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Hamiltonian Decompositions of Cayley Graphs on Abelian Groups of Odd Order
Journal of Combinatorial Theory, Series B 1645 journal of combinatorial theory, Series B 66, 7586 (1996) article no. 0007
Hamiltonian Decompositions of Cayley Graphs on Abelian Groups of Odd Order Jiuqiang Liu Department of Mathematics, Eastern Michigan University, Ypsilanti, Michigan 48197 Received July 28, 1992
Alspach has conjectured that any 2k-regular connected Cayley graph cay(A, S) on a finite abelian group A can be decomposed into k hamiltonian cycles. In this paper, the conjecture is shown to be true if S=[s 1 , s 2 , ..., s k ] is a minimal generating set of an abelian group A of odd order (where a generating set S of a group G is minimal if no proper subset of S can generate G ). 1996 Academic Press, Inc.
1. Introduction Let (A, +) be a finite group and S be a subset of A with 0 not in S. The Cayley graph cay(A, S) is defined to be the graph G with V(G)=A and E(G )=[xy | x, y # A, x&y # S or y&x # S]. We say an edge xy in cay(A, S ) is generated by s # S if x&y=s or y&x=s and the subgraph Q of cay(A, S) is generated by s if E(Q) consists of all the edges generated by s.
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Remark 1. From the definition, it is clear that any element of S with order 2 generates a 1-factor of cay(A, S ) while any element of S with order at least 3 generates a 2-factor of cay(A, S ). Furthermore, cay(A, S ) is connected if and only if S generates A. It is known that any connected Cayley graph on a finite abelian group is hamiltonian [9]. In [1], Alspach conjectured that any 2k-regular connected Cayley graph on a finite abelian group has a hamiltonian decomposition. Bermond, Favaron, and Maheo [4] proved that every 4-regular connected Cayley graph cay(A, S) on a finite abelian group A can be decomposed into two hamiltonian cycles. Liu [8] showed that the conjecture is true provided that either cay(A, S ) is 2m-regular and S=[s 1 , s 2 , ..., s k ] is a generating set of A such that gcd(ord(s i ), ord(s j ))=1 for i{j or S= [s 1 , s 2 , s 3 ] is a minimal generating set of A with |A| being odd. Here our main purpose is to prove the following theorem which establishes the conjecture for a more general case. 75 0095-895696 12.00 Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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Theorem 1. If A is an abelian group of odd order and S=[s 1 , s 2 , ..., s k ] is a minimal generating set of A, then cay(A, S) has a hamiltonian decomposition. By convention, terminology and notation not mentioned here are referred to as in [5] and [7].
2. Preliminary Results First, we recall a definition and some results. Definition 1. The cartesian product G=G 1_G 2 has vertex set V(G )=V(G 1 )_V(G 2 ) and edge set E(G )=[(u 1 , u 2 )(v 1 , v 2 ) | u 1 =v 1 and u 2 v 2 # E(G 2 ) or u 2 =v 2 and u 1 v 1 # E(G 1 )]. Theorem A (Alspach et al. [2]). If C i is a cycle for each i=1, 2, ..., n, then the cartesian product G=C 1_C 2_ } } } _C n can be decomposed into n hamiltonian cycles. Theorem B (Bermond et al. [4]). Every 4-regular connected Cayley graph cay(A, S ) on a finite abelian group A can be decomposed into two hamiltonian cycles. Theorem C (Liu [8]). Let A be a finite abelian group of odd order and S=[s 1 , s 2 , s 3 ] be a minimal generating set of A. Then, cay(A, S ) has a hamiltonian decomposition. The following proposition is Lemma 2.5 in [8]. Proposition 1. Let A be a finite abelian group which is generated by S=[s 1 , s 2 , ..., s k ], A 1 be the subgroup of A which is generated by S$=[s 1 , s 2 , ..., s k&1 ] and J=( s k ). If A 1 & J=[0], then cay(A, S )= cay(A 1 , S$)_cay(J, [s k ]). From now on, we let C 1 =a 1 a 2 } } } a n a 1 and C 2 =b 1 b 2 } } } b m b 1 be two cycles. By convention, the subscripts of a are expressed modulo n and the subscripts of b are expressed modulo m.
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Definition 2. For 0 r m&1, the r-pseudo-cartesian product of C 1 and C 2 , denoted by C 1_r C 2 , is the graph which is obtained from C 1_C 2 by replacing the edge set [(a 1 , b i )(a n , b i ) | 1 i m] by the edge set [(a 1 , b i+r )(a n , b i ) | 1 i m].
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From the definition, it is easy to see that C 1_0 C 2 =C 1_C 2 =C 1_m C 2 . For convenience, we call the vertex set [(a i , b j ) | 1 i n] the b j -row and the vertex set [(a i , b j ) | 1 j m] the a i -column. Also, we call the edges whose two end-vertices have the same first component vertical edges and the edges with different first components horizontal-type edges in an r-pseudo-cartesian product. Remark 2. If gcd(r, m)=t in an r-pseudo-cartesian product C 1_r C 2 , then the horizontal-type edges form a 2-factor H which consists of t cycles of length (mn)t and any consecutive t rows of C 1_r C 2 are on t different cycles of H. Before proceeding further we give two simple facts. Fact 1. If u 1 u 2 # E(Q 1 ) and v 1 v 2 # E(Q 2 ), where Q 1 and Q 2 are two vertex-disjoint cycles, then (Q 1 _ Q 2 &[u 1 u 2 , v 1 v 2 ]) _ [u 1 v 1 , u 2 v 2 ] is a cycle. Fact 2. Given a cycle C, let u 1 u 2 and v 1 v 2 be edges of C which are separated by at least two edges. Then (C&[u 1 u 2 , v 1 v 2 ]) _ [u 1 v 1 , u 2 v 2 ] is a 2-factor containing at most two cycles. For the following discussions, we color all horizontal-type edges in C 1_r C 2 by one color, say blue, and all vertical edges by another color, say red. Definition 3. A color switching in C 1_r C 2 means that for some 1 i n&1 and some 1 s m we interchange the colors between the edge sets [(a i , b s )(a i , b s+1 ), (a i+1 , b s )(a i+1 , b s+1 )] and [(a i , b s )(a i+1 , b s ), (a i , b s+1 )(a i+1 , b s+1 )], and we call it the [a i , a i+1 , b s , b s+1 ]-color switching. The next proposition is Lemma 3.13 in [8] which follows easily from Remark 2 and Fact 1. Proposition 2. If gcd(r, m)=2t+1 3, then, by making the color switchings 1, 2, ..., 2t in C 1_r C 2 shown in Fig. 1, we obtain a blue hamiltonian cycle and connect three red cycles in the a j -columns for j=1, 2, 3 to a single red cycle. Also, by Remark 2 and Fact 1, the following lemma can be derived easily.
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Lemma 1. Suppose n 5 and gcd(r, m)=2t+1 3. Then, by making the color switchings in C 1_r C 2 shown in Fig. 2, we obtain a blue hamiltonian cycle and connect the red cycles in the a j -columns for 1 j y to a single red cycle, where y=3 if 2t+1=3, and y=5 otherwise.
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Fig. 1. Color switchings in C 1_r C 2 , I, where 0 x m&1.
Lemma 2. If A is an abelian group of odd order and S=[s 1 , s 2 , ..., s k ] is a minimal generating set of A, then |A| 3 k. Proof. We proceed by induction on k. For k=1, the result is clear. Assume the result for k
Fig. 2. Color switchings in C 1_r C 2 , II, where 0 x m&1.
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Lemma 3. If G is a graph of order n 26 which can be decomposed into three hamiltonian cycles, then, for any hamiltonian decomposition H 1 , H 2 , H 3 of G, there are three paths P 1 =u 1 u 2 u 3 , P 2 =v 1 v 2 v 3 v 4 v 5 , and P 3 =w 1 w 2 w 3 w 4 w 5 such that P i is on H i for 1 i 3, where either P 1 and P 2 have at most one common vertex u 3 =v 1 and V(P 3 ) & [V(P 1 ) _ V(P 2 )]=< or P 1 and P 3 have at most one common vertex u 3 =w 1 and V(P 2 ) & [V(P 1 ) _ V(P 3 )]=<. Proof. Suppose P 3 =w 1 w 2 w 3 w 4 w 5 is on H 3 . Then the vertices w j for 1 j 5 divide H 2 into five subpaths. Since n>25, at least one of such subpath has length at least 6, so we obtain a path P 2 =v 1 v 2 v 3 v 4 v 5 on H 2 such that P 2 and P 3 are vertex-disjoint. Now, the vertices v i for 1 i 5 and w j for 1 j 5 divide H 1 into ten subpaths. If there is one such subpath of length at least 4, then we obtain a path P 1 =u 1 u 2 u 3 on H 1 which is vertex-disjoint with P 2 and P 3 . Thus we assume that all subpaths have length at most 3. Since n 26, there are at least six of the subpaths of length 3. The four vertices v 1 , v 5 , w 1 , w 5 are the ends of at least five subpaths on H 1 so that one of them is an end of a subpath of length 3. By reordering the v-path or the w-path if necessary, we conclude that there is a desired subpath P 1 =u 1 u 2 u 3 on H 1 with either v 1 =u 3 or w 1 =u 3 . K Similar to the proof of Lemma 3, we can derive the following result. Lemma 4. If G is a 2k-regular graph of order n>5(5k&7), where k 2, and G has a hamiltonian decomposition, then for any hamiltonian decomposition H 1 , H 2 , ..., H k of G, there exist k vertex-disjoint paths P 1 =u 1 v 1 w 1 and P j =u j v j w j x j y j for 2 j k such that P j is on H j for 1 j k. Proof. If 05(5k&t&7) vertices of H k&t into 5t segments. The longest one, say P, has at least five vertices since 4(5t)
3. Proof of Theorem 1
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Throughout this section we let C 1 =a 1 a 2 } } } a n a 1 and C 2 =b 1 b 2 } } } b m b 1 be two cycles of order n and m, respectively. The subscripts of a are expressed modulo n and the subscripts of b are expressed modulo m. By convention, we give a natural direction to a drawing of C 1_r C 2 on a torus as in Fig. 1 so that the a i -column is to the left of the a i+1-column for 1 i n&1.
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Let A be a finite abelian group, S=[s 1 , s 2 , ..., s k ] be a minimal generating set of A and m=ord(s k ) 3. Let F be the 2-factor of cay(A, S ) which is generated by s k . Then F consists of n= |A|m cycles of length m. Let J=( s k ) and S =[s 1 , s 2 , ..., s k&1 ] A 1 =AJ, where we use x to represent the coset x+J. Definition 4. For any edge xy of cay(A 1 , S ), where x &y =s i # S, we call the edge set [u 1 u 2 | u 1 =x, u 2 =y and u 1 &u 2 =s i ] of cay(A, S) the lifting edge set of the edge xy. Definition 5. For any subgraph Q of cay(A 1 , S ), the spanning subgraph Q of cay(A, S ) with the edge set being the union of the lifting edge sets of the edges of Q is called the subgraph lifted by Q and we say Q lifts to Q. It is easy to see from the above definitions that edge-disjoint subgraphs of cay(A 1 , S ) lift to edge-disjoint subgraphs of cay(A, S ). The following proposition is Lemma 3.7 in [8]. Proposition 3. Any hamiltonian cycle H =g 1 g 2 } } } g n g 1 of cay(A 1 , S ) lifts to a 2-factor H of cay(A, S ). Furthermore, under the map 8: a i +b j (a i , b j ), the union of F and H is isomorphic to an r-pseudo-cartesian product C 1_r C 2 in which the edges of H are horizontal-type and the edges of F are vertical, where 0 r m&1, a i =g i for 1 i n, and b j =( j&1) s k for 1 j m. Next we introduce a special class of graphs which plays a central role in our discussion. Definition 6. For k 2, define D k to be a 2k-regular graph satisfying: (1)
V(D k )=[(a i , b j ) | 1 i n and 1 j m],
(2)
E(D k ) can be decomposed into 2-factors H 1 , H 2 , ..., H k&1 and F,
(3) (a i , b 1 ),
F= ni=1 F i , where each F i is the cycle (a i , b 1 )(a i , b 2 ) } } } (a i , b m )
(4)
H 1 _ F=C 1_r1 C 2 , and
(5) for 2 j k&1, H j _ F=C 1j _rj C 2 , where 0 r j m&1 and C =a (?j)j (1) a (?j)j (2) } } } a (?j)j (n) a (?j)j (1) with ? j being a permutation of [1, 2, ..., n] and (a (i j) , b t )=(a i , b t+hi, j ) for 1 i n. j 1
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The example shown in Fig. 3 is a graph D 3 with n=6, m=5, r 1 =1, r 2 =2, ? 2 =(124653), and h i, 2 #3i&1 (mod 5) for 1 i 6.
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Fig. 3. A graph D 3 .
Clearly, from the definition it is easy to see that in each H j _ F= C 1j _rj C 2 the edges of F are vertical while the edges of H j are horizontaltype (in the sense that they are not vertical), where C 11 =C 1 . Thus, by Remark 2, H j is the union of t j =gcd(r j , m) cycles of equal length (mn)t j . For example, both H 1 and H 2 of the graph D 3 in Fig. 3 are hamiltonian cycles. From Proposition 3 and the definition for D k the following lemma can be derived easily. Lemma 5. If cay(A 1 , S ) can be decomposed into k&1 hamiltonian cycles H j =g ?j (1) g ?j (2) } } } g ?j (n) g ?j (1) for 1 j k&1, where ? 1 =I (the identity) and ? j is a permutation of [1, 2, ..., n] for 2 j k&1, then cay(A, S )$D k with each H j being the 2-factor lifted by H j and F being the 2-factor generated by s k . Proof. By Proposition 3, for each 1 j k&1, H j lifts to a 2-factor H j and the union H j _ F is isomorphic to C 1j _rj C 2 with C 1j =a (?j)j (1) a (?j)j (2) } } } a (?j)j (n) a (?j)j (1) under the map 8 j : a (i j) +b t (a (i j) , b t ), where a (i j) =g i and b t =(t&1) s k . For convenience, we denote a (1) i =a i for 1 i n, and so C 11 =C 1 =a 1 a 2 } } } a n a 1 and H 1 _ F=C 1_r1 C 2 . Since a (i j) =a i for 2 j k&1 and 1 i n, we let a (i j) =a i +h i, j s k . Then a (i j) +b t =a (i j) + (t & 1) s k = a i + (t + h i, j & 1) s k = a i + b t+hi, j and we have (a (i j) , b t ) =
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(a i , b t+hi, j ). Recall that edge-disjoint subgraphs of cay(A 1 , S ) lift to edgedisjoint subgraphs of cay(A, S). We conclude that cay(A, S)$D k . K By Lemma 5, we see that an important step to prove Theorem 1 by induction on k is to prove that D k can be decomposed into k hamiltonian cycles. Thus, our next major effort is to show that with certain restrictions D k has a hamiltonian decomposition. For the following discussions, we color the edges of D k so that all edges of F are of red color and for 1 j k&1, all edges of H j are of color c j . In order to decompose D k into hamiltonian cycles, our main idea here is to find some edge-disjoint color switchings so that making those color switchings results in k monochromatic hamiltonian cycles. In each of the following lemmas, we will find the necessary color switchings in two steps: First, we make some color switchings in each H j _ F=C 1j _rj C 2 to obtain a hamiltonian cycle of color c j for 1 j k&1; then, we concentrate on H 1 _ F=C 1_r1 C 2 and make some additional color switchings between red edges and edges of color c 1 to obtain a red hamiltonian cycle meanwhile the edges of color c 1 still form a hamiltonian cycle. Also, for the following discussions it is very helpful to visualize H j _ F= C 1j _rj C 2 in a way similar to (form B) of Fig. 3. Lemma 6. Let m>0 and n>0 be odd integers, and let k 2. Assume each H j in D k consists of t j cycles with t 1 t 2 } } } t k&1 . If 2 t 1
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Proof. Recall that in each H j _ F=C 1j _rj C 2 , the vertical edges are red, the horizontal-type edges are of color c j , and each row (in the sense that we visualize H j _ F as in (form B) of Fig. 3) is in the same cycle of H j . By Remark 2, each H j consists of t j =gcd(r j , m)=2t$j +1 cycles. If t j =1 for some j, then H j is already a hamiltonian cycle, so we can disregard it and work on the remaining graph obtained by removing the edges of H j from D k . Thus, we may assume that t j =2t$j +1 3 for 1 j k&1. Since t1 | m, t 1
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that for each 2 j k&1, by choosing x properly, we can apply Proposition 2 to H j _ F=C 1j _rj C 2 without using any vertical edges from the paths P i =(a i , b 1 )(a i , b 2 )(a i , b 3 ) for 4 i n in H 1 _ F=C 1_r1 C 2 to obtain a hamiltonian cycle of color c j and connect the red cycles in the a ?j (i)-columns for 1 i 3 to a single red cycle. Clearly, we now have n&2(k&1) red cycles and each column is in a single red cycle. Since n is odd, n&2(k&1)=2n 1 +1 for some n 1 >0. Now, we will find some additional edge-disjoint color switchings between red edges and edges of color c 1 so that we can connect all red cycles to a red hamiltonian cycle while we still have a hamiltonian cycle of color c 1 . For that purpose, we focus on H 1 _ F=C 1_r1 C 2 . Let 0
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Lemma 7. Let m=2h+1 5, n be odd, and k 2. Assume each H j in D k consists of t j cycles with t 1 t 2 } } } t k&1 . If t 1 =m and the sets K 1 =[1, 2, 3] and K j =[? j (i) | 1 i 5] for 2 j k&1 are mutually disjoint except for K 1 and K 2 having at most one common element ? 2(1)=3, then D k has a hamiltonian decomposition.
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Proof. We proceed in a way similar to the proof of Lemma 6. Again, we may assume that each H j consists of t j =gcd(r j , m)=2t$j +1 3 cycles for 1 j k&1. Since t 1 =m, we must have r 1 =0 or r 1 =m which implies that H 1 _ F=C 1_r1 C 2 =C 1_C 2 . We first apply Proposition 2 to H 1 _ F= C 1_C 2 with x=0 to obtain a hamiltonian cycle of color c 1 and connect the red cycles in the a i-columns for 1 i 3 to a single red cycle Q 1 . In the a 1-column, note that the edges of color c 1 form a matching consisting of all odd vertical edges, i.e., M=[(a 1 , b 2i&1 )(a 1 , b 2i ) | 1 i h], and all odd edges are red in the a 3-column. For each 2 j k&1, we call the a lj -column leftmost, where l j =min K j . Since K j & (K 1 _ K 2 )=< for 3 j k&1 and the sets K 3 , K 4 , ..., K k&1 are mutually disjoint, for each 2 j k&1 we apply Lemma 1 to H j _ F=C 1j _rj C 2 , with x being chosen properly, so that for j=2 and ? 2(1)=3 we use only odd edges in the a 3-column and for j 3 or ? 2(1)>3 we can locate two unused adjacent red edges anywhere as we wish in the leftmost a lj -column, to obtain a hamiltonian cycle of color c j and connect the involved columns to a single red cycle Q j . Clearly, for ? 2(1)=3, Q 1 is connected to Q 2 via the a 3-column. Note that each column is in a single red cycle and each red cycle consists of 3 columns, 5 columns, or 7 columns. We now have 2n 1 +1 red cycles for some n 1 >0 as n is odd. Since m 5, each a i -column for 3 i n has at least one red odd edge and each a i -column for 4 i n has property P: for any 1 f m one of the vertical edges (a i , b f )(a i , b f+1 ) and (a i , b f+2 )(a i , b f+3 ) is red (see Fig. 2). Similar to the proof of Lemma 6, we will make some additional edge-disjoint color switchings between red edges and edges of color c 1 so that we obtain a red hamiltonian cycle and a hamiltonian cycle of color c 1 without destroying the existing hamiltonian cycles of color c j for 2 j k&1. For that purpose, we focus on H 1 _ F= C 1_C 2 . Let 0
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vertical edges e 1 =(a z2i , b yi &1 )(a z2i , b yi ) and e 2 =(a z2i , b yi +1 )(a z2i , b yi +2 ) is red, say e 1 . Recall that both the a z2i&1 +1-column (it might happen that z 2i&1 +1=z 2i ) and the a z2i +1-column are either free or leftmost, we define X 2i&1 to be the [a z2i&1 , a z2i&1 +1 , b yi , b yi +1 ]-color switching and X 2i to be the [a z2i , a z2i +1 , b yi &1 , b yi ]-color switching. Since X 2i&1 uses only odd vertical edges while X 2j does not use odd vertical edges, we conclude that X 1 , X 2 , ..., X 2n1 are edge-disjoint. Now, from the choice of the z j 's it follows that by making the color switchings X 1 , X 2 , ..., X 2n1 , we obtain a red hamiltonian cycle. To see that the edges of color c 1 still form a hamiltonian cycle, we make the color switchings in the order X 1 , X 2 , ..., X 2n1 with one pair X 2i&1 and X 2i at each time. It follows from Fact 2 that, by making the color switching X 2i&1 , the existing hamiltonian cycle of color c 1 is separated into two cycles with one being (a z2i&1 +1 , b yi )(a z2i&1 +2 , b yi ) } } } (a n , b yi )(a 1 , b yi )(a 1 , b yi +1 ) (a n , b yi +1 )(a n&1 , b yi +1 ) } } } (a z2i&1 +1 , b yi +1 )(a z2i&1 +1 , b yi ). Then it follows from Fact 1 that by making the color switching X 2i those two cycles are connected to a hamiltonian cycle of color c 1 again. Thus, together with the existing monochromatic hamiltonian cycles of color c j for 2 j k&1, we obtain a hamiltonian decomposition of D k . K We are ready to prove our main result.
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Proof of Theorem 1. We proceed by induction on k. For k=1, the result is trivial. For k=2 or 3, the result follows from Theorems B and C. Assume the result for k
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H j cay(A, S$). By Lemma 2, |A 1 | 3 k&1. Thus, |A 1 | 27 for k=4, and |A 1 | >5[5(k&1)&7] for k 5. By applying Lemmas 3 and 4 to cay(A 1 , S ), we may assume that all ? i , for 1 i k&1, satisfy the conditions in Lemmas 6 and 7. Now, it follows from Lemmas 6 and 7 that cay(A, S )$D k has a hamiltonian decomposition. K I believe that the techniques used here for abelian groups of odd order are useful in obtaining a similar result for abelian groups of even order.
Acknowledgments The author is grateful to the referees whose many comments have significantly helped to improve the presentation of this article. In particular, the author thanks the referees for providing a proof for Lemma 4.
References
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1. B. Alspach, Research Problem 59, Discrete Math. 50 (1984), 115. 2. B. Alspach, J.-C. Bermond, and D. Sotteau, Decomposition into cycles. I. Hamilton decompositions, in ``Cycles and Rays'' (G. Hahn, G. Sabidussi, and R. Woodrow, Eds.), pp. 918, Kluwer, Amsterdam, 1990. 3. J. Aubert and B. Schneider, Decomposition de la somme cartesienne d'un cycle et de l'union de deux cycles hamiltonens en cycles hamiltoniens, Discrete Math. 38 (1982), 716. 4. J.-C. Bermond, O. Favaron, and M. Maheo, Hamiltonian decomposition of Cayley graphs of degree 4, J. Combin. Theory Ser. B 46 (1989), 142153. 5. G. Chartrand and L. Lesniak, ``Graphs and Digraphs,'' 2nd ed., Wadsworth and BrooksCole, Monterey, CA, 1986. 6. M. F. Foregger, Hamiltonian decompositions of products of cycles, Discrete Math. 24 (1978), 251260. 7. J. A. Gallian, ``Contemporary Abstract Algebra,'' Heath, Lexington, MA, 1986. 8. J. Liu, Hamiltonian decompositions of Cayley graphs on abelian groups, Discrete Math. 131 (1994), 163171. 9. D. Marusic, Hamiltonian circuits in Cayley graphs, Discrete Math. 46 (1983), 4954. 10. R. Stong, Hamilton decompositions of cartesian products of graphs, Discrete Math. 90 (1991), 169190.
Hamiltonian Decompositions of Cayley Graphs on Abelian Groups of Odd Order Jiuqiang Liu Department of Mathematics, Eastern Michigan University, Ypsilanti, Michigan 48197 Received July 28, 1992
Alspach has conjectured that any 2k-regular connected Cayley graph cay(A, S) on a finite abelian group A can be decomposed into k hamiltonian cycles. In this paper, the conjecture is shown to be true if S=[s 1 , s 2 , ..., s k ] is a minimal generating set of an abelian group A of odd order (where a generating set S of a group G is minimal if no proper subset of S can generate G ). 1996 Academic Press, Inc.
1. Introduction Let (A, +) be a finite group and S be a subset of A with 0 not in S. The Cayley graph cay(A, S) is defined to be the graph G with V(G)=A and E(G )=[xy | x, y # A, x&y # S or y&x # S]. We say an edge xy in cay(A, S ) is generated by s # S if x&y=s or y&x=s and the subgraph Q of cay(A, S) is generated by s if E(Q) consists of all the edges generated by s.
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Remark 1. From the definition, it is clear that any element of S with order 2 generates a 1-factor of cay(A, S ) while any element of S with order at least 3 generates a 2-factor of cay(A, S ). Furthermore, cay(A, S ) is connected if and only if S generates A. It is known that any connected Cayley graph on a finite abelian group is hamiltonian [9]. In [1], Alspach conjectured that any 2k-regular connected Cayley graph on a finite abelian group has a hamiltonian decomposition. Bermond, Favaron, and Maheo [4] proved that every 4-regular connected Cayley graph cay(A, S) on a finite abelian group A can be decomposed into two hamiltonian cycles. Liu [8] showed that the conjecture is true provided that either cay(A, S ) is 2m-regular and S=[s 1 , s 2 , ..., s k ] is a generating set of A such that gcd(ord(s i ), ord(s j ))=1 for i{j or S= [s 1 , s 2 , s 3 ] is a minimal generating set of A with |A| being odd. Here our main purpose is to prove the following theorem which establishes the conjecture for a more general case. 75 0095-895696 12.00 Copyright 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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Theorem 1. If A is an abelian group of odd order and S=[s 1 , s 2 , ..., s k ] is a minimal generating set of A, then cay(A, S) has a hamiltonian decomposition. By convention, terminology and notation not mentioned here are referred to as in [5] and [7].
2. Preliminary Results First, we recall a definition and some results. Definition 1. The cartesian product G=G 1_G 2 has vertex set V(G )=V(G 1 )_V(G 2 ) and edge set E(G )=[(u 1 , u 2 )(v 1 , v 2 ) | u 1 =v 1 and u 2 v 2 # E(G 2 ) or u 2 =v 2 and u 1 v 1 # E(G 1 )]. Theorem A (Alspach et al. [2]). If C i is a cycle for each i=1, 2, ..., n, then the cartesian product G=C 1_C 2_ } } } _C n can be decomposed into n hamiltonian cycles. Theorem B (Bermond et al. [4]). Every 4-regular connected Cayley graph cay(A, S ) on a finite abelian group A can be decomposed into two hamiltonian cycles. Theorem C (Liu [8]). Let A be a finite abelian group of odd order and S=[s 1 , s 2 , s 3 ] be a minimal generating set of A. Then, cay(A, S ) has a hamiltonian decomposition. The following proposition is Lemma 2.5 in [8]. Proposition 1. Let A be a finite abelian group which is generated by S=[s 1 , s 2 , ..., s k ], A 1 be the subgroup of A which is generated by S$=[s 1 , s 2 , ..., s k&1 ] and J=( s k ). If A 1 & J=[0], then cay(A, S )= cay(A 1 , S$)_cay(J, [s k ]). From now on, we let C 1 =a 1 a 2 } } } a n a 1 and C 2 =b 1 b 2 } } } b m b 1 be two cycles. By convention, the subscripts of a are expressed modulo n and the subscripts of b are expressed modulo m.
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Definition 2. For 0 r m&1, the r-pseudo-cartesian product of C 1 and C 2 , denoted by C 1_r C 2 , is the graph which is obtained from C 1_C 2 by replacing the edge set [(a 1 , b i )(a n , b i ) | 1 i m] by the edge set [(a 1 , b i+r )(a n , b i ) | 1 i m].
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From the definition, it is easy to see that C 1_0 C 2 =C 1_C 2 =C 1_m C 2 . For convenience, we call the vertex set [(a i , b j ) | 1 i n] the b j -row and the vertex set [(a i , b j ) | 1 j m] the a i -column. Also, we call the edges whose two end-vertices have the same first component vertical edges and the edges with different first components horizontal-type edges in an r-pseudo-cartesian product. Remark 2. If gcd(r, m)=t in an r-pseudo-cartesian product C 1_r C 2 , then the horizontal-type edges form a 2-factor H which consists of t cycles of length (mn)t and any consecutive t rows of C 1_r C 2 are on t different cycles of H. Before proceeding further we give two simple facts. Fact 1. If u 1 u 2 # E(Q 1 ) and v 1 v 2 # E(Q 2 ), where Q 1 and Q 2 are two vertex-disjoint cycles, then (Q 1 _ Q 2 &[u 1 u 2 , v 1 v 2 ]) _ [u 1 v 1 , u 2 v 2 ] is a cycle. Fact 2. Given a cycle C, let u 1 u 2 and v 1 v 2 be edges of C which are separated by at least two edges. Then (C&[u 1 u 2 , v 1 v 2 ]) _ [u 1 v 1 , u 2 v 2 ] is a 2-factor containing at most two cycles. For the following discussions, we color all horizontal-type edges in C 1_r C 2 by one color, say blue, and all vertical edges by another color, say red. Definition 3. A color switching in C 1_r C 2 means that for some 1 i n&1 and some 1 s m we interchange the colors between the edge sets [(a i , b s )(a i , b s+1 ), (a i+1 , b s )(a i+1 , b s+1 )] and [(a i , b s )(a i+1 , b s ), (a i , b s+1 )(a i+1 , b s+1 )], and we call it the [a i , a i+1 , b s , b s+1 ]-color switching. The next proposition is Lemma 3.13 in [8] which follows easily from Remark 2 and Fact 1. Proposition 2. If gcd(r, m)=2t+1 3, then, by making the color switchings 1, 2, ..., 2t in C 1_r C 2 shown in Fig. 1, we obtain a blue hamiltonian cycle and connect three red cycles in the a j -columns for j=1, 2, 3 to a single red cycle. Also, by Remark 2 and Fact 1, the following lemma can be derived easily.
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Lemma 1. Suppose n 5 and gcd(r, m)=2t+1 3. Then, by making the color switchings in C 1_r C 2 shown in Fig. 2, we obtain a blue hamiltonian cycle and connect the red cycles in the a j -columns for 1 j y to a single red cycle, where y=3 if 2t+1=3, and y=5 otherwise.
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Fig. 1. Color switchings in C 1_r C 2 , I, where 0 x m&1.
Lemma 2. If A is an abelian group of odd order and S=[s 1 , s 2 , ..., s k ] is a minimal generating set of A, then |A| 3 k. Proof. We proceed by induction on k. For k=1, the result is clear. Assume the result for k
Fig. 2. Color switchings in C 1_r C 2 , II, where 0 x m&1.
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Lemma 3. If G is a graph of order n 26 which can be decomposed into three hamiltonian cycles, then, for any hamiltonian decomposition H 1 , H 2 , H 3 of G, there are three paths P 1 =u 1 u 2 u 3 , P 2 =v 1 v 2 v 3 v 4 v 5 , and P 3 =w 1 w 2 w 3 w 4 w 5 such that P i is on H i for 1 i 3, where either P 1 and P 2 have at most one common vertex u 3 =v 1 and V(P 3 ) & [V(P 1 ) _ V(P 2 )]=< or P 1 and P 3 have at most one common vertex u 3 =w 1 and V(P 2 ) & [V(P 1 ) _ V(P 3 )]=<. Proof. Suppose P 3 =w 1 w 2 w 3 w 4 w 5 is on H 3 . Then the vertices w j for 1 j 5 divide H 2 into five subpaths. Since n>25, at least one of such subpath has length at least 6, so we obtain a path P 2 =v 1 v 2 v 3 v 4 v 5 on H 2 such that P 2 and P 3 are vertex-disjoint. Now, the vertices v i for 1 i 5 and w j for 1 j 5 divide H 1 into ten subpaths. If there is one such subpath of length at least 4, then we obtain a path P 1 =u 1 u 2 u 3 on H 1 which is vertex-disjoint with P 2 and P 3 . Thus we assume that all subpaths have length at most 3. Since n 26, there are at least six of the subpaths of length 3. The four vertices v 1 , v 5 , w 1 , w 5 are the ends of at least five subpaths on H 1 so that one of them is an end of a subpath of length 3. By reordering the v-path or the w-path if necessary, we conclude that there is a desired subpath P 1 =u 1 u 2 u 3 on H 1 with either v 1 =u 3 or w 1 =u 3 . K Similar to the proof of Lemma 3, we can derive the following result. Lemma 4. If G is a 2k-regular graph of order n>5(5k&7), where k 2, and G has a hamiltonian decomposition, then for any hamiltonian decomposition H 1 , H 2 , ..., H k of G, there exist k vertex-disjoint paths P 1 =u 1 v 1 w 1 and P j =u j v j w j x j y j for 2 j k such that P j is on H j for 1 j k. Proof. If 0
3. Proof of Theorem 1
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Throughout this section we let C 1 =a 1 a 2 } } } a n a 1 and C 2 =b 1 b 2 } } } b m b 1 be two cycles of order n and m, respectively. The subscripts of a are expressed modulo n and the subscripts of b are expressed modulo m. By convention, we give a natural direction to a drawing of C 1_r C 2 on a torus as in Fig. 1 so that the a i -column is to the left of the a i+1-column for 1 i n&1.
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Let A be a finite abelian group, S=[s 1 , s 2 , ..., s k ] be a minimal generating set of A and m=ord(s k ) 3. Let F be the 2-factor of cay(A, S ) which is generated by s k . Then F consists of n= |A|m cycles of length m. Let J=( s k ) and S =[s 1 , s 2 , ..., s k&1 ] A 1 =AJ, where we use x to represent the coset x+J. Definition 4. For any edge xy of cay(A 1 , S ), where x &y =s i # S, we call the edge set [u 1 u 2 | u 1 =x, u 2 =y and u 1 &u 2 =s i ] of cay(A, S) the lifting edge set of the edge xy. Definition 5. For any subgraph Q of cay(A 1 , S ), the spanning subgraph Q of cay(A, S ) with the edge set being the union of the lifting edge sets of the edges of Q is called the subgraph lifted by Q and we say Q lifts to Q. It is easy to see from the above definitions that edge-disjoint subgraphs of cay(A 1 , S ) lift to edge-disjoint subgraphs of cay(A, S ). The following proposition is Lemma 3.7 in [8]. Proposition 3. Any hamiltonian cycle H =g 1 g 2 } } } g n g 1 of cay(A 1 , S ) lifts to a 2-factor H of cay(A, S ). Furthermore, under the map 8: a i +b j (a i , b j ), the union of F and H is isomorphic to an r-pseudo-cartesian product C 1_r C 2 in which the edges of H are horizontal-type and the edges of F are vertical, where 0 r m&1, a i =g i for 1 i n, and b j =( j&1) s k for 1 j m. Next we introduce a special class of graphs which plays a central role in our discussion. Definition 6. For k 2, define D k to be a 2k-regular graph satisfying: (1)
V(D k )=[(a i , b j ) | 1 i n and 1 j m],
(2)
E(D k ) can be decomposed into 2-factors H 1 , H 2 , ..., H k&1 and F,
(3) (a i , b 1 ),
F= ni=1 F i , where each F i is the cycle (a i , b 1 )(a i , b 2 ) } } } (a i , b m )
(4)
H 1 _ F=C 1_r1 C 2 , and
(5) for 2 j k&1, H j _ F=C 1j _rj C 2 , where 0 r j m&1 and C =a (?j)j (1) a (?j)j (2) } } } a (?j)j (n) a (?j)j (1) with ? j being a permutation of [1, 2, ..., n] and (a (i j) , b t )=(a i , b t+hi, j ) for 1 i n. j 1
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The example shown in Fig. 3 is a graph D 3 with n=6, m=5, r 1 =1, r 2 =2, ? 2 =(124653), and h i, 2 #3i&1 (mod 5) for 1 i 6.
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Fig. 3. A graph D 3 .
Clearly, from the definition it is easy to see that in each H j _ F= C 1j _rj C 2 the edges of F are vertical while the edges of H j are horizontaltype (in the sense that they are not vertical), where C 11 =C 1 . Thus, by Remark 2, H j is the union of t j =gcd(r j , m) cycles of equal length (mn)t j . For example, both H 1 and H 2 of the graph D 3 in Fig. 3 are hamiltonian cycles. From Proposition 3 and the definition for D k the following lemma can be derived easily. Lemma 5. If cay(A 1 , S ) can be decomposed into k&1 hamiltonian cycles H j =g ?j (1) g ?j (2) } } } g ?j (n) g ?j (1) for 1 j k&1, where ? 1 =I (the identity) and ? j is a permutation of [1, 2, ..., n] for 2 j k&1, then cay(A, S )$D k with each H j being the 2-factor lifted by H j and F being the 2-factor generated by s k . Proof. By Proposition 3, for each 1 j k&1, H j lifts to a 2-factor H j and the union H j _ F is isomorphic to C 1j _rj C 2 with C 1j =a (?j)j (1) a (?j)j (2) } } } a (?j)j (n) a (?j)j (1) under the map 8 j : a (i j) +b t (a (i j) , b t ), where a (i j) =g i and b t =(t&1) s k . For convenience, we denote a (1) i =a i for 1 i n, and so C 11 =C 1 =a 1 a 2 } } } a n a 1 and H 1 _ F=C 1_r1 C 2 . Since a (i j) =a i for 2 j k&1 and 1 i n, we let a (i j) =a i +h i, j s k . Then a (i j) +b t =a (i j) + (t & 1) s k = a i + (t + h i, j & 1) s k = a i + b t+hi, j and we have (a (i j) , b t ) =
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(a i , b t+hi, j ). Recall that edge-disjoint subgraphs of cay(A 1 , S ) lift to edgedisjoint subgraphs of cay(A, S). We conclude that cay(A, S)$D k . K By Lemma 5, we see that an important step to prove Theorem 1 by induction on k is to prove that D k can be decomposed into k hamiltonian cycles. Thus, our next major effort is to show that with certain restrictions D k has a hamiltonian decomposition. For the following discussions, we color the edges of D k so that all edges of F are of red color and for 1 j k&1, all edges of H j are of color c j . In order to decompose D k into hamiltonian cycles, our main idea here is to find some edge-disjoint color switchings so that making those color switchings results in k monochromatic hamiltonian cycles. In each of the following lemmas, we will find the necessary color switchings in two steps: First, we make some color switchings in each H j _ F=C 1j _rj C 2 to obtain a hamiltonian cycle of color c j for 1 j k&1; then, we concentrate on H 1 _ F=C 1_r1 C 2 and make some additional color switchings between red edges and edges of color c 1 to obtain a red hamiltonian cycle meanwhile the edges of color c 1 still form a hamiltonian cycle. Also, for the following discussions it is very helpful to visualize H j _ F= C 1j _rj C 2 in a way similar to (form B) of Fig. 3. Lemma 6. Let m>0 and n>0 be odd integers, and let k 2. Assume each H j in D k consists of t j cycles with t 1 t 2 } } } t k&1 . If 2 t 1
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Proof. Recall that in each H j _ F=C 1j _rj C 2 , the vertical edges are red, the horizontal-type edges are of color c j , and each row (in the sense that we visualize H j _ F as in (form B) of Fig. 3) is in the same cycle of H j . By Remark 2, each H j consists of t j =gcd(r j , m)=2t$j +1 cycles. If t j =1 for some j, then H j is already a hamiltonian cycle, so we can disregard it and work on the remaining graph obtained by removing the edges of H j from D k . Thus, we may assume that t j =2t$j +1 3 for 1 j k&1. Since t1 | m, t 1
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that for each 2 j k&1, by choosing x properly, we can apply Proposition 2 to H j _ F=C 1j _rj C 2 without using any vertical edges from the paths P i =(a i , b 1 )(a i , b 2 )(a i , b 3 ) for 4 i n in H 1 _ F=C 1_r1 C 2 to obtain a hamiltonian cycle of color c j and connect the red cycles in the a ?j (i)-columns for 1 i 3 to a single red cycle. Clearly, we now have n&2(k&1) red cycles and each column is in a single red cycle. Since n is odd, n&2(k&1)=2n 1 +1 for some n 1 >0. Now, we will find some additional edge-disjoint color switchings between red edges and edges of color c 1 so that we can connect all red cycles to a red hamiltonian cycle while we still have a hamiltonian cycle of color c 1 . For that purpose, we focus on H 1 _ F=C 1_r1 C 2 . Let 0
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Lemma 7. Let m=2h+1 5, n be odd, and k 2. Assume each H j in D k consists of t j cycles with t 1 t 2 } } } t k&1 . If t 1 =m and the sets K 1 =[1, 2, 3] and K j =[? j (i) | 1 i 5] for 2 j k&1 are mutually disjoint except for K 1 and K 2 having at most one common element ? 2(1)=3, then D k has a hamiltonian decomposition.
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Proof. We proceed in a way similar to the proof of Lemma 6. Again, we may assume that each H j consists of t j =gcd(r j , m)=2t$j +1 3 cycles for 1 j k&1. Since t 1 =m, we must have r 1 =0 or r 1 =m which implies that H 1 _ F=C 1_r1 C 2 =C 1_C 2 . We first apply Proposition 2 to H 1 _ F= C 1_C 2 with x=0 to obtain a hamiltonian cycle of color c 1 and connect the red cycles in the a i-columns for 1 i 3 to a single red cycle Q 1 . In the a 1-column, note that the edges of color c 1 form a matching consisting of all odd vertical edges, i.e., M=[(a 1 , b 2i&1 )(a 1 , b 2i ) | 1 i h], and all odd edges are red in the a 3-column. For each 2 j k&1, we call the a lj -column leftmost, where l j =min K j . Since K j & (K 1 _ K 2 )=< for 3 j k&1 and the sets K 3 , K 4 , ..., K k&1 are mutually disjoint, for each 2 j k&1 we apply Lemma 1 to H j _ F=C 1j _rj C 2 , with x being chosen properly, so that for j=2 and ? 2(1)=3 we use only odd edges in the a 3-column and for j 3 or ? 2(1)>3 we can locate two unused adjacent red edges anywhere as we wish in the leftmost a lj -column, to obtain a hamiltonian cycle of color c j and connect the involved columns to a single red cycle Q j . Clearly, for ? 2(1)=3, Q 1 is connected to Q 2 via the a 3-column. Note that each column is in a single red cycle and each red cycle consists of 3 columns, 5 columns, or 7 columns. We now have 2n 1 +1 red cycles for some n 1 >0 as n is odd. Since m 5, each a i -column for 3 i n has at least one red odd edge and each a i -column for 4 i n has property P: for any 1 f m one of the vertical edges (a i , b f )(a i , b f+1 ) and (a i , b f+2 )(a i , b f+3 ) is red (see Fig. 2). Similar to the proof of Lemma 6, we will make some additional edge-disjoint color switchings between red edges and edges of color c 1 so that we obtain a red hamiltonian cycle and a hamiltonian cycle of color c 1 without destroying the existing hamiltonian cycles of color c j for 2 j k&1. For that purpose, we focus on H 1 _ F= C 1_C 2 . Let 0
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vertical edges e 1 =(a z2i , b yi &1 )(a z2i , b yi ) and e 2 =(a z2i , b yi +1 )(a z2i , b yi +2 ) is red, say e 1 . Recall that both the a z2i&1 +1-column (it might happen that z 2i&1 +1=z 2i ) and the a z2i +1-column are either free or leftmost, we define X 2i&1 to be the [a z2i&1 , a z2i&1 +1 , b yi , b yi +1 ]-color switching and X 2i to be the [a z2i , a z2i +1 , b yi &1 , b yi ]-color switching. Since X 2i&1 uses only odd vertical edges while X 2j does not use odd vertical edges, we conclude that X 1 , X 2 , ..., X 2n1 are edge-disjoint. Now, from the choice of the z j 's it follows that by making the color switchings X 1 , X 2 , ..., X 2n1 , we obtain a red hamiltonian cycle. To see that the edges of color c 1 still form a hamiltonian cycle, we make the color switchings in the order X 1 , X 2 , ..., X 2n1 with one pair X 2i&1 and X 2i at each time. It follows from Fact 2 that, by making the color switching X 2i&1 , the existing hamiltonian cycle of color c 1 is separated into two cycles with one being (a z2i&1 +1 , b yi )(a z2i&1 +2 , b yi ) } } } (a n , b yi )(a 1 , b yi )(a 1 , b yi +1 ) (a n , b yi +1 )(a n&1 , b yi +1 ) } } } (a z2i&1 +1 , b yi +1 )(a z2i&1 +1 , b yi ). Then it follows from Fact 1 that by making the color switching X 2i those two cycles are connected to a hamiltonian cycle of color c 1 again. Thus, together with the existing monochromatic hamiltonian cycles of color c j for 2 j k&1, we obtain a hamiltonian decomposition of D k . K We are ready to prove our main result.
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Proof of Theorem 1. We proceed by induction on k. For k=1, the result is trivial. For k=2 or 3, the result follows from Theorems B and C. Assume the result for k
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H j cay(A, S$). By Lemma 2, |A 1 | 3 k&1. Thus, |A 1 | 27 for k=4, and |A 1 | >5[5(k&1)&7] for k 5. By applying Lemmas 3 and 4 to cay(A 1 , S ), we may assume that all ? i , for 1 i k&1, satisfy the conditions in Lemmas 6 and 7. Now, it follows from Lemmas 6 and 7 that cay(A, S )$D k has a hamiltonian decomposition. K I believe that the techniques used here for abelian groups of odd order are useful in obtaining a similar result for abelian groups of even order.
Acknowledgments The author is grateful to the referees whose many comments have significantly helped to improve the presentation of this article. In particular, the author thanks the referees for providing a proof for Lemma 4.
References
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1. B. Alspach, Research Problem 59, Discrete Math. 50 (1984), 115. 2. B. Alspach, J.-C. Bermond, and D. Sotteau, Decomposition into cycles. I. Hamilton decompositions, in ``Cycles and Rays'' (G. Hahn, G. Sabidussi, and R. Woodrow, Eds.), pp. 918, Kluwer, Amsterdam, 1990. 3. J. Aubert and B. Schneider, Decomposition de la somme cartesienne d'un cycle et de l'union de deux cycles hamiltonens en cycles hamiltoniens, Discrete Math. 38 (1982), 716. 4. J.-C. Bermond, O. Favaron, and M. Maheo, Hamiltonian decomposition of Cayley graphs of degree 4, J. Combin. Theory Ser. B 46 (1989), 142153. 5. G. Chartrand and L. Lesniak, ``Graphs and Digraphs,'' 2nd ed., Wadsworth and BrooksCole, Monterey, CA, 1986. 6. M. F. Foregger, Hamiltonian decompositions of products of cycles, Discrete Math. 24 (1978), 251260. 7. J. A. Gallian, ``Contemporary Abstract Algebra,'' Heath, Lexington, MA, 1986. 8. J. Liu, Hamiltonian decompositions of Cayley graphs on abelian groups, Discrete Math. 131 (1994), 163171. 9. D. Marusic, Hamiltonian circuits in Cayley graphs, Discrete Math. 46 (1983), 4954. 10. R. Stong, Hamilton decompositions of cartesian products of graphs, Discrete Math. 90 (1991), 169190.