Generalized Cayley Digraphs

Pure Mathematical Sciences, Vol. 1, 2012, no. 1, 1-12 Generalized Cayley Digraphs Anil Kumar V. Department of Mathematics, University of Calicut Malappuram, Kerala, India 673 635 anilashwin2003@yahoo.com Abstract In this paper we introduce a new class of digraphs induced by groups. These digraphs can be considered as the generalization of Cayley digraphs. Moreover, various graph properties are expressed in terms of algebraic properties. This did not attract much attention in the literature. Mathematics Subject Classification: 05C25 Keywords: Cayley digraph, vertex-transitive graph 1 Introduction A binary relation on a set V is a subset E of V V.Adigraph is a pair (V,E) where V is a non-empty set (called vertex set) and E is a binary relation on V. The elements of E are the edges of the digraph. An edge of the from (x, x) is called a loop. A digraph (V,E) is called vertex-transitive if, given any two vertices a and b of V, there is some automorphism f : V V such that f(a) = b. In other words, a digraph is vertex-transitive if its automorphism group acts transitively upon its vertices [4]. Whenever the word graph is used in this paper it will be referring to a digraph unless otherwise stated. Let G be a group and let D be a subset of G. The Cayley digraph X = C(G, D) is the digraph with vertex set G, and the vertex x is adjacent to the vertex y if and only if x 1 y D [6]. The subset D is called the connection set of X. That is, Cayley digraph C(G; D) has as its vertex-set and arc-set, respectively, V = G and E = {x z : x G and z D} = {(x, xz) :z D} = {(x, y) :y = xz for some z D}.

2 Anil Kumar V. Arc x z joins vertex x to vertex y. Let S 3 denote the set of all one-to-one functions from {1, 2, 3} to itself. Then S 3, under function composition, is a group with six elements. The six elements are 0 = μ 1 = ( 1 2 3 1 2 3 ), 1 = ( ) 1 2 3, μ 1 3 2 2 = ( 1 2 3 2 3 1 ), 2 = ( ) 1 2 3,μ 3 2 1 3 = ( 1 2 ) 3 3 1 2 ( 1 2 ) 3 2 1 3 The group S 3 is called permutation group of degree three. Observe that S 3 is a non- commutative group. In this paper we introduce a new class of digraphs called generalised cayley digraphs induced by groups. These digraphs can be considered as a generalization of cayley digraphs defined in [6]. Moreover, various graph properties are expressed in terms of algebraic properties. 2 Generalized Cayley Graphs In this section, we first introduce the following notations: Let A and B be subsets of a group G. Then (1) [A l B l ] = {x G : x = z 1 xz 2 for some z 1 A, z 2 B 2 }. (2) [A] = the semigroup generated by A. (3) [A B] ={a 1 (a 2 (a n 1 (a n b n )b n 1 ) b 2 )b 1 : a i A, b i B,n =1, 2, 3,...}. Definition 2.1. Let G be a group and let D and D be subsets of G. Let R D,D = {(x, y) :y = z 1 xz 2 for some z 1 D, z 2 D } Then the digraph (G, R D,D ) is called the generalized Cayley Digraph of G with connection sets D and D. Observe that when D = {1}, then (G, R D,D ) reduces to the well known Cayley digraph X = C(G, D ). The following are some examples of Generalized Cayley digraphs. Example 2.2. Let G be the permutation group S 3, D = { 1,μ 1 } and D = {μ 1,μ 2,μ 3 }. Then the generalized Cayley digraph for S 3 with connection sets D and D is a complete digraph as shown in figure 1. Example 2.3. Let G be the permutation group S 3, D = { 1 } and D = {μ 1,μ 2,μ 3 }. Then the generalized Cayley digraph for S 3 with connection sets D and D is a complete bipartite digraph as shown in figure 2.

Generalized Cayley digraphs 3 μ μ 0 1 0 μ 2 1 2 Figure 1: Generalised Cayley digraph with connection sets D = { 1,μ 1 } and D = {μ 1,μ 2,μ 3 } μ μ μ 1 2 3 0 1 2 Figure 2: Generalised Cayley digraph with connection sets D = { 1 } and D = {μ 1,μ 2,μ 3 }

4 Anil Kumar V. μ3 μ μ 1 2 0 2 1 Figure 3: Generalised Cayley digraph with connection sets D = {μ 1 } and D = {μ 1,μ 2,μ 3 } Example 2.4. The generalized Cayley digraph (S 3,R {μ1 },{μ 1,μ 2,μ 3 }) is shown in figure 3. Observe that this graph is the complement of the generalized Cayley digraph shown in figure 2. Next, we prove that many graph properties can be expressed in terms of algebraic properties. Proposition 2.5. (G, R D,D ) is an empty graph if and only if D = or D =. Proposition 2.6. (G, R D,D ) is a reflexive graph if and only if G =[D l D l ]. Proof. Suppose that (G, R D,D ) is a reflexive graph and let x G. (x, x) R D,D. This implies that Then. This implies that x = z 1 xz 2 for some z 1 D, z 2 D. G =[D l D l ]. Conversely, assume that G =[D l D l ]. We will show that (x, x) R D,D for all x G. Observe that x G x [D l D l ] x = z 1 xz 2, for somez 1 D, z 2 D (x, x) R D,D.

Generalized Cayley digraphs 5 Proposition 2.7. (G, R D,D ) is a symmetric graph, then DD = D 1 (D ) 1. Proof. First, assume that (G, R D,D ) is symmetric. Observe that x DD x = z 1 z 2, for some z 1 D, z 2 D (1,x) R D,D This implies that DD = D 1 (D ) 1. (x, 1) R D,D 1=t 1 xt 2, for some t 1 D, t 2 D x = t 1 1 t 1 2 D 1 (D ) 1. Proposition 2.8. If D = D 1 and D =(D ) 1, then (G, R D,D ) is symmetric. Proof. Let x and y be any two vertices of (G, R D,D ) such that (x, y) R D,D. But then y = z 1 xz 2 for some z 1 D, z 2 D. This implies that x = z1 1 yz 1 2 Since D = D 1 and D =(D ) 1, it follows that (y, x) R D,D. Proposition 2.9. (G, R D,D ) is a transitive graph, then D 2 (D ) 2 DD. Proof. Assume that (G, R D,D ) is a transitive graph. Then for all z 1,z 2 D, z 3,z 4 D, we have (1,z 1 z 3 ) and(z 1 z 3,z 2 z 1 z 3 z 4 ) R D,D This implies that (1,z 2 z 1 z 3 z 4 ) R D,D The above statement tells us that z 2 z 1 z 3 z 4 = t 1 1t 2 for some t 1 D, t 2 D. Equivalently, D 2 (D ) 2 DD. Proposition 2.10. If D 2 D and (D ) 2 D, then (G, R D,D ) is a transitive graph

6 Anil Kumar V. Proof. Let x, y and z G such that (x, y) R D,D and (y, z) R D,D. Then This implies that y = z 1 xz 2 for some z 1 D, z 2 D and z = z 3 yz 4 for some z 3 D, z 4 D. z =(z 3 z 1 )x(z 2 z 4 ). Since D 2 D and (D ) 2 D, it follows that (x, z) R D,D. Hence (G, R D,D ) is a transitive graph. Proposition 2.11. (G, R D,D ) is a complete graph, then G = DD. Proof. Assume that (G, R D,D ) is a complete graph and let x G. Then (1,x) R D,D. This implies that x = z 1 z 2, for some z 1 D and z 2 D. That is, x DD. Since x is an arbitrary element of G, G = DD. Proposition 2.12. If G = D D, then (G, R D,D ) is a complete graph. Proposition 2.13. If (G, R D,D ) is connected, then G =[D D ]. Proof. Suppose that (G, R D,D ) is connected and let x G. Then there is a path from 1 to x, say: (1,x 1,x 2,,x n,x) Then we have x 1 = z 1 t 1,x 2 = z 2 x 1 t 2,...,x n = z n x n 1 t n,x= z n+1 x n t n+1 for some z i D and t i D. This implies that x = z n+1 z n...z 1 t 1 t 2...t n+1 Consequently, G =[D D ]. Proposition 2.14. If G =[D] =[D ] and 1 D D, then (G, R D,D ) is connected. Proof. Let x and y be ant two elements in G. Let y = z 1 xz 2. Then z 1,z 2 G. Without loss of generality assume that z 1 [D] and z 2 [D ]. Let z 1 = t 1 t 2...t n+1,t i D and z 2 = w 1 w 2...w n+1,w i D.

Generalized Cayley digraphs 7 Then we have, Let y = t 1 t 2...t n+1 xw 1 w 2...w n+1 We find that x 1 = t n+1 xw 1,x 2 = t n x 1 w 2,...,x n = t 2 x n 1 w n,x n+1 = t 1 x n w n+1 (x, x 1 ) R D,D, (x 1,x 2 ) R D,D,...,(x n,y) R D,D Hence, (x, x 1,...,x n,y) is a path from x to y and hence (G, R D,D ) is connected. Proposition 2.15. If (G, R D,D ) is locally connected, then [D D ]=[D 1 D 1 ] Proof. Assume that (G, R D,D ) is locally connected. Let x [D D ]. Then for some z i D and t i D. Let x = z 1 z 2...z n t n t n 1...t 2 t 1 x 1 = z n t n,x 2 = z n 1 x 1 t n 1,...,x n = z 1 x n 1 t 1 Then (1,x 1,...,x n ) is a path from 1 to x. Since (G, R D,D ) is locally connected, there exits a path from x to 1, say: (x, y 1,...,y n, 1) This implies that x [D 1 D 1 ]. Hence [D D ] [D 1 D 1 ]. Similarly, [D 1 D 1 ] [D D ]. Proposition 2.16. If D = D 1 and D = D 1, then (G, R D,D ) is locally connected. Proposition 2.17. If (G, R D,D ) is semi connected, then G =[D D ] [D 1 D 1 ]. Proof. Assume that (G, R D,D ) is semi connected and let x G. then there exits a path from 1 to x, say or a path from x to 1, say (1,x 1,...,x n,x) (x, y 1,...,y m, 1) This implies that x [D D ] [D 1 D 1 ]. Since x is arbitrary, it follows that G =[D D ] [D 1 D 1 ].

8 Anil Kumar V. Proposition 2.18. If (G, R D,D ) is a quasi ordered set, then (i) G =[D l D l ] (ii) D 2 (D ) 2 DD. Proposition 2.19. If G = [D l Dl ], D 2 D and (D ) 2 D, then (G, R D,D ) is a quasi ordered set. Proposition 2.20. If (G, R D,D ) is a partially ordered set, then (i)dd D 1 D 1 = {1} (ii)d 2 (D ) 2 DD. Proposition 2.21. If (G, R D,D ) is a linearly ordered set, then (i) DD D 1 D 1 = {1} (ii) D 2 (D ) 2 DD. (iii) DD D 1 D 1 = G Proposition 2.22. (G, R D,D ) is a hasse- diagram, if and only if D n D = or (D ) n D =,n 2. Proof. First, assume that (G, R D,D ) is a hasse- diagram. then for any x 0,x 1,...,x n G with (x i,x i+1 B) R D,D for all i =0, 1, 2,...,n 1 implies that (x 0,x n ) / R D,D. Observe that (x i,x i+1 ) R D,D for all i =0, 1, 2,...,n 1 implies that x n = z 1 x 0z 2 for some z 1 D n and z 2 (D ) n. Since (x 0,x n ) / R D,D, therefore Conversely, assume that D n D = or (D ) n D =. D n D = or (D ) n D =,n 2. We will show that (G, R D,D ) is a hasse-diagram. Let x 0,x 1,...,x n be any (n + 1) elements of G with n 2, and (x i,x i+1 ) R D,D i =0, 1, 2,...,n 1. Then we have for all x n = z n z n 1...z 2 z 1 x 0 t 1 t 2...t n for some z i D and t i D This implies that x n = z1 x 0z2 for some z 1 Dn and z2 (D ) n. Since D n D = or (D ) n D =, (x 0,x n ) / R D,D. Hence (G, R D,D ) is a hasse-diagram.

Generalized Cayley digraphs 9 Proposition 2.23. (G, R D,D ) is self dual if and only if D and D commutes with every elements of G. Proof. Define a mapping θ : G G by (i) θ is one-to-one. For if θ(x) =x 1 θ(x) =θ(y) x 1 = y 1 x = y (ii) (x, y) R D,D (θ(x),θ(y)) R 1 D,D (x, y) R D,D y = z 1 xz 2 for some z 1 D, z 2 D y 1 = z 1 2 x 1 z 1 1 x 1 = z 2 y 1 z 1 (θ(x),θ(y)) R 1 D,D (iii) Obviously, θ is onto. Hence (G, R D,D ) is isomorphic to (G, R 1 D,D ). Proposition 2.24. (G, R D,D ) is a forest if and only if [D n l (D ) n l ] = for all n =1, 2,... Proof. Assume that (G, R D,D ) is a forest. If possible, suppose that [D n l (D ) n l ] for some n. Let x [D n l (D ) n l ]. Then x = z 1 z 2...z n xt 1 t 2...t n for some z i D and t i D. Let x 1 = z n xt 1,x 2 = z n 1 x 1 t 2,...,x n = z 1 x n 1 t n = x. Then obviously (x, x 1,x 2,...,x n 1,x) is a circuit in (G, R D,D ). This contradicts the assumption that (G, R D,D )is a forest. Hence [Dl n (D ) n l ] = for all n =1, 2,... Conversely, assume that [Dl n (D ) n l ] = for all n =1, 2,... We will show that (G, R D,D ) is a forest. Suppose this does not hold. Then (G, R D,D ) must contain a circuit, say (x, y 1,y 2,...,y n,x) This implies that y 1 = z 1 xt 1,y 2 = z 2 y 1 t 2,...,y n = z n y n 1 t n,x= z n+1 y n t n+1

10 Anil Kumar V. for some z i D, t i D. That is, Equivalently, x = z n+1 z n...z 1 xt 1 t 2...t n+1 x [D n l (D ) n l ] This contradiction completes the proof. Proposition 2.25. Suppose (G, R D,D ) is a finite graph with n vertices and [D n l (D ) n l ] [D k l (D ) k l ] = for k =1, 2,...,n 1. Then (G, R D,D ) is Hamiltonian if and only if G =[D n l (D ) n l ]. Proof. Suppose that (G, R D,D ) is a Hamiltonian graph. Let x be any vertex in G. Then there exits a circuit, say (x, x 1,x 2,...,x n 1,x) containing all the n vertices of (G, R D,D ). Then we have x 1 = z 1 xt 1,x 2 = z 2 x 1 t 2,x 3 = z 3 x 2 t 3,...,x n 1 = z n 1 x n 2 t n 1,x= z n x n 1 t n for some z i D, t i D. That is, This implies that x = z n z n 1...z 2 z 1 xt 1 t 2...t n x [D n l (D ) n l ]. Since x is an arbitrary element of G, it follows that G =[D n l (D ) n l ]. Conversely assume that G =[D n l (D ) n l ]. We will show that (G, R D,D )is Hamiltonian. Let x G. Then for some z i D, t i D. Let x = z n z n 1...z 2 z 1 xt 1 t 2...t n x 1 = z 1 xt 1,x 2 = z 2 x 1 t 2,x 3 = z 3 x 2 t 3,...,x n 1 = z n 1 x n 2 t n 1 Then x, x 1,x 2,...,x n 1 distinct elements in G. For if x i = x j (i<j) x = z n z n 1...z j x j t j t j+1...t n x = z n z n 1...z j x i t j t j+1...t n x = z n z n 1...z j (z i z i+1...z 1 xt 1 t 2...t i )t j t j+1...t n x [D n j+i l (D ) n j+i l ]

Generalized Cayley digraphs 11 x [D n l (D ) n l ] [D n j+i l (D ) n j+i l ] This contradiction shows that the elements x, x 1,x 2,...,x n 1 are all distinct. It is easily seen that (x, x 1,x 2,...,x n 1,x) is a circuit containing all the vertices of (G, R D,D ). Hence (G, R D,D )isa Hamiltonian graph. Proposition 2.26. The in-degree of the vertex 1 is the cardinal number D 1 D 1. Proof. Let Observe that Hence (1) = D 1 D 1. (1) = {x G :(x, 1) R D,D } x (1) (x, 1) R D,D 1=z 1 xz 2 x = z1 1 z 1 2 x D 1 D 1 Proposition 2.27. The out-degree of the vertex 1 is the cardinal number DD. Proof. Let Observe that Hence σ(1) = DD. σ(1) = {x G :(1,x) R D,D } x σ(1) (1,x) R D,D x = z 1 z 2 for some z 1 D, z 2 D x DD It is well known that all Cayley digraphs are vertex transitive digraphs [4]. So it is natural to think whether (G, R D,D ) is vertex- transitive. If G is commutative or A, B, D and D commutes with every elements of G, then (G, R D,D ) is vertex transitive. We conclude this paper with the following problem: Problem 2.28. Let G be a non- commutative group and let D and D be subsets of G such that they do not commute with any elements of G. Prove or disprove that (G, R D,D ) is vertex-transitive.

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