] as an open problem. However, it seems difficult to obtain a general answer to this question. So the strategy for solving this question is finding various kinds of conditions of regularity for various kinds of graphs. In [19], connected bipartite End-regular graphs were explicitly found. A family of circulant complete graphs K(n, 3) whose endomorphism monoids are regular was described in [6]. Hou, Luo and Cheng [8] explored the endomorphism monoid of Pn, the complement of a path Pn with n vertices. It was shown that Pn is End-orthodox. The split graphs with regular endomorphism monoids were characterized in [14]. The joins of split graphs with regular endomorphism monoids were studied in [7]. The joins of bipartite graphs with regular endomorphism monoids were studied in [9]. In this paper, we will determine the End-regular and the End-orthodox generalized lexicographic products of connected bipartite graphs.

The graphs X 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 x1 and x2 are adjacent in graph X, the edge connecting x1 and x2 is denoted by {x1, x2} and write {x1, x2} ∊ E(X). A graph X is called bipartite if X has no odd cycle. It is known that, if a graph X is bipartite, then its vertex set V(X) can be partitioned into two disjoint non-empty subsets, such that no edge joins two vertices in the same set.

Let X1 and X2 be two graphs. The join of X1 and X2, denoted by X1 + X2, is a graph such that V(X1 + X2) = V(X1) U V(X2) and E(X1 + X2) = E(X1) U E(X2) U {{x1, x2}|x1V(X1),x2V(X2)}. The lexicographic product of X1 and X2, denoted by X1 [X2], is a graph with vertex set V(X1[X2]) = V(X1) × V(X2), and edge set E(X1[X2]) = {{(x, y),(x1, y1)}|{x, x1} ∈ E(X1), or x = x1 and{y, y1} ∈ E(X2)}. The generalized lexicographic product of a graph G with a family of graphs {Bi|iV(G)}, denoted by G[Bi]iV(G), is defined as a graph whose vertex set V(G[Bi]iV(G)) = {(x, yx)|xV(G), yxV(Bx)}, and {(x, y),(x1, y1)} ∈ E (G[Bi]iV(G)) if and only if {x, x1} ∈ E(G), or x = x1 and {y, y1} ∈ E(Bx).

Let X and Y be graphs. A mapping f from V(X) to V(Y) is called a homomorphism (from X to Y) if {x1, x2} ∈ E(X) implies that {f(x1), f(x2)} ∈ E(Y). A homomorphism f from X to itself is called an endomorphism of X. Denote by End(X) the set of all endomorphisms of X. A retraction of a graph X is a homomorphism f from X to a subgraph Y of X such that the restriction f|Y of f to V(Y) is the identity mapping on V(Y). It is known that the idempotents of End(X) are retractions of X Denote by Idpt(X) the set of all idempotents of End(X). Let fEnd(X). A subgraph of X is called the endomorphic image of X under f, denoted by If, if V(If) = f(V(X)) and {f(a), f(b)} ∈ E(If) if and only if there exist cf–1(f(a)) and df–1(f(b)) such that {c, d} ∈ E(X). By ρf we denote the equivalence relation on V(X) induced by f, i.e., for a; bV(X), (a, b) ∈ ρf if and only if f(a) = f(b). Denote by [a]ρf the equivalence class containing aV(X) with respect to ρf.

An element a of a semigroup S is said to be regular if there exists xS such that axa = a. A semigroup S is called regular if all its elements are regular. A semigroup S is called orthodox if S is regular and the set of all idempotents forms a subsemigroup, that is, a regular semigroup is orthodox if the product of its any two idempotents is still an idempotent. A graph X is said to be End-regular(orthodox) if its endomorphism monoid End(X) is regular( orthodox). Clearly, End-orthodox graphs are End-regular.

For undefined notation and terminology in this paper the reader is referred to [1, 4, 5]. We list some known results which will be used in the sequel.

Lemma 1.1([16]). Let X be a graph and let fEnd(X). Then f is regular if and only if there exists g, hIdpt(X) such that ρg = ρf and Ih = If

Lemma 1.2([3]). If X [Y] is End-regular, then both X and Y are End-regular

Lemma 1.3([19]). If X is End-regular and Y is a retract of X, then Y is End-regular

Lemma 1.4([19]). Let X be a connected bipartite graph. Then X is End-regular if and only if X is one of the following graphs:

(1) Complete bipartite graph,

(2) Trees T with d(T) = 3,

(3) Cycle C6 and C8,

(4) Path with 5 vertices, i.e. P5.

Lemma 1.5([9]). Let B1 and B2 be two connected bipartite graphs. Then B1 + B2 is End-regular if and only if

(1) One of them is End-regular and the other is K2, or

(2) B1 + B2 = T1 + T2, where where T1 and T2 are trees with diameter 2, or

(3) B1 + B2 = Km1,n1 + Km2,n2, where Kmi,ni (i = 1; 2) denotes complete bipartite graphs, or

(4) B1 + B2 = T + Km, n, where Km;n denotes a complete bipartite graph and T denotes a tree with diameter 2.

Lemma 1.6([10]). Let X and Y be two connected bipartite graphs. Then X[Y] is End-regular if and only if

(1) X = K2 and Y = Km, n, where Km, n denotes complete bipartite graphs, or

(2) X is End-regular and Y = K2.

Lemma 1.7([3]). Let X be a connected bipartite graph. Then X is End-orthodox if and only if X is one of the following graphs : K2, P3, P4, C4.

Lemma 1.8([15]). Let G be a graph and let {Bi| iV(G)} be a family of graphs. If G[Bi]i∈(G) is End-regular, then for any iV(G), Bi is End-regular

Lemma 1.9([10]). Let X and Y be two graphs. If X [Y] is End-orthodox, then both X and Y are End-orthodox

Lemma 1.10([9]). Let B1 and B2 be two connected bipartite graphs, then B1 + B2 is End-orthodox if and only if

(1) One of them is End-orthodox and the other is K2,

(2) B1 + B2 = P3C4.

## 2 End-regular generalized lexicographic products of bipartite graphs

Recall that End-regular bipartite graphs are characterized in Lemma 1.4 and End-regular lexicographic products of bipartite graphs are determined in Lemma 1.6. In this section, we characterize the End-regular generalized lexicographic products of connected bipartite graphs. The following theorem is our main result.

Theorem2.1. Let G be a connected bipartite graph and let {Bi| iV(G)} be a family of connected bipartite graphs. Then G[Bi]iV(G) is End-regular if and only if one of the following conditions hold:

(1) G = K2 with V(G) = {1,2}, B1 is End-regular and B2 = K2;

(2) G = K2 with V(G) = {1, 2}, B1 and B2 are trees of diameter 2;

(3) G = K2 with V(G) = {1,2}, B1 and B2 are complete bipartite graphs;

(4) G = K2 with V(G) = {1;2}, B1 is a tree of diameter 2 and B2 is a complete bipartite graph;

(6) GK2 is End-regular and Bi = K2for any iV(G).

To prove our main result, we need the following characterizations of the regular endomorphisms of lexicographic products of bipartite graphs.

Lemma 2.2Let G be a bipartite graph and let {Bi|iV(G)} be a family of connected bipartite graphs. If G[Bi]iV(G) is End-regular, then G is End-regular

Proof Since Bi is bipartite, its vertex set V(Bi) can be partitioned into two disjoint non-empty subsets Ai and Ci, such that no edge joins two vertices in the same set. Take yi0Ai and zi0Ci such that {yi0, zi0} ∈ E(Bi) for any iV(G). Define the mapping on the vertex set V(Bi) as follows:

$gi(x)={yi0,x∈Ai,zi0,x∈Ci.$

Let fEnd(G). Define a mapping F from V(G[Bi]iV(G)) to itself by

$F((i,x))={(f(i),yf(i)0),x∈Ai,(f(i),zf(i)0),x∈Ci.$

Let (i, xi), (j, x2) ∈ V(G[Bi]iV(G)) with {(i, xi), (j, x2)} e E(G[Bi]isViG)). If {i, j} ∈ E(G), then {f(i), f(j)} ∈ E(G) and so {F((i, x1)), F((j, x2))} ∈ E(G[Bi]iV(G)). If i = j and{x1, x2} ∈ E(Bi), then {F((i, x1)), F((j, x2))} = {(f (i), yf(i)0),(f(i), zf(i)0)} ∈ E(G[Bi]iV(G)). Hence FEnd(G[Bi]iV(G)). Note that F is regular. Thus there exists pseudoinverse KEnd(G[Bi]iV(G)) such that FKF D F.

Let (i, x) ∈ V(G[Bi]iV(G)) and K((i, x)) = (j, x′) for some jV(G) and x′ ∈ V(Bj). If xAi, then x′ ∈ Aj; If xCi, then x′ ∈ Cj Define a mapping k from V(G) to itself by k(i) = j, where j satisfies K((i, x)) = (j, x′) for some x′ ∈ V(Bj). We now prove that kEnd(G). Let i, jV(G) be such that {i, j} ∈ E(G). Then {(i, x1), (j, x2)} ∈ E(G[Bi]iV(G)) for any x1Ai and x2Aj Since KEnd{G[Bi]iEV(G)), there exists ${x}_{1}^{\text{'}}$V(Bk(i)) and ${x}_{2}^{\text{'}}$V(Bk(j)) such that {K((i, x1)), K((j, x2))} = {{k{i), ${x}_{1}^{\text{'}}$), (k(j), ${x}_{2}^{\text{'}}$)} ∈ E(G[Bi]iV(G)). If {k(i), k(j)} ∉ E(G), then k(i) = k(j) and {${x}_{1}^{\text{'}}$, ${x}_{2}^{\text{'}}$} ∈ E(Bk(i)). Note that ${x}_{1}^{\text{'}}$, ${x}_{2}^{\text{'}}$Ak(i). This is a contradiction. Hence {k(i), k(j)} ∉E(G) and so kEnd(G). And by the definition of k, we get that there exists x′ ∈ V(Bfkf(i)) such that FKF((i, x)) = (fkf(i), x′). Hence fkf = f This means that k is a pseudoinverse of f and so f is regular. Therefore G is End-regular. □

Lemma 2.3Let G be a bipartite graph and let {Bi|iV(G)} be a family of connected bipartite graphs. If G1 is a retract of G and {Hi|iV(G)} are retracts of {Bi|iV(G)}, then G1[Hi]iV(G1) is a retract of G[Bi]iV(G).

Proof. We only need to show that there exists a retraction from G[Bi]iV(G) to G1[Hi]iV(G1). Since G1 and {Hi|iV(G)} are retracts of G and {Bi| iV(G)} respectively, there exist retractions f from G to G1 and hi from Bi to Hi For any jV(G1), take yj1AjIhj and Zj1CjIhj.

Let F be the mapping from G[Bi]iV(G) to itself defined as follows:

$F((k,y))={(k,hk(y)),if k∈V(G1),(f(k),yf(k)1),if k∉V(G1)and y∈Ak,(f(k),zf(k)1),if k∉V(G1)and y∈Ck.$

Let (j, x1), (k, x2) ∈ V(G[Bi]iV(G)) be such that {(j, x1), (k, x2)} ∈ E(G[Bi]iV(G)). Then {j, k} ∈ or j = k and {x1, x2} ∈ E(Bj). If {j, k} ∈ E(i), then there exist ${x}_{1}^{\text{'}}$V(Bf(j)) and ${x}_{2}^{\text{'}}$V(Bf(k)) such that {F((j, x1)), F((k, x2))} = {(f(j), ${x}_{1}^{\text{'}}$, (f(k), ${x}_{2}^{\text{'}}$} ∈ E(G[Bi]iV(G)). If j = k and {x1,x2} ∈ E(Bj), without loss of generality, suppose x1Aj and x2Cj. If jV(G1), then {F((j, x1)), F((k, x2))} = {(j, hj (x1)), (j, hj (x2))}. Since hj is a retraction of Bj, {hj (x1), hj (x2)} ∈ E(Bj). Hence {(j, hj (x1)), (j, hj (x2))} ∈ E(G[Bi]iV(G)). If jV(G1), then {F((j, x1)), F((k, x2))} = {(f(j), yf(j)1), (f(j), zf(j)1)} ∈ E(G[Bi] iV(G)). Therefore FEnd(G[Bi]iV(G)). It is easy to check that IF = G1[Hi]iV(G1). Let (j, x) ∈ V(G1[Hi]iV(G1). Then F((j, x)) = (j, hj(x)) = (j, x) and so F is a retraction from G[Bi]iV(G) to G1[Hi]iV(G1). □

Next we start to seek the conditions for a generalized lexicographic product of bipartite graphs under which G[Bi]iV(G) is End-regular.

Lemma 2.4Let G = K2 with V(G) = {1,2} and let {Bi|iV(G)} be two connected bipartite graphs. Then G[Bi]iV(G) is End-regular if and only if one of the following conditions hold:

(1) B1 is End-regular and B2 = K2,

(2) Both B1 and B2 are trees of diameter 2,

(3) Both B1 and B2 are complete bipartite graphs,

(4) B1 is a tree of diameter 2 and B2 is a complete bipartite graph

Proof Note that G[Bi]iV(G) is isomorphic to B1 + B2. Then the result follows directly from Lemma 1.5. □

Lemma 2.5Both P3 [P3, K2, K2] and P3 [K2, P3, K2] are not End-regular

Fig. 1

Graph P3[P3, K2, K2]

Fig. 2

Graph P3[K2, P3, K2]

Proof Let

$f=(x1x2x3y1y2z1z2x1y1x1y2x2x3y1).$

Then fEnd(P3[P3, K2, K2]). It is easy to see that ρf = {[x1, x3], [y2], [z1], [x2, z2], [y1]}. Suppose that there exists an idempotent endomorphism g of P3[P3, K2, K2] such that ρg = ρf Then g(y1) = y1, g(y2) = y2 and g(z1) = z1. Since z2 is adjacent to every vertices of {y1, y2, z1}, g(z2) is adjacent to every vertex of g({y1, y2, z1}) = {y1, y2, z1}. Note that z2 is the only vertex in V(P3[P3, K2, K2]) adjacent to every vertices of {y1, y2, z1}. Then g(x2) = g(z2) = z2. Since {x1, x2} ∈ E(P3[P3, K2, K2]), {g(x1) g(x2)} = {g(x1), z2)} ∈ E(P3[P3, K2, K2]). Thus g(x1) ∈ {y1, y2, z1)}. This is a contradiction. By Lemma 1.1 f is not regular and so P3[P3, K2, K2] is not End-regular.

Let

$h=(a1a2b1b2b3c1c2a1b1a2b2a2b3a1).$

Then hEnd(P3[K2, P3, K2]). It is easy to see that ρh = {[a1, c2], [a2], [b1, b3], [b2], [c1]}. Suppose that there exists an idempotent endomorphism k of P3[K2, P3, K2] such that ρh = ρh Then k(a2) = a2, k(b2) = b2 and k(c1) = c1. Since b1 is adjacent to every vertex of {a2, b2, c1}, k(b1) is adjacent to every vertex of k({a2, b2,c1}) = {a2, b2, c1}. Note that {a1, a2} ∈ E(P3[K2, P3, K2]), {c2, b2} ∈ E(P3[K2, P3, K2]) and {c2, c1} ∈ E(P3[K2, P3, K2]). Then k(a1) = k(c2) is adjacent to every vertex of k({a2, b2, c1}) = {a2, b2, c1}. It follows from {a1, b1} ∈ E(P3[K2, P3, K2]) that {f(a1), f(b1)} ∈ E(P3[K2, P3, K2]). Note that there are no two adjacent vertices in V(P3[K2, P3, K2]) adjacent to every vertex of {a2, b2, c1}. This is a contradiction. By Lemma 1.1 h is not regular and so P3[K2, P3, K2] is not End-regular. □

Lemma 2.6Let GK2 be a connected End-regular bipartite graph and let {Bi|iV(G)} be a family of connected bipartite graphs. Then G[Bi]iV(G) is End-regular if and only if Bi = K2 for any iV(G).

Proof. Necessity. Suppose that G[Bi]iV(G) is End-regular. Then G is End-regular by Lemma 2.2 and Bi is End-regular for any iV(G) by Lemma 1.8. Note that G is a connected bipartite graph and GK2. Then G contains at least two edges. Thus P3 is a retract of G Let {Bi|iV(G)} be a family of connected bipartite graphs. If there exists jV(G) such that BjK2, then P3 is a retract of Bj Note that K2 is a retract of any bipartite graphs. By Lemma 2.3, P3[P3, K2, K2] or P3[K2, P3, K2] is a retract of G[Bi]iV(G). Since both P3[P3, K2, K2] and P3[K2, P3, K2] are not End-regular, by Lemma 1.3, G[Bi]iV(G) is not End-regular. This is a contradiction.

Sufficiency. If Bi = K2 for any iV(G), then G[Bi]iV(G) = G[K2], the lexicographic products of G and K2. By Lemma 1.6, G[Bi]iV(G) is End-regular. □

With these preparations, the proof of our main result is straightforward.

Proof of Theorem 2.1 This follows directly from Lemmas 2.4 and 2.6. □

## 3 End-orthodox generalized lexicographic products of bipartite graphs

In this section, we characterize the End-orthodox generalized lexicographic products of bipartite graphs. Since End-orthodox graphs are End-regular, we always assume our graphs are End-regular in this section. The following theorem is our main result.

Theorem 3.1Let G be a connected bipartite graph and let {Bi|iV(G)} be a family of connected bipartite graphs. Then G[Bi]iV(G) is End-orthodox if and only if

(1) G = K2, B1 = K2 and B2 ∈ {K2, P3, P4, C4}, or

(2) G = K2, B1 = P3 and B2 = C4.

To prove our main result, we need the following characterizations of the endomorphism monoids of lexicographic products of bipartite graphs.

Lemma 3.2If X is End-orthodox and Y is a retract of X, then Y is End-orthodox

Proof. Since X is End-orthodox, X is End-regular. Let Y be a retract of X By Lemma 1.3, Y is End-regular. To show that Y is End-orthodox, we only need to prove that the composition of any two idempotent endomorphisms of Y is also idempotent.

Since Y is a retract of X, there exists a retraction f from X to Y Let g1,g2Idpt(Y). Then g1f, g2fIdpt(X). Moreover, (g1f)(y) = g1(y) and (g2f)(y) = g2(y) for any yV(Y). Now it is easy to see that [(g1f)(g2f)](y) = (g1f)[(g2f)(y)] = (g1f)(g2(y)). Note that g2(y) ∈ V(Y). Thus (g1f)(g2(y)) = g1(g2(y)) = (g1g2)(y) ∈ V(Y). Hence [(g1f)(g2f)]2(y) = [(g1f)(g2f)]([(g1f)(g2f)](y)) = [(g1f)(g2f)][(g1g2)(y)] = (g1g2)[(g1g2)(y)] = (g1g2)2(y). Note that X is End-orthodox. Then [(g1f)(g2f)]2 = (g1f)(g2f). Thus (g1g2)2(y) = (g1g2)(y) for any yV(Y). Hence g1g2Idpt(Y) and so Y is End-orthodox. □

Lemma 3.3Let G be a bipartite graph and let {Bi|iV(G)} be a family of bipartite graphs. If G[Bi]i∈V(G) is End-orthodox, then G and Bi are End-orthodox for any iV(G).

Proof Since G[Bi]iV(G) is End-orthodox, G[Bi]iV(G) is End-regular. Then G is End-regular by Lemma 2.2 and Bi are End-regular for any iV(G) by Lemma 1.8.

Since Bi is bipartite, its vertex set V(Bi) can be partitioned into two disjoint non-empty subsets Ai and Ci, such that no edge joins two vertices in the same set. Take yi0Ai and zi0Ci for any iV(G). Let F be a mapping from V(G[Bi]iV(G)) to itself defined by

$F((i,x))={(i,yi0),x∈Ai,(i,zi0),x∈Ci.$

Then it is easy to check that FIdpt(G[Bi]i∈V(G)) and IFG[K2]. Note that G[K2] is a retract of G[Bi]i∈V(G). By Lemma 3.3, G[K2] is End-orthodox. By Lemma 1.9, G is End-orthodox.

Let iV(G) and g1 , g2Idpt(Bi). Define two mappings G1 and G2 from V(G[Bi]i∈V(G)) to itself by

$G1((i,x))={(i,g1(x)),x∈V(Bi),(i,x),others. and G2((i,x))={(i,g2(x)),x∈V(Bi),(i,x),others.$

Then it is easy to check that G1, G2Idpt(G[Bi]i∈V(G)) and so G1G2 is also an idempotent of End(G[Bi]i∈V(G)) since G[Bi]iV(G) is End-orthodox. For any xV(Bj), we have

$(G1G2)2((i,x))=(G1G2)((i,(g1g2)(x)))=(i,(g1g2)2(x))=(G1G2)((i,x))=(i,(g1,g2)(x)).$

Clearly, (g1g2)2 = g1g2. Hence g1g2Idpt(Bi) and so Bi is End-orthodox.

Proof of Theorem 3.1 Sufficiency. In case (1), G[Bi]iV(G) is isomorphic to K2 + K2, K2 + P3, K2 + P4, or K2 + C4. By Lemma 1.10, we have immediately that G[Bi]iV(G) is End-orthodox. In case (2), G[Bi]iV(G) is isomorphic to P3 + C4. Also by Lemma 1.10, G[Bi]iV(G) is End-orthodox.

Necessity. We only need to show that G[Bi]iV(G) is not End-orthodox for the following cases:

Case 1. G = K2, B1 and B2 do not satisfy the conditions (1) and (2). Then G[Bi]iV(G) = B1 + B2. By Lemma 1.10, G[Bi]iV(G) is not End-orthodox.

Fig. 3

Graphs P3[K2], P4[K2] and C4[K2]

Case 2. G = P3 and Bi = K2 for any i ∈ V(G). Then G[Bi]i∈V(G)P3 [K2] (see Fig. 3). We will show that there exist f, gIdpt(G[Bi]iV(G)) such that fgIdpt(G[Bi]iV(G)). Let

$f=(x1x2y1y2z1z2z2z1y1y2z1z2)$

and

$g=(x1x2y1y2z1z2x1x2y1y2x1x2).$

Then f, gIdpt(G[Bi]iV(G)). But

$fg=(x1x2y1y2z1z2z2z1y1y1z2z1).$

It is easy to check fgIdpt(G[Bi]iV(G)). Hence G[Bi]iV(G) is not End-orthodox.

Case 3.G = P4 and Bi = K2 for any iV(G). Then G[Bi]i∈V(G)P4[K2] (see Fig.3). Let

$f=(a1a2b1b2c1c2d1d2c1c2d1d2c1c2d1d2)$

and

$g=(a1a2b1b2c1c2d1d2a1a2b1b2a2a1b1b2).$

Then f, gIdpt(G[Bi]iV(G)). But

$fg=(a1a2b1b2c1c2d1d2c1c2d1d2c2c1d1d2).$

It is easy to check fgIdpt(G[Bi]iV(G)). Hence G[Bi]iV(G) is not End-orthodox.

Case 4. G = C4 and Bi = K2 for any iV(G). Then G[Bi]iV(G)C4[K2] (see Fig.3). Let

$f=(a1a2b1b2c1c2d1d2a1a2b1b2a1a2b1b2)$

and

$g=(a1a2b1b2c1c2d1d2a1a2d2d1a1a2d1d2).$

Then f, gIdpt(G[Bi]iV(G)). Bu

$fg=(a1a2b1b2c1c2d1d2a1a2b2b1a1a2b1b2).$

It is easy to check fgIdpt(G[Bi]iV(G)). Hence G[Bi]iV(G) is not End-orthodox. □

## Acknowledgement

The authors wish 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 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).

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## About the article

Accepted: 2016-02-03

Published Online: 2016-04-06

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, Export Citation