Vertex-Transitive Graphs Of Prime-Squared Order Are Hamilton-Decomposable

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1 Vertex-Transitive Graphs Of Prime-Squared Order Are Hamilton-Decomposable Brian Alspach School of Mathematical and Physical Sciences University of Newcastle Callaghan, NSW 2308, Australia Darryn Bryant Department of Mathematics University of Queensland, Qld 4072, Australia Donald L. Kreher Department of Mathematical Sciences Michigan Technological University Houghton, Michigan Abstract We prove that all connected vertex-transitive graphs of order p 2, p a prime, can be decomposed into Hamilton cycles. Keywords: Hamilton cycle, Hamilton decomposition, vertex-transitive graph, Cayley graph AMS Classification: 05C25, 05C70 1 Introduction Let X be a graph that is regular of valency d. If the edge set of X can be partitioned into d/2 Hamilton cycles when d is even, or into (d 1)/2 Hamilton cycles and a perfect matching when d is odd, then we say that X is Hamilton-decomposable or admits a Hamilton decomposition. There has been an interest in Hamilton-decomposable graphs from the infancy of graph theory. We are interested in a particular family of graphs with respect to this property. 1

2 Let G be a finite group and S an inverse-closed subset of G such that 1 S. The Cayley graph on G with connection set S is the graph whose vertices are labelled with the elements of G, where the vertex labelled g is adjacent to all vertices of the form gs as s runs through S. We denote the graph by Cay(G; S). When G is abelian, we normally use additive notation so that g is adjacent to g +s as s runs through S. Cayley graphs on cyclic groups are usually called circulant graphs and have the special notation circ(n; S), where n is the order of the cyclic group. The first author [1] asked whether all connected Cayley graphs on abelian groups admit Hamilton decompositions. There have been some nice partial results obtained since the problem was first posed, and it has been transformed into a conjecture over the intervening years. One approach to the conjecture has been based on valency. Bermond, Favaron, and Maheo [5] showed that the conjecture holds when the graph has valency 4. Dean [6] studied circulant graphs of valency 6 and proved that odd order connected circulant graphs of valency 6 are Hamilton-decomposable. He also verified that many even order connected circulant graphs of valency 6 have Hamilton decompositions, but there was a small class of such circulants for which he could not verify the conjecture. Recently, Westland, Liu, and Kreher [15] have extended Dean s results to connected Cayley graphs on abelian groups. They had a similar outcome in that they could do it for all connected Cayley graphs of valency 6 on odd order abelian groups, and for most connected Cayley graphs of valency 6 on even order abelian groups but not all. Liu approached the conjecture from the standpoint of structural properties of the connection set. In a series of three very nice papers [9, 10, 11], he proved that if the connection set S is a minimal Cayley generating set, then Cay(G; S) is Hamilton-decomposable with one exceptional case for G even. The connection set S is a minimal Cayley generating set when S = G but S {s, s 1 } is a proper subgroup of G for every s S. Liu s result subsumes several older special cases for which the conjecture was known to hold. For example, his result covers the cartesian product of any number of cycles and the even dimensional cubes. The object of this paper is to approach the conjecture from the standpoint of the order of the graph. It is a trivial observation that every vertextransitive graph of prime order is a circulant graph. It is easy to see that a prime order circulant graph circ(p; S) has a Hamilton decomposition because all the edges arising from a fixed element s S generate a Hamilton cycle. We aim to prove that all connected vertex-transitive graphs of order 2

3 p 2, p a prime, are Hamilton-decomposable. 2 Preliminaries As observed above, a vertex-transitive graph of prime order is a circulant graph. Marušič [13] proved that every vertex-transitive graph of order p 2, p a prime, is a Cayley graph. To within isomorphism there are two groups of order p 2, Z p 2 and Z p Z p, and both are abelian. Thus, every vertextransitive graph of order p 2 is a Cayley graph on an abelian group. We now move to definitions and results that are required to attain the objective of this paper. Let σ be a permutation of the vertex set V (X) of a graph X. If Y is a subgraph of X, then σ(y ) denotes the subgraph of X induced by the set of edges {σ(u)σ(v) : uv E(Y )} provided that σ(u)σ(v) actually is an edge of X for every uv E(Y ). If the latter is not the case, then σ(y ) simply is undefined. 2.1 Definition. Let F be a 2-factorization of a 2d-regular graph X with 2-factors F 1, F 2,..., F d. A d-matching M in X is said to be orthogonal to F if M shares exactly one edge with each F i, 1 i d. We frequently say M is an orthogonal matching. There is a standard near-1-factorization of an odd order complete graph based on an orthogonal matching. Think of the complete graph of order 2m+1 as the circulant graph circ(2m+1; S), where S = {1, 2,..., 2m}. Let F be the 2-factorization obtained by letting F i, 1 i m, consist of the edges generated by i S. If we now let M be the m-matching consisting of the edges u 1 u 2m, u 2 u 2m 1,..., u m u m+1, it is clear that M is orthogonal to F. The near-1-factorization is M, ρ(m ),..., ρ 2m (M ), where ρ is the permutation (u 0 u 1 u 2m ). 2.2 Definition. Let X = circ(n; S) and n be odd. If S = 2d, then the d-matching M(0) = M E(X) is orthogonal to the 2-factorization of X, where each 2-factor consists of the edges generated by a single element s S. We call M(0) the canonical orthogonal matching centered at u 0. We obtain the canonical orthogonal matching M(i) centered at u i by taking M(i) = ρ i (M(0)), where ρ is the same permutation as in the preceding paragraph. 3

4 The edge space of a graph X, denoted E(X), is the vector space over F 2, the field of order 2, coordinatized by the edges of X. Hence, the elements of E(X) are the subgraphs of X. We use for the sum of vectors in E(X). Hence, if X 1 and X 2 are subgraphs of X, then the edge set of X 1 X 2 is the symmetric difference of the edge sets of X 1 and X 2. y x y x w z w z = y x w z Figure 1 3 The Group Z p Z p The next two results set the stage for an approach based on partitioning the connection set of a Cayley graph. 3.1 Lemma. Let G be a group of order p 2, where p is a prime. If g, h G both have order p and g h, then X = Cay(G; {±g, ±h}) is Hamiltondecomposable. Proof. In the special case of p = 2, X is a 4-cycle. So the result holds for p = 2 and we assume p is an odd prime for the rest of the proof. The edges generated by h produce a 2-factor F composed of p cycles C(1), C(2),..., C(p) all of length p. Because g h, the edges generated by g join cycles of F together. Without loss of generality, we may assume the cycles are labelled so that there is an edge joining a vertex of C(1) to a vertex of C(2). Because G is abelian, it is easy to see that there is a perfect matching joining vertices of C(1) and C(2). In fact, the subgraph induced 4

5 by X on V (C(1)) V (C(2)) is isomorphic to the cartesian product of a p-cycle and K 2. The cycles C(1), C(2),..., C(p) must be joined cyclically, and because G is abelian and g has order p, it then follows that X is isomorphic to the cartesian product of two p-cycles. Thus, X has a Hamilton decomposition by [5]. When working in Z n, we use the residues 0, 1, 2,..., n 1 (unless specifically mentioned to the contrary), and we think of the residues being ordered 0 < 1 < < n 1. The notion of one residue being smaller than another is in reference to the order just given. In a circulant graph X with vertices labelled v 0, v 1,..., v n 1, the length of an edge v i v j is defined to be the smaller of {j i, i j} in Z n. 3.2 Theorem. If X is a connected circulant graph of prime order p > 2 and C p denotes the cycle of length p, then the cartesian product C p X has a Hamilton decomposition. Proof. Let the vertices of the cartesian product be denoted {u i,j : 0 i, j p 1}. Let the connection set for X be S = {±s 1, ±s 2,..., ±s t }, where s 1 < s 2 < < s t < p/2. It is obvious that all the edges of any constant length yield a Hamilton cycle in X. This provides a trivial Hamilton decomposition of X. For simplicity of language, we refer to the vertices u i,j with i fixed as a column. Thus, the subgraphs of C p X induced on the columns are copies of X. For each column i, 0 i p 1, consider the canonical matching M(i) for X centered at u i,i and orthogonal to the constant length Hamilton decomposition of X. For each s k {s 1,..., s t }, we obtain a Hamilton cycle in C p X as follows. Let C(i, s k ) denote the cycle spanning column i generated by the element s k. For each i satisfying 0 i p 1, let e i,k denote the edge of length k in the canonical matching M i. Then let F i,j be the 4-cycle obtained by taking e i,j, the two horizontal edges from the ends of e i,j to vertices of column i + 1, and the edge of length s k joining them in column i + 1. We claim that C(0, s k ) F 0,k C(1, s k ) F 1,k C(p 1, s k ) F p 1,k is a Hamilton cycle. This is seen easily by thinking of the sum as an iterative process of adding a 4-cycle and the next column cycle to obtain a single cycle (see Figure 1). After the column cycle for p 1 has been included, we have a Hamilton cycle C, but we still have F p 1,k to include in the sum. If we 5

6 traverse the cycle C, the two edges of length s k in F p 1,k are oriented in the same direction. Thus, C F p 1,k is a Hamilton cycle. To conclude the proof of the theorem, we need to show that edges left over, after carrying out the preceding construction for each s 1,..., s t, themselves form a Hamilton cycle. So let Y denote the subgraph induced by the remaining edges. Note that Y is regular of valency 2. Because of the pattern used for forming the k Hamilton cycles, it is clear that the permutation f defined by f(u i,j ) = u i+1,j+1, where the subscript arithmetic is carried out modulo p, is an automorphism of Y. The group f generated by f has p orbits each of length p. If we show that a component of Y intersects each orbit, then we know that Y either consists of p cycles of length p or is a Hamilton cycle because the restriction of f to each component is a p-cycle (as a permutation). In order to facilitate the understanding of the following argument, the reader is advised to consider Figure 2 where the case for p = 11 and s i = 1, s 2 = 2, s 3 = 4 is illustrated. We introduce an auxiliary graph M (the graph on the left in Figure 2). The vertices of M are labelled 0, 1, 2,..., p 1. We then insert the edges for the canonical orthogonal matchings M(0) and M(p 1). At this point we have a collection of paths in M. There is then a unique way to form a Hamilton cycle from these paths using edges joining successively labelled vertices. We add these edges to obtain M. We use M to describe a subpath of Y of length p whose vertices intersect all the orbits such that the first and last vertices are different and lie in the same orbit. This then implies that Y is a Hamilton cycle. For the general argument, let O(i) denote the orbit of f containing u 0,i for i = 0, 1,..., p 1. Consider the edges of the orthogonal matching M 0 centered at 0. Every edge of M 0 has end vertices of the form i and p i for some i. There is a corresponding edge e of the same length in column 0 whose end vertices are u 0,i and u 0,p i, that is, its end vertices lie in O(i) and O(p i). Because of the action of the permutation f, f k (e) always has its end vertices in O(i) and O(p i). When we form the 4- cycle u 0,p i u 0,i u 1,i u 1,p i u 0,p i to combine the p-cycles forming column 0 and column 1, note that the edge u 1,i u 1,p i has its end vertices in O(i 1) and O(p i 1), but these are the labels of the end vertices of the corresponding edge in M(p 1). 6

7 Figure 2 Therefore, the labels on the vertices of the auxiliary graph M tell us which orbit the vertex belongs to as we trace out a path in the graph Y. We may start the subpath at many locations, but to be specific, choose the edge of M(0) with end vertices i and p i, i < p/2, so that i is as small as possible. Start traversing the auxiliary graph M by traversing the edge from p i to i and continue along the Hamilton cycle that forms M. The corresponding edge in Y we use is the edge u 0,p i u 0,i and we traverse it from u 0,p 1 to u 0,i. Whenever we traverse an edge of either M(0) or M(p 1) in M, we traverse the corresponding edge in Y in the vertical direction. When we traverse an edge of M that is not in M(0) or M(p 1), we use the corresponding horizontal edge in Y between the appropriate orbits. We obtain a path of length p in Y and we are done if we show that when we return to a vertex of O(p i), it is not the beginning vertex u 0,p i. However, this is easy to see because M has an odd number of edges of which an even number are matching edges. Thus, the number of horizontal edges in the path of length p in Y is odd and lies strictly between 0 and p. This implies that the path cannot terminate in column Corollary. Every connected Cayley graph on the group Z p Z p, p a prime, is Hamilton-decomposable. 7

8 Proof. Let X be a connected Cayley graph on the group Z p Z p. The graph X is either K 4 or a cycle of length 4 when p = 2. Both of these graphs are Hamilton-decomposable so we assume that p is an odd prime for the rest of the proof. Let S be the connection set for X. Every element of S is a scalar multiple of precisely one element in {(0, 1), (1, 0), (1, 1),..., (1, p 1)}. Partition S into parts S 0, S 1,..., S p, where S i contains the scalar multiples of (1, i) for i = 0, 1,..., p 1 and S p contains the scalar multiples of (0, 1). At least two of the parts are non-empty because X is connected. Consider two non-empty parts S a and S b. Without loss of generality assume S a S b. Because S is inverse-closed, both S a and S b contain an even number of elements. If S a = 2, then the subgraph Cay(Z p Z p ; S a S b ) is isomorphic to the cartesian product of a circulant graph of order p and C p. It is Hamilton-decomposable by Theorem 3.2. On the other hand, if S a > 2, then choose ±s 1 from S a and ±s 2 from S b. These two elements generate a connected subgraph of valency 4 that has a Hamilton decomposition by Lemma 3.1. Continue removing two elements at a time from both S a and S b until there are two elements of S a remaining. Use the argument of the preceding paragraph at this point. If there are an even number of non-empty parts, use the above scheme on pairs of non-empty parts to obtain a Hamilton decomposition of X. If there are an odd number of non-empty parts, choose three non-empty parts S a, S b, S c and let S a S b S c. Because there are an even number of non-empty parts left over (possibly zero), it suffices to show that the subgraph generated by the elements of S a S b S c is Hamilton-decomposable. If either inequality is strict, it is easy to see how to obtain a Hamilton decomposition. Simply remove pairs of elements (using Lemma 3.1) until one set is empty. Then complete by working with the leftover pair of parts. If all three parts have the same cardinality, remove elements alternately from different parts. This either terminates with all three parts being left empty, or all three parts end up with cardinality 2. In the latter case, these elements generated a connected Cayley graph on Z p Z p of valency 6. This subgraph can be decomposed into three Hamilton cycles by the main result of [15]. 8

9 4 The Group Z p 2 We now consider Cayley graphs on the group Z p 2, that is, circulant graphs of order p 2. The following lemma reduces the scope of the proof of the main result of this section. 4.1 Lemma. If the circulant graph of order p 2 and with connection set S = {±1, ±s 1, ±s 2,..., ±s t }, where s i is a multiple of p for i = 1, 2,..., t, is Hamilton-decomposable, then every connected circulant graph of order p 2 is Hamilton-decomposable. Proof. Let X be a connected circulant graph of order p 2 with connection set S. If gcd(a, p 2 ) = 1, then it is easy to see that the circulant graph of order p 2 with connection set as is isomorphic to X. A circulant graph of order p 2 whose connection set contains only multiples of p clearly is disconnected. Thus, S must contain an element b such that gcd(b, p 2 ) = 1. The circulant graph Y with connection set S = b 1 S is isomorphic to X and contains ±1 in its connection set. Any element s S that is not a multiple of p and is different from ±1 generates a Hamilton cycle on its own. Hence, we may remove ±s, for all such elements in S, leaving us with ±1 and all the multiples of p in S. If this circulant subgraph is Hamiltondecomposable, then Y is Hamilton-decomposable. This, of course, implies that X is Hamilton-decomposable. We employ a special path in the proof of the next theorem and here is its description. Let X be a circulant graph of order n with connection set S = {±s 1, ±s 2,..., s t } satisfying s 1 < s 2 < < s t < n/2. Suppose the vertices of X are labelled u 0, u 1,..., u n 1. We introduce the notion of a zig-zag path based on the elements of the connection set. If the path starts at a vertex u i, the first edge joins u i to u i+s1. The next edge then joins u i+s1 to u i+s1 s 2. We continue in the obvious way using edges of increasing lengths until using an edge of length s t. It is easy to see why it is called a zig-zag path. Let P be the path of length 2t in X obtained as follows. We first construct the zig-zag path of length t using the lengths in the connection set for X starting at a fixed vertex u i. Suppose this path terminates at vertex u j. Place the n vertices of X on a regular polygon with n vertices. For each vertex of the polygon there is the standard reflection about the line through the vertex and the midpoint of the opposite line segment, when n is odd, or the line joining opposite vertices, when n is even. Reflect the zig-zag path 9

10 through this line passing through u j. The resulting path of length 2t is what we call the reflected zig-zag path rooted at u i. The preceding path has one desirable property we exploit. The path is composed of two t-matchings M 1, M 2 obtained by letting alternate edges form a matching. The desirable property is that both matchings contain one edge of each of the lengths s 1, s 2,..., s t. 4.2 Theorem. If X is the circulant graph of order p 2 with connection set S = {±1, ±s 1, ±s 2,..., ±s t }, where s i is a multiple of p for i = 1, 2,..., t, then X is Hamilton decomposable. Proof. Let X be a circulant graph satisfying the hypotheses. If S = {±1}, then X is a Hamilton cycle and we are done. So we assume that S contains ±s 1, ±s 2,..., ±s t, t 1, all of which are multiples of p. The vertices of X normally are labelled u 0, u 1,..., u p 2 1 so that the permutation (u 0 u 1 u p 2 1) belongs to Aut(X) and u i u i+1 is an edge because ±1 is in the connection set. We represent the vertices as a p p array. The vertex v i,j in the array is the vertex u i+jp, 0 i, j p 1. We refer to vertices v i,j with i fixed as column i and with j fixed as row j. There is an edge joining v i,j and v i+1,j for all i = 0, 1,..., p 2, and an edge joining v p 1,j and v 0,j+1 for all j = 0, 1,..., p 1, where subscript arithmetic is carried out modulo p, because ±1 S. Because all remaining elements of S are multiples of p, it is easy to see that the rest of the edges in X join vertices in the same column. In fact, the subgraph induced on each of the columns is the circulant graph Y of order p with connection set {±s 1 /p, ±s 2 /p,..., ±s t /p}. We know that Y has a decomposition into Hamilton cycles H 1,..., H t, where H k is composed of the edges of length s k /p in Y. Let H k (i), 0 i p 1, denote the copy of H k in column i. Without loss of generality, we assume that the elements of the connection set for Y satisfy 0 < s 1 /p < s 2 /p < < s t /p < p/2. For each i satisfying 0 i p 2, let P i denote the reflected zig-zag path rooted at v i,0 in column i. Let M 1 and M 2 be the two matchings of P i, where M 1 contains the edge incident with v i,0. We don t place additional notation on the matchings because it always will be clear in which column we are working. Suppose v 0,j1 v 0,j2 is an edge of P 0 belonging to the matching M 1 10

11 and that the edge belongs to the Hamilton cycle H k (0). If we let C be the 4-cycle v 0,j1 v 0,j2 v 1,j2 v 1,j1 v 0,j1, then H k (0) C H k (1) is a cycle of length 2p spanning the vertices of columns 0 and 1 (see Figure 1). We call this the 4-cycle switch on the edge u 0,j1 u 0,j2. Continuing in the same way, if we perform the 4-cycle switches using the edge of M 1 in H k (i) between columns i and i + 1, i even and 0 i < p 1, and using the edge of M 2 in H k (i) for i odd and 0 < i < p, it is easy to see that we obtain a Hamilton cycle in X that has tied together the p copies of H k in the columns. Doing this for each of H 1, H 2,..., H t, we have t edge-disjoint Hamilton cycles so far. Exactly one edge of H k in columns 0 and p 1 has been removed, and two edges of H k have been removed in each of the remaining columns. It is clear that the edges left over form a regular graph of valency 2, that is, they form a 2-factor. We call this 2-factor the residue 2-factor and denote it F. If we show that F is a Hamilton cycle, then the proof is complete. We now describe the edges of F. In column 0 there is a copy of the matching M 1, in column p 1 there is a copy of the matching M 2, and in each of the remaining columns there is a copy of P rooted at the vertex in row 0. If the vertex v 0,j is not in M 1 and the vertex v(1, j) is not in the copy of M 2 in column 1, then no vertex in row j lies in any copy of P or on an edge of M 2. Thus, the path Q j = v 0,j v 1,j v 2,j v p 1,j lies in F. Consider next edges of F that tie together the subpaths described in the preceding paragraph. Consider the path P 1 in column 1. It is rooted at v 1,0 and the edge v 1,0 v 1,s1 /p incident with it in column 1 is in F because this edge is in the matching M 1 and we performed a 4-cycle switch on v 0,0 v 0,s1 /p. Hence, the other edge of F incident with v 1,0 is v 2,0. Similarly, for every odd i satisfying 0 < i < p, the edge v i,0 v i+1,0 lies in F. If v 1,l is the other end of P 1 in column 1, a similar analysis establishes that the edge v 0,l v 1,l is in F. Hence, for every even i satisfying 0 i < p 1, the edge v i,l v i+1,l lies in F. These edges then tie the copies of P in columns 1, 2,..., p 2 into a single path P whose terminal edges are v 0,l v 1,l and v p 2,0 v p 1,0. Finally, all of the edges of the form v p 1,j v 0,j+1 lie in F because no 4- cycle switches were carried out between column p 1 and column 0. We now have the essential ingredients to verify that F is a Hamilton cycle. We introduce an auxiliary multigraph aux(p ) to help us do so. Label the vertices of aux(p ) with 0, 1,..., p 1. Join i to i + 1 for i = 0, 1,..., p 1, where we treat the operation modulo p, giving us a Hamilton cycle. Then put in the edges of P rooted at 0. If s 1 /p = 1, then aux(p ) has two edges with multiplicity 2. (Figure 3 illustrates aux(p ) when p = 7 and the connection 11

12 set is {±1, ±3}.) 0 = w = w 1 2 = w 5 = w = w 3 4 Figure 3 We are going to describe a closed walk in aux(p ) that demonstrates that F is a Hamilton cycle. Before doing so, we introduce some useful interval notation. Orient the Hamilton cycle in aux(p ) in the direction vertex 0 followed by vertex 1 and so on. The interval [0, 4] consists of the vertices 0, 1, 2, 3 and 4. The interval (0, 4] consists of the vertices 1, 2, 3 and 4. The interval (p 2, 3) consists of the vertices p 1, 0, 1 and 2. It should be clear from these three examples how we are using interval notation. The vertices of the path P in aux(p ) provide a collection of 2t + 1 closed intervals that overlap only at their endpoints and contain all the vertices. Label the vertices of P with w 0, w 1,..., w t following the orientation of the Hamilton cycle. Note that w 0 = 0 and w 1 = s 1 /p. The intervals then are [w 0, w 1 ], [w 1, w 2 ],..., [w 2t, w 0 ], where [w 0, w 1 ] = [0, s 1 /p], [w t, w t+1 ] = [l s 1 /p, l] (recall that above we let rows 0 and l contain the ends of the copies of the reflected zig-zag path) and [w 2t, w 0 ] = [s 1 /p s 2 /p, 0]. Consider the following closed walk W in aux(p ). Start at the vertex w 0 = 0 and take the edge of P to w 1 = s 1 /p. Now traverse the interval [w 0, w 1 ] from w 1 back to w 0. Now traverse the edges of P until reaching the other end at vertex w t+1 = l. Continue as follows. The vertex l is one end of the interval [w t, w t+1 ] = [l s 1 /p, l]. Traverse the interval in reverse orientation from w t+1 to w t. Then traverse the edge w t w t+1 of M 2 to w t+1. Now traverse the interval [w t+1, w t+2 ] to w t+2. Now add the edge from w t+2 to w t noting that this is an edge of M 1. Then traverse the interval [w t 1, w t ] from w t to w t 1. The edge w t 1 w t+2 belongs to M 2 and we add this edge to the walk. We then 12

13 traverse the interval [w t+2, w t+3 ] from w t+2 to w t+3. We continue in this way and because the interval [0, l] is composed of t + 1 intervals arising from the vertices of P, whereas, the interval [l, 0] is composed of only t intervals from P, we eventually reach w 0 along the interval [w 2t, w 0 ] traversing it from w 2t to w 0. The walk W terminates at this point and is closed as claimed. We now translate W into a traversal of F starting at v 0,0 with the edge of M 1 from v 0,0 to v 0,s1 /p. Traversing the interval [w 0, w 1 ] from w 1 back to w 0 in W corresponds to taking the edge from v 0,s1 /p to v p 1,s1 /p 1 in F and continuing as follows. If s 1 /p 1 > 0, then we traverse the path Q s1 /p 1 until reaching v 0,s1 /p 1. We then drop down to row s 1 /p 2 using the edge to v p 1,s1 /p 2 and continue in the obvious way until reaching v p 1,0. If s 1 /p 1 = 1, then we are at v p 1,0 already. This corresponds to being back at 0 in W. So far the traversal of F has used all vertices in rows 1, 2,..., s 1 /p 1 when s 1 /p > 1, and when s 1 /p = 1, there are no such rows. When we reach v p 1,0, we are at one end of the path P in F. For W we now traversed the path P and finish at l. This corresponds in F to travelling along P until reaching its other end at v 0,l. So far the traversal of F has used all vertices in any rows lying strictly between row 0 and row s 1 /p, and all the vertices of P. Our traversal of the 2-factor F currently rests at the vertex v 0,l. The walk W continues by traversing the interval [w t, w t+1 ] = [l s 1 /p, l] in the direction from w t+1 to w t. For the 2-factor F, this corresponds to taking the edge from v 0,l to v p 1,l 1. If w t < l 1, then F contains the path Q l 1 and we traverse it from v p 1,l 1 to v 0,l 1. We then do the same thing with the edge from v 0,l 1 to v p 1,l 2. It is then easy to see that for each vertex j in (w t, w t+1 ), we cover the path Q j from column p 1 to column 0. When we finish the interval our traversal of F has reached the vertex v p 1,l s1 /p. The walk W now uses the edge from w t to w t+2 and this is precisely the edge of M 2 in column p 1 incident with the vertex v p 1,l s1 /p. So we traverse that edge next in F. The walk W now traverses [w t+1, w t+2 ] from w t+1 to w t+2. This corresponds to traversing all paths Q j for j (w t+1, w t+2 ) from column 0 to column p 1. The last edge takes us to the end vertex of the next edge of M 1 in column 0. It is clear that as we continue the traversal of F we eventually reach v 0,0 along the edge from v p 1,p 1 to v 0,0. Every vertex is included because the walk W traverses all intervals. Thus, F is a Hamilton cycle. This concludes the proof. 13

14 4.3 Theorem. Every connected vertex-transitive graph of order p 2, p a prime, has a Hamilton decomposition. Proof. When p = 2, the only possible graphs are the cycle of length 4 and K 4. Both of them have Hamilton decompositions. So we assume that p is an odd prime. Every vertex-transitive graph of order p 2 is a Cayley graph. There are two groups of order p 2 : The elementary abelian group Z p Z p of rank 2 and the cyclic group Z p 2. Corollary 3.3 establishes the result for Cayley graphs on Z p Z p. Lemma 4.1 and Theorem 4.2 establish the result for circulant graphs of order p 2. Acknowledgements The second author acknowledges the support of the Australian Research Council via grants DP , DP and DP All three authors express their gratitude for support from the School of Mathematical and Physical Sciences (MAPS) and the Priority Research Centre for Computer-Assisted Research Mathematics and its Applications (CARMA) both of which are located at the University of Newcastle. References [1] B. Alspach, Research Problem 59, Discrete Math. 50 (1984), 115. [2] B. Alspach, J.-C. Bermond and D. Sotteau, Decompositions into cycles I: Hamiliton decompositions, in Cycles and Rays, edited by G. Hahn, G. Sabidussi, and R. Woodrow, Kluwer Acad. Publ., Dordrecht, (1990) [3] J. Aubert and B. Schneider, Décomposition de la somme cartésienne d un cycle et de l union de 2 cycles Hamiltoniens, Discrete Math. 38 (1982), [4] J.-C. Bermond, Hamilton decomposition of graphs, directed graphs, and hypergraphs, Advances in Graph Theory, Annals of Discrete Math. 3 (1978), [5] J.-C. Bermond, O. Favaron, and M. Maheo, Hamiltonian decomposiiton of Cayley graphs of degree four, J. Combin. Theory Ser. B 46, (1990),

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