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We prove that a Cayley digraph on the direct product of dihedral groups
D_{2n} × D_{2m} with outdegree two is Hamiltonian if and only if it is connected.

For a finite group G and a subset S of G, the Cayley digraph is the directed graph with vertex set G and arcs from to for each and. The set S is often called the connection set of the digraph, and this digraph is connected if and only if is a generating set for G. The connection set S is said to be minimal if it is a minimal generating set of G, and it is said to be minimum if it is a minimal connection set of minimum cardinality. A Hamilton cycle (path) in a digraph of with n vertices is a directed cycle (path) with n vertices. A digraph is said to be Hamiltonian if it has a Hamilton cycle.

Each arc in of the form is labelled, and called an s-arc. A Hamilton cycle in can be specified by the sequence of vertices encountered or by the sequence of arcs traversed. In the latter case, it is often more convenient to list the labels of the arcs, rather than the arcs themselves, since for each vertex there is exactly one out-arc with label for each. An ordered sequence of the arc labels encountered in a Hamilton cycle is called a Hamiltonian arc sequence. Since Cayley digraphs are vertextransitive, any cyclic shift of a Hamiltonian arc sequence of a Cayley digraph is also a Hamiltonian arc sequence of the digraph, and traversing a Hamiltonian arc sequence of a Cayley digraph starting from any vertex will yield a Hamilton cycle of the digraph. For convenience and brevity, we sometimes omit the commas and brackets from an arc sequence. For an arc sequence x, the symbol denotes the concatenation of copies of. If for some and, we sometimes write

to denote the fact that there is an s-arc from to in.

For an integer, the symbol denotes the dihedral group of order. For this is the group of symmetries of the regular -gon under the operation of function composition, and it has the presentation, where R is the counterclockwise rotation of and is a reflection across any axis of symmetry. For n = 2 the same presentation can be used to define D_{4}. Note that.

One fundamental problem is that of determining which Cayley digraphs are Hamiltonian. This is a longstanding problem which can be traced back to bell ringing, or campanology, since the orders in which a set of church bells may be rung form a group, and a Hamilton cycle in a Cayley digraph of this group gives a sequence of these bell ringing orders which is pleasing to the ear. The problem is longstanding mainly due to its difficulty. There are several good surveys on the problem, including [1-3], which discusses recent progress and current directions in the more general related problem of finding Hamilton cycles and paths in vertex-transitive graphs.

One of the first elegant results on the problem of the Hamiltonicity of Cayley digraphs is due to Rankin [

Theorem 1.1. A Cayley digraph on with outdegree two is Hamiltonian if and only if it is connected.

This is a new result since such digraphs do not satisfy the hypothesis of Rankin’s result. We first prove that if is generated by two elements then both and m are odd, and the proof makes use of the following result due to Gaschütz in 1955 [

Proposition 1.2. (Gaschütz [

In this section we prove Theorem 1.1. We make use of the following lemma.

Lemma 2.1. If is generated by two elements, then both and are odd.

Proof: Since any dihedral group is generated by two elements, Proposition 1.2 implies that is generated by two elements if and only if is generated by two elements, where and denote the intersections of the maximal normal subgroups of and, respectively. If is even and, then the normal subgroups of

are

Thus the intersection of the maximal normal subgroups of is. On the other hand, if is odd, then the normal subgroups of are all of the form for, and so in this case there is only one maximal normal subgroup, namely. Note that and . Thus Proposition 1.2 implies that if and are both even then is generated by two elements if and only if is generated by two elements, and if exactly one of n or m is even, say n is even, then is generated by two elements if and only if is generated by two elements. Using induction and the fact that, we conclude that if or is even and is generated by two elements, then either or is generated by two elements, a contradiction. Hence both and must be odd.

Proof of Theorem 1.1: If a Cayley digraph is Hamiltonian, then it is certainly connected. Conversely, if a Cayley digraph on of outdegree two is connected, then is generated by two elements. Let be a generating set for. By Lemma 2.1, both and must be odd. If and are both rotations in for some, then S does not generate, a contradiction. Hence at least one of and is a reflection for. If are all reflections, then every element in is an ordered pair of reflections or an ordered pair of rotations, a contradiction. If a rotation in does not generate the cyclic subgroup of rotations of, or if a rotation in does not generate the cyclic subgroup of rotations of, then S does not generate all of, a contradiction. Thus any 2-element generating set must have one of the following two forms:

1) for reflections and, and rotations and or orders and, respectively.

We will show that

is a Hamiltonian arc sequence in. Each element of may be written uniquely in the form where, , and. For convenience, we will represent the element by the ordered string. Following the arc labelled from a given vertex of the digraph increases the value of by 1 modulo if, decreases the value of by 1 modulo if, fixes the value of, increases the value of by 1 modulo 2, and fixes the value of. Similarly, following the arc labelled from a given vertex of the digraph increases the value of by 1 modulo if, decreases the value of by 1 modulo if, fixes the value of, increases the value of by 1 modulo 2, and fixes the value of.

Starting for the identity vertex 0000 and following the sequence, we form a path which visits each vertex of the form, for which and have the same parity, exactly once. Now following we extend this path to visit each vertex of the form where. Again following, we visit each vertex of the form where. Continuing in this way, starting from 0000 and following the arc sequence, we form a path which visits each vertex of the form where both and. Starting from the last vertex on this path and following the arc sequence, we visit each vertex of the form where and. In total, starting from the identity vertex 0000 and following arc sequence, we form a path which visits each vertex of the form exactly once, where. Now following arc from the last vertex on this path, we land on the first vertex of our path of the form with. Now repeating the arc sequence, we visit each vertex of the form with exactly once, and finish on the vertex. Thus we have formed a Hamilton path. Finally, following arc, we land back on the identity vertex 0000. Hence the complete arc sequenceis a Hamiltonian arc sequence. This arc sequence is shown in

2) for reflections and, and rotations and or orders and, respectively.

In this case, we show that

is a Hamiltonian arc sequence in. First we will prove that this arc sequence traces out a walk in which visits all vertices, then we will show that this walk is closed. Note that each vertex of this graph can be written uniquely in the form where, and.

To see that the arc sequence visits all vertices of the digraph, notice that starting at any vertex and following arc sequence, we visit all vertices in the coset of which contains. Hence, starting from, if the vertex reached by arc sequence and the vertex reached by arc sequence lie in different cosets of whenever, we can conclude that the arc sequence traces out a walk which visits all vertices of the digraph. Suppose, for the sake of contradiction, that and lie in the same coset of. We have

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It is easy to see that traveling by a sequence of -arcs doesn’t change the exponent of. Also, each time a vertex travels by an a-arc, its first coordinate alternates between and, and each time a vertex travels by the arc sequence, its second coordinate alternates between and. Hence

and

.

Since and are in the same coset of, can be reached from through a sequence of -arcs, which does not change the exponent of in the second coordinate. Thus

and so

Also, since the exponent of in and are equal, can be reached from by a sequence of aarcs of even length. This implies that

and thus

Since, this implies, so

But then since is odd we have

which contradicts (1). We conclude that and lie in different cosets of, and so the walk traced by the arc sequence visits every vertex of the digraph.

To show the walk is closed, we choose an initial vertex and observe that

which reduces to the initial vertex v.

Finally, since the arc sequence traces out a walk of length which is closed and visits every vertex of the digraph, we conclude that it is a Hamiltonian arc sequence. This arc sequence is shown in