C 7 -DECOMPOSITIONS OF THE TENSOR PRODUCT OF COMPLETE GRAPHS

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1 Discussiones Mathematicae Graph Theory 37 (2017) doi: /dmgt.1936 C 7 -DECOMPOSITIONS OF THE TENSOR PRODUCT OF COMPLETE GRAPHS R.S. Manikandan Department of Mathematics Bharathidasan University Constituent College Lalgudi , India manirs2004@yahoo.co.in and P. Paulraja Department of Mathematics Kalasalingam University Krishnankoil , India ppraja56@gmail.com Abstract In this paper we consider a decomposition of K m K n, where denotes the tensor product of graphs, into cycles of length seven. We prove that for m,n 3, cycles of length seven decompose the graph K m K n if and only if (1) either m or n is odd and (2) 14 m(m 1)n(n 1). The results of this paper together with the results of [C p -Decompositions of some regular graphs, Discrete Math. 306 (2006) ] and [C 5 -Decompositions of the tensor product of complete graphs, Australasian J. Combinatorics 37 (2007) ], give necessary and sufficient conditions for the existence of a p- cycle decomposition, where p 5 is a prime number, of the graph K m K n. Keywords: cycle decomposition, tensor product Mathematics Subject Classification: 05C Introduction All graphs considered here are simple and finite. Let C n denote the cycle of length n. We write G = H 1 H 2 H k if H 1,H 2,...,H k are edge-disjoint subgraphs of G and E(G) = E(H 1 ) E(H 2 ) E(H k ). If the edge set of

2 524 R.S. Manikandan and P. Paulraja the graph G can be partitioned into cycles C n1,c n2,...,c nr, then we say that C n1,c n2,...,c nr decompose G. If n 1 = n 2 = = n r = k, then we say that G has a C k -decomposition and in this case we write C k G. We may also call a cycle of length k a k-cycle. If G has a 2-factorization and each 2-factor of it has only cycles of length k, then we say that G has a C k -factorization (we use the notation C k G.) The complete graph on m vertices is denoted by K m and its complement is denoted by K m. For some positive integer k, the graph kh denotes k disjoint copies of H. For two graphs G and H their wreath product, G H, has the vertex set V(G) V(H) in which (g 1,h 1 ) and (g 2,h 2 ) are adjacent whenever g 1 g 2 E(G) or g 1 = g 2 and h 1 h 2 E(H). Similarly, G H, the tensor product of the graphs G and H has the vertex set V(G) V(H) in which two vertices (g 1,h 1 ) and (g 2,h 2 ) are adjacent whenever g 1 g 2 E(G) and h 1 h 2 E(H). Clearly the tensor product is distributive over edge-disjoint union of graphs; that is, if G = H 1 H 2 H k, then G H = (H 1 H) (H 2 H) (H k H). For h V(H), V(G) h = {(v,h) v V(G)} is called the column of vertices in G H correspondingtoh.further, forx V(G), x V(H) = {(x,v) v V(H)} is called the layer of vertices in G H corresponding to x. Similarly we can define column and layer for wreath product of graphs also. We can easily observe that K m K n is isomorphic to the complete m-partite graph in which each partite set has exactly n vertices. A latin square of order n is an n n array, each cell of which contains exactly one of the symbols in {1,2,...,n}, such that each row and each column of the array contains each of the symbols in {1,2,...,n} exactly once. A latin square is said to be idempotent if the cell (i,i) contains the symbol i, 1 i n. Let (X,Y) be the bipartition of the complete bipartite graph K n,n, where X = {x 1,x 2,...,x n } and Y = {y 1,y 2,...,y n }. Let F i (X,Y) = {x 1 y 1+i, x 2 y 2+i,...,x n y n+i }, 0 i n 1, where the addition in the suffixes are taken modulo n with residues 1,2,...,n. We call F i (X,Y) as the 1-factor of distance i from X to Y in K n,n. Note that, in general, F i (X,Y) need not be equal to F i (Y,X) as F i (Y,X) = {y 1 x 1+i,y 2 x 2+i,...,x n y n+i }. From the definition of the tensor product and the wreath product, it is clear that K m K n = K m K n i j F 0(X i,x j ),wherex j sarethepartititesetsofthecompletem-partitegraph in which each of the partite sets has cardinality n. In fact, i j F 0(X i,x j ) consists of n disjoint copies of K m. It is known that if n is odd and m ( (( n 2) or n is even and m n ) 2 n 2), then C m K n or C m K n I, where I is a 1-factor of K n ; see [1, 13]. A similar problem can also be considered for regular complete multipartite graphs; Billington and Cavenagh [6] and Mahamoodian and Mirzakhani [9] have considered C 5 -decompositions of complete tripartite graphs. Moreover, Billington [3] has studied the decompositions of complete tripartite graphs into cycles of length 3

3 C 7 -Decompositions of the Tensor Product of Complete Graphs 525 and 4. Further, Billington and Cavenagh [5] have studied the decompositions of complete multipartite graphs into cycles of length 4, 6 and 8. The present authors, in [10, 11], have proved that the necessary conditions for the existence of a C 5 -decomposition and a C p -decomposition for p 11, where p is a prime number, of K m K n are sufficient. Smith, in [14, 15, 16], proved that the obvious necessary conditions for the existence of a C k -decomposition of K m K n are sufficient when k = 2p,3p and p 2, where p 3 is a prime number. Later, in [17], he has obtained the conditions under which λ(k m K n ) can be decomposed into cycles of length p, where p is a prime; his approach is different from[10,11]. Forrelatedworkseealso[4]. Liu, in[8], studiedthec t -factorization of K m K n, m 3. One can easily observe that the graph K m K n is obtained from the regular complete multipartite graph K m K n, by deleting a suitable set of n disjoint copies of K m. In this paper, the obvious necessary conditions for K m K n, m,n 3, to admit a C 7 -decomposition are proved to be sufficient. In this context, it is pertinent to point out that the existence of C p, p being a prime, decomposition of K m K n played a significant role in establishing the existence of C p - decomposition of K m K n, see [10, 11]. We give below the main theorem obtained here. Theorem 1. For m, n 3, C 7 K m K n if and only if (1) 14 nm(m 1)(n 1), and (2) either m or n is odd. For our future reference we list below some known results. Theorem A [2]. Let s be an odd integer and t be a prime so that 3 s t. Then C s K t has a 2-factorization so that each 2-factor is composed of s cycles of length t. Theorem B [1]. If n 1 or 7 (mod 14), then C 7 K n. Theorem C [7]. Let m be an odd integer, m 3. (1) If m 1 or 3 (mod 6), then C 3 K m. (2) If m 5 (mod 6), then K m can be decomposed into (m(m 1) 20)/6 3-cycles and two 5-cycles. 2. C 7 -Decompositions of C 3 K m We quote the following lemma for our future reference.

4 526 R.S. Manikandan and P. Paulraja Lemma 2 [10]. For any odd integer t 3, C t C 3 K t. Lemma 3. C 7 C 3 K 8. Proof. Let the partite sets of the tripartite graph C 3 K 8 be {u 1,u 2,...,u 8 }, {v 1,v 2,...,v 8 } and {w 1,w 2,...,w 8 }, where we assume that the vertices having the same subscript are the corresponding vertices of the partite sets. Now the cycle (u 1 v 3 w 6 v 2 w 7 v 1 w 8 u 1 ) under the permutation (u 1 u 2 u 8 )(v 1 v 2 v 8 ) (w 1 w 2 w 8 ) and its powers give us eight 7-cycles. These eight 7-cycles under the permutation (u 1 v 1 w 1 )(u 2 v 2 w 2 ) (u 8 v 8 w 8 ) and its powers give us the required twenty four 7-cycles. Remark 4. Let the partite sets of the complete tripartite graph C 3 K m, m 1, be {u 1,u 2,...,u m }, {v 1,v 2,...,v m } and {w 1,w 2,...,w m }. Consider a latin square L of order m. We associate a triangle of C 3 K m with each entry of L as follows: if k is the (i,j) th entry of L, then the triangle of C 3 K m corresponding to k is (u i v j w k u i ). Clearly the triangles corresponding to the entries of L decompose C 3 K m, see [3]. ThenecessaryconditionfortheexistenceofdecompositionofC 3 K m,m 3, into C 7 is m 0 or 1 (mod 7). We prove that it is also sufficient. Theorem 5. C 7 C 3 K m if and only if m 0 or 1 (mod 7). Proof. The necessity is obvious. We prove the sufficiency in two cases. Case 1. m 1 (mod 7). Let m = 7k +1. Subcase( 1.1. k 2. Let partite sets of the tripartite graph C 3 K m be k ) ( U = {u 0 } i=1 {ui 1,ui 2,...,ui 7 } k ), V = {v 0 } i=1 {vi 1,vi 2,...,vi 7 } and W = ( k ) {w 0 } i=1 {wi 1,wi 2,...,wi 7 } ; we assume that the vertices having the same subscript and superscript are the corresponding vertices of the partite sets. By the definition of the tensor product,{u 0,v 0,w 0 } and {u i j,vi j,wi j }, 1 j 7, are independent sets and the subgraph induced by each of the sets U V, V W and W U is isomorphic to K m,m F 0, where F 0 is the 1-factor of distance zero in K m,m. We obtain a new graph out of H = (C 3 K m ) {u 0,v 0,w 0 } = C 3 K 7k as follows: for each i, 1 i k, identify the sets of vertices {u i 1,ui 2,...,ui 7 }, {v1 i,vi 2,...,vi 7 } and {wi 1,wi 2,...,wi 7 } into new vertices ui,v i and w i respectively; two new vertices are adjacent if and only if the corresponding sets of vertices in H induce a complete bipartite subgraph K 7,7 or a K 7,7 F, where F is a 1-factor of K 7,7. This defines the graph C 3 K k with partite sets {u 1,u 2,...,u k }, {v 1,v 2,...,v k } and {w 1,w 2,...,w k }.

5 C 7 -Decompositions of the Tensor Product of Complete Graphs 527 ConsideranidempotentlatinsquareLoforderk,k 2(whichexists,see[7]). To complete the proof of this subcase, we associate with entries of L edge-disjoint subgraphs of C 3 K m which are decomposable by C 7. The i th diagonal entry of L corresponds to the triangle (u i v i w i u i ), 1 i k, of C 3 K k, see Remark 4. The subgraph of H corresponding to the triangle of C 3 K k is isomorphic to C 3 K 7. For each triangle (u i v i w i u i ), 1 i k, of C 3 K k corresponding to the i th diagonal entry of L, associate the subgraph of C 3 K m induced by vertices {u 0,u i 1,ui 2,...,ui 7 } {v 0,v1 i,vi 2,...,vi 7 } {w 0,w1 i,wi 2,...,wi 7 }; as this subgraph is isomorphic to C 3 K 8, it can be decomposed into 7-cycles, by Lemma 3. Again, if we consider the subgraph of H corresponding to the triangle of C 3 K k, which corresponds to a non-diagonal entry of L, then it is isomorphic to C 3 K 7. By Theorem A, C 3 K 7 can be decomposed into 7-cycles. Thus we have decomposed C 3 K m into 7-cycles when k 2. Subcase 1.2. k = 2. By Theorem B, C 7 K 15 and hence we write C 3 K 15 = K 15 C 3 = (C 7 C 3 ) (C 7 C 3 ) (C 7 C 3 ). Each copy of C 7 C 3 can be decomposed into 7-cycles, see Figure 1. This proves that C 7 C 3 K 15. x 1 x 2 x 3 x 4 x 5 x 6 x 7 y 1 y 2 y 3 y 4 y5 y 6 y 7 z 1 z 2 z 3 z 4 z 5 z 6 z 7 C 3 C 7 Figure 1. A 7-cycle decomposition of C 3 C 7. Different types of edges give different 7-cycles. Case 2. m 0 (mod 7). Let m = 7k. Subcase 2.1. k 2. As in the previous case, let the partite sets of the tripartite graph C 3 K m be U = k i=1 {ui 1,ui 2,...,ui 7 }, V = k i=1 {vi 1,vi 2,...,vi 7 } and W = k i=1 {wi 1,wi 2,...,wi 7 }. We assume that the vertices having the same subscript and superscript are the corresponding vertices of the partite sets. As in

6 528 R.S. Manikandan and P. Paulraja the proof of Subcase 1.1, from C 3 K m = C 3 K 7k we obtain the graph C 3 K k with partite sets {u 1,u 2,...,u k },{v 1,v 2,...,v k } and {w 1,w 2,...,w k }. Consider an idempotent latin square L of order k, k 2. The diagonal entries of L correspond to the triangles (u i v i w i u i ), 1 i k, of C 3 K k. If we consider the subgraph of C 3 K m corresponding to a triangle of C 3 K k, which corresponds to a diagonal entry of L, then it is isomorphic to C 3 K 7. By Lemma 2, C 7 C 3 K 7. Again, as in the previous case, the triangle of C 3 K k corresponding to a non-diagonal entry of L gives a subgraph of C 3 K m isomorphic to C 3 K 7 ; by Theorem A, C 7 C 3 K 7. Subcase 2.2. k = 2. Let the partite sets of the tripartite graph C 3 K 14 be X = {x 1,x 2,...,x 14 },Y = {y 1,y 2,...,y 14 } and Z = {z 1,z 2,...,z 14 }; we assume that the vertices having the same subscript are the corresponding vertices of the partite sets. By the definition of the tensor product, {x i,y i,z i }, 1 i 14, are independent sets and the subgraph induced by each of the subsets of vertices X Y, Y Z and Z X are isomorphic to K 14,14 F 0, where F 0 is the 1-factor of distance zero in K 14,14. We obtain a new graph out of C 3 K 14 as follows: for each i, 1 i 7, identify the subsets of vertices {x 2i 1,x 2i },{y 2i 1,y 2i } and {z 2i 1,z 2i } into new vertices x i,y i and z i, respectively, and two of these vertices are adjacent if and only if the corresponding subsets of vertices in C 3 K 14 induce a K 2,2. The resulting graph is isomorphic to C 3 K 7 with partite sets X = {x 1,x 2,...,x 7 },Y = {y 1,y 2,...,y 7 } and Z = {z 1,z 2,...,z 7 }; note that {x i,y i,z i }, 1 i 7, are independent sets of C 3 K 7. Now C 3 K 7 = C3 (C 7 C 7 C 7 ) = (C 3 C 7 ) (C 3 C 7 ) (C 3 C 7 ). The graph C 3 C 7 can be decomposed into 7-cycles, see Figure 1, and hence C 7 C 3 K 7. By lifting back these 7-cycles of C 3 K 7 to C 3 K 14, we get edge-disjoint subgraphs isomorphic to C 7 K 2. But C 7 K 2 can be decomposed into cycles of length 7, see [12]. Thus the subgraphs of C 3 K 14 obtained by lifting back the 7-cycles of C 3 K 7 to C 3 K 14 can be decomposed into cycles of length 7. The edges of C 3 K 14 which are not covered by these 7-cycles are shown in Figure 2. To complete the proof we fuse some of the 7-cycles obtained above with the graph of Figure 2 and decompose the resulting graph into cycles of length 7. Let H be the graph obtained by the union of the graph of Figure 2 and the subgraph of C 3 K 14 which is obtained by lifting back two 7-cycles of C 3 K 7, namely, (x 1 y 2 x 3 y 4 x 5 z 6 y 7 x 1 ) and (z 1 y 2 z 3 y 4 z 5 x 6 y 7 z 1 ) shown in Figure 1. The subgraph H of C 3 K 14 is shown in Figure 3. A 7-cycle decomposition of H is given below: (x 1 y 2 z 1 x 2 y 4 z 6 y 3 x 1 ), (x 1 z 2 y 1 x 2 y 3 z 5 y 4 x 1 ), (x 5 y 6 z 5 y 8 z 9 y 7 z 6 x 5 ), (x 6 y 5 z 6 y 8 z 10 y 7 z 5 x 6 ), (x 9 y 10 z 9 x 11 y 14 x 12 z 10 x 9 ), (x 10 y 9 z 10 x 11 y 13 x 12 z 9 x 10 ), (x 14 y 13 z 1 y 4 z 2 y 14 z 13 x 14 ), (x 13 y 14 z 1 y 3 z 2 y 13 z 14 x 13 ), (x 3 y 4 x 6 y 8 x 5 y 3 z 4 x 3 ),

7 C 7 -Decompositions of the Tensor Product of Complete Graphs 529 (x 4 z 3 y 4 x 5 y 7 x 6 y 3 x 4 ), (x 7 y 8 x 10 z 11 x 9 y 7 z 8 x 7 ), (x 8 y 7 x 10 z 12 x 9 y 8 z 7 x 8 ), (x 11 y 12 z 11 y 13 x 2 y 14 z 12 x 11 ) and (x 12 y 11 z 12 y 13 x 1 y 14 z 11 x 12 ). This completes the proof. x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 y 10 y 11 y 13 y 12 y 14 z 1 z 2 z 3 z 4 z 5 z 6 z 7 z 8 z 9 z 10 z 11 z 12 z 13 z 14 Figure 2 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 y 2 y 1 y 4 y 3 y 6 y 5 y 8 y 7 y 10 y 9 y 11 y 12 y 14 y 13 z 1 z 2 z 3 z 4 z 5 z 6 z 7 z 8 z 9 z 10 z 11 z 12 z 13 z 14 Figure 3 3. C 7 -Decomposition of C 5 K m For our future reference we quote the following results. Theorem 6 [11]. For m 3, k 1, C 2k+1 C 2k+1 K m. Theorem 7 [11]. For m, k 1, C 2k+1 C 2k+1 K m. Lemma 8 [10]. For any odd integer t 5, C t C 5 K t. Lemma 9. C 7 C 5 K 8. Proof. Let the partite sets of the 5-partite graph C 5 K 8 be {u 1,u 2,...,u 8 }, {v 1,v 2,...,v 8 },{w 1,w 2,...,w 8 },{x 1,x 2,...,x 8 }and{y 1,y 2,...,y 8 }.Weassume that the vertices having the same subscript are the corresponding vertices of the partite sets. Now the cycle (u 1 v 3 w 6 x 2 y 7 x 1 y 8 u 1 ) under the permutation

8 530 R.S. Manikandan and P. Paulraja (u 1 u 2 u 8 )(v 1 v 2 v 8 ) (w 1 w 2 w 8 )(x 1 x 2 x 8 )(y 1 y 2 y 8 ) and its powers give us eight 7-cycles. These eight 7-cycles under the permutation (u 1 v 1 w 1 x 1 y 1 ) (u 2 v 2 w 2 x 2 y 2 ) (u 8 v 8 w 8 x 8 y 8 ) and its powers give us the required forty 7-cycles. Remark 10. Let the vertex set of the 5-partite graph C 5 K m, m 2, be {u 1,u 2,...,u m }, {v 1,v 2,...,v m }, {w 1,w 2,...,w m }, {x 1,x 2,...,x m } and {y 1,y 2,...,y m }.FromTheorems6and7, C 5 K m hasa5-cycledecompositioncontaining the m 5-cycles {(u i v i w i x i y i u i ) 1 i m}, since the edge set of C 5 K m and C 5 K m differ only by these m disjoint 5-cycles. Theorem 11. For m 3, C 7 C 5 K m if and only if m 0 or 1 (mod 7). Proof. The proof of the necessity is obvious. We prove the sufficiency in two cases. Case 1. m 1 (mod 7). Let m = 7k +1. Subcase 1.1. k 2. Let the partite sets of the 5-partite graph C 5 K m be U = {u 0 } ( ( k i=1 {ui 1,ui 2,...,ui 7 }), V = {v k ) 0} i=1 {vi 1,vi 2,...,vi 7 }, W = ( k ) ( {w 0 } i=1 {wi 1,wi 2,...,wi 7 } k ), X = {x 0 } i=1 {xi 1,xi 2,...,xi 7 } and Y = ( k ) {y 0 } i=1 {yi 1,yi 2,...,yi 7 }, where we assume that the vertices having the same subscript and superscript are the corresponding vertices of the partite sets. From the definition of the tensor product, in C 5 K m, {u 0,v 0,w 0,x 0,y 0 } and {u i j,vi j,wi j, xi j,yi j }, 1 j 7, 1 i k, are independent sets and the subgraph induced by each of the sets U V, V W, W X, X Y and Y U is isomorphic to K m,m F 0, where F 0 is the 1-factor of distance zero. WeobtainanewgraphoutofH = (C 5 K m ) {u 0,v 0,w 0,x 0,y 0 } = C 5 K 7k as follows: for each i, 1 i k, identify the subsets of vertices {u i 1,ui 2,...,ui 7 }, {v1 i,vi 2,...,vi 7 }, {wi 1,wi 2,...,wi 7 }, {xi 1,xi 2,...,xi 7 } and {yi 1,yi 2,...,yi 7 } into new vertices u i,v i,w i,x i and y i, respectively, and two new vertices are adjacent if and only if the corresponding sets of vertices in H induce a complete bipartite subgraph K 7,7 or a complete bipartite subgraph minus a 1-factor K 7,7 F, where F is a 1-factor of K 7,7. The new graph thus obtained is isomorphic to the graphc 5 K k withpartitesets{u 1,u 2,...,u k },{v 1,v 2,...,v k },{w 1,w 2,...,w k }, {x 1,x 2,...,x k } and {y 1,y 2,...,y k }. The graph C 5 K k has a C 5 -decomposition containing the 5-cycles (u i v i w i x i y i u i ), 1 i k, by Remark 10. The subgraph of H corresponding to these k 5-cycles of the graph C 5 K k consists of k vertex disjoint copies of C 5 K 7. To each of these k 5-cycles (u i v i w i x i y i u i ), 1 i k, associate the 5-partite subgraph of C 5 K m induced by {u 0,u i 1,ui 2,...,ui 7 } {v 0,v1 i,vi 2,...,vi 7 } {w 0,w1 i,wi 2,...,wi 7 } {x 0,x i 1,xi 2,...,xi 7 } {y 0,y1 i,yi 2,...,yi 7 }; as this induced subgraph is isomorphic to C 5 K 8, it can be decomposed into 7-cycles, by Lemma 9. Again, the subgraphs of C 5 K m corresponding to the

9 C 7 -Decompositions of the Tensor Product of Complete Graphs 531 other 5-cycles in the decomposition of C 5 K k are isomorphic to C 5 K 7, and they can be decomposed into 7-cycles, by Theorem A. Thus we have decomposed C 5 K m into 7-cycles when k 2. Subcase 1.2. k = 2. By Theorem B, C 7 K 15 and hence C 5 K 15 = K15 C 5 = (C7 C 5 ) (C 7 C 5 ) (C 7 C 5 ). Further, C 7 C 5 can be decomposed into 7-cycles, see Figure 4. Case 2. m 0 (mod 7). Let m = 7k. Subcase2.1. k 2.Asinthepreviouscase, letthepartitesetsofthe5-partite graph C 5 K m be U = k i=1 {ui 1,ui 2,...,ui 7 }, V = k i=1 {vi 1,vi 2,...,vi 7 }, W = k i=1 {wi 1,wi 2,...,wi 7 }, X = k i=1 {xi 1,xi 2,...,xi 7 } and Y = k i=1 {yi 1,yi 2,...,yi 7 }. We assume that the vertices having the same subscript and superscript are the corresponding vertices of the partite sets. As in the proof of Subcase 1.1, we obtain the graph C 5 K k with partite sets {u 1,u 2,...,u k }, {v 1,v 2,...,v k }, {w 1,w 2,...,w k }, {x 1,x 2,...,x k } and {y 1,y 2,...,y k }, by suitable identification of vertices of C 5 K m. By Remark 10, the graph C 5 K k has a C 5 -decomposition containing the 5-cycles (u i v i w i x i y i u i ), 1 i k. Corresponding to each of these k 5-cycles, associate the corresponding 5-partite subgraph of C 5 K m induced by {u i 1,ui 2,...,ui 7 } {vi 1,vi 2,...,vi 7 } {wi 1,wi 2,...,wi 7 } {xi 1,xi 2,...,xi 7 } {y1 i,yi 2,...,yi 7 }; as this subgraph is isomorphic to C 5 K 7, it can be decomposed into 7-cycles, by Lemma 8. Corresponding to each of the other 5-cycles of the C 5 - decomposition of C 5 K k if we associate the corresponding subgraph of C 5 K m, then we get a subgraph isomorphic to C 5 K 7, and it can be decomposed into 7-cycles, by Theorem A. Thus we have decomposed C 5 K m into 7-cycles when k 2. Subcase 2.2. k = 2. Let the partite sets of the 5-partite graph C 5 K 14 be U = {u 1,u 2,...,u 14 },V = {v 1,v 2,...,v 14 },W = {w 1,w 2,...,w 14 },X = {x 1,x 2,...,x 14 }, and Y = {y 1,y 2,...,y 14 }; we assume that the vertices having the same subscript are the corresponding vertices of the partite sets. From the definition of the tensor product, in C 5 K 14, {u i,v i,w i,x i,y i }, 1 i 14, are independent setsandthesubgraphinducedbyeachofthesetsu V,V W,W X,X Y and Y U is isomorphic to K 14,14 F 0, where F 0 is the 1-factor of distance zero. As in Subcase 1.1 above, we obtain a new graph out of C 5 K 14 as follows: for each i, 1 i 7, identify the set of vertices {u 2i 1, u 2i }, {v 2i 1, v 2i }, {w 2i 1, w 2i }, {x 2i 1, x 2i } and {y 2i 1, y 2i } into new vertices u i, v i, w i, x i and y i, respectively, and two of these vertices are adjacent if and only if the corresponding sets of vertices in C 5 K 14 induce the subgraph isomorphic to K 2,2 in C 5 K 14. The resulting graph is isomorphic to C 5 K 7 with partite sets U = {u 1,u 2,...,u 7 }, V = {v 1,v 2,...,v 7 }, W = {w 1,w 2,...,w 7 }, X = {x 1,x 2,...,x 7 } and Y = {y 1,y 2,...,y 7 }, where {u i,v i,w i,x i,y i }, 1 i 7, are independent sets

10 532 R.S. Manikandan and P. Paulraja of C 5 K 7. Clearly, C 5 K 7 = (C 5 C 7 ) (C 5 C 7 ) (C 5 C 7 ). The graph C 5 C 7 can be decomposed into 7-cycles, see Figure 4. u 1 u 2 u 3 u 4 u 5 u 6 u 7 v 1 v 2 v 3 v 4 v 5 v 6 v 7 w 1 w 2 w 3 w 4 w 5 w 6 w 7 x 1 x 2 x 3 x 4 x 5 x 6 x 7 y 1 y 2 y 3 y 4 y 5 y 6 y 7 C 5 C 7 Figure 4. A 7-cycle decomposition of C 5 C 7. u 1 u 2 u 3 u 4 u 5 u 6 u 7 u 8 u 9 u 10 u 11 u 12 u 13 u 14 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 v 11 v 12 v 13 v 14 w 1 w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w 10 w 11 w 12 w 13 w 14 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 y 10 y 11 y 12 y 13 y 14 Figure 5 By lifting back each of these 7-cycles in a C 7 -decomposition of C 5 C 7 to C 5 K 14, the corresponding subgraph is isomorphic to C 7 K 2 and this graph can be decomposed into cycles of length 7, see [12]. Thus the subgraph of C 5 K 14 obtained by the lifting of the 7-cycles of C 5 K 7 can be decomposed into cycles

11 C 7 -Decompositions of the Tensor Product of Complete Graphs 533 of length 7. The edges of C 5 K 14 which are not covered by these 7-cycles are shown in Figure 5. To complete the proof we fuse with the graph of Figure 5 some of the 7-cycles obtained above and decompose the resulting graph, say, H, into 7-cycles. Let H be the graph obtained by the union of the graph of Figure 5 and the subgraph of C 5 K 14 whichcorrespondstothe7-cycleofc 5 K 7,namely,(u 1 v 2 w 3 x 4 y 5 u 6 v 7 u 1 ) shown by solid line in Figure 4. The graph H is shown in Figure 6. A 7-cycle decomposition of H is given below: (u 1 v 2 w 1 x 2 y 1 u 2 v 4 u 1 ), (u 1 y 2 x 1 w 2 v 1 u 2 v 3 u 1 ), (u 3 y 4 x 3 w 4 v 3 w 6 v 4 u 3 ), (u 4 y 3 x 4 w 3 v 4 w 5 v 3 u 4 ), (u 6 y 5 x 6 w 5 x 8 w 6 v 5 u 6 ), (u 5 y 6 x 5 w 6 x 7 w 5 v 6 u 5 ), (u 7 y 8 x 7 y 10 x 8 w 7 v 8 u 7 ), (u 8 y 7 x 8 y 9 x 7 w 8 v 7 u 8 ), (u 9 y 10 u 12 y 9 x 10 w 9 v 10 u 9 ), (u 10 y 9 u 11 y 10 x 9 w 10 v 9 u 10 ),(u 11 v 12 w 11 x 12 y 11 u 12 v 14 u 11 ),(u 11 y 12 x 11 w 12 v 11 u 12 v 13 u 11 ), (u 13 y 14 x 13 w 14 v 13 u 2 v 14 u 13 ) and (u 14 y 13 x 14 w 13 v 14 u 1 v 13 u 14 ). u 1 u 2 u 3 u 4 u 5 u 6 u 7 u 8 u 9 u 10 u 11 u 12 u 13 u 14 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 v 11 v 12 v 13 v 14 w 1 w 2 w 3 w 4 w 5 w 6 w 7 w 8 w 9 w 10 w 11 w 12 w 13 w 14 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 y 10 y 11 y 12 y 13 y 14 Figure 6 4. Proof of the Main Theorem Proof of Theorem 1. The proof of the neccessity is obvious and we prove the sufficiency in two cases. Since the tensor product is commutative, we may assume that m is odd and so m 1,3 or 5 (mod 6). Case 1. n 0 or 1 (mod 7). Subcase 1.1. m 1 or 3 (mod 6). By Theorem C, C 3 K m and hence K m K n = (C 3 K n ) (C 3 K n ) (C 3 K n ). As C 7 C 3 K n, by Theorem 5, C 7 K m K n.

12 534 R.S. Manikandan and P. Paulraja Subcase 1.2. m 5 (mod 6). By Theorem C, K m = C 3 C 3 C }{{} 3 (C 5 C 5 ). (m(m 1) 20)/6 times Now K m K n = ((C 3 K n ) (C 3 K n ) (C 3 K n )) ((C 5 K n ) (C 5 K n )). As C 7 C 3 K n, by Theorem 5, and C 7 C 5 K n, by Theorem 11, C 7 K m K n. Case 2. n 0 or 1 (mod 7). As n(n 1) 0 (mod 7), condition (1) implies that m 0 or m 1 (mod 7). As m is odd we have m 1 or m 7(mod 14). As C 7 K m, by Theorem B, K m K n = (C 7 K n ) (C 7 K n ) (C 7 K n ). C 7 C 7 K n, by Theorem 6, and so C 7 K m K n. This completes the proof. References [1] B. Alspach and H. Gavlas, Cycle decompositions of K n and K n I, J. Combin. Theory Ser. B 81 (2001) doi: /jctb [2] B. Alspach, P.J. Schellenberg, D.R. Stinson and D. Wagner, The Oberwolfach problem and factors of uniform odd length cycles, J. Combin. Theory Ser. A 52 (1989) doi: / (89) [3] E.J. Billington, Decomposing complete tripartite graphs into cycles of length 3 and 4, Discrete Math. 197/198 (1999) doi: /s x(98) [4] E.J. Billington, D.G. Hoffman and B.M. Maenhaut, Group divisible pentagon systems, Util. Math. 55 (1999) [5] N.J. Cavenagh and E.J. Billington, Decompositions of complete multipartite graphs into cycles of even length, Graphs Combin. 16 (2000) doi: /s [6] N.J. Cavenagh and E.J. Billington, On decomposing complete tripartite graphs into 5-cycles, Australas. J. Combin. 22 (2000) [7] C.C. Lindner and C.A. Rodger, Design Theory (CRC Press New York, 1997). [8] J. Liu, A Generalization of the Oberwolfach problem and C t -factorizations of complete equipartite graphs, J. Combin. Designs 8 (2000) doi: /(sici) (2000)8:1 42::AID-JCD6 3.0.CO;2-R [9] E.S. Mahmoodian and M. Mirzakhani, Decomposition of complete tripartite graphs into 5-cycles, in: Combinatorics and Advances, C.J. Colbourn and E.S. Mahmoodian (Ed(s)), (Kluwer Academic Publ., 1995) [10] R.S. Manikandan and P. Paulraja, C p -decompositions of some regular graphs, Discrete Math. 306 (2006) doi: /j.disc

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