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International Journal of Combinatorics
Volume 2013 (2013), Article ID 725809, 7 pages
On Cayley Digraphs That Do Not Have Hamiltonian Paths
Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, AB, Canada T1K 3M4
Received 23 June 2013; Accepted 4 November 2013
Academic Editor: Jun-Ming Xu
Copyright © 2013 Dave Witte Morris. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We construct an infinite family of connected, -generated Cayley digraphs that do not have hamiltonian paths, such that the orders of the generators and are unbounded. We also prove that if is any finite group with , then every connected Cayley digraph on has a hamiltonian path (but the conclusion does not always hold when or ).
Definition 1. For a subset of a finite group , the Cayley digraph is the directed graph whose vertices are the elements of and with a directed edge for every and . The corresponding Cayley graph is the underlying undirected graph that is obtained by removing the orientations from all the directed edges.
It has been conjectured that every (nontrivial) connected Cayley graph has a hamiltonian cycle. (See the bibliography of  for some of the literature on this problem.) This conjecture does not extend to the directed case, because there are many examples of connected Cayley digraphs that do not have hamiltonian cycles. In fact, infinitely many Cayley digraphs do not even have a hamiltonian path.
Proposition 2 (attributed to Milnor [2, p. 201]). Assume the finite group is generated by two elements and , such that . If , then the Cayley digraph does not have a hamiltonian path.
The examples in the above proposition are very constrained, because the order of one generator must be exactly 2 and the order of the other generator must be exactly 3. In this note, we provide an infinite family of examples in which the orders of the generators are not restricted in this way. In fact, and can both be of arbitrarily large order.
Theorem 3. For any , there is a connected Cayley digraph , such that(1) does not have a hamiltonian path,(2) and both have order greater than .Furthermore, if is any prime number such that and , then we may construct the example so that the commutator subgroup of has order . More precisely, is a semidirect product of two cyclic groups, so is metacyclic.
Remark 4. Here are some related open questions and other comments.(1)The above results show that connected Cayley digraphs on solvable groups do not always have hamiltonian paths. On the other hand, it is an open question whether connected Cayley digraphs on nilpotent groups always have hamiltonian paths. (See  for recent results on the nilpotent case.)(2)The above results always produce a digraph with an even number of vertices. Do there exist infinitely many connected Cayley digraphs of odd order that do not have hamiltonian paths?(3)We conjecture that the assumption “ (mod 4)” can be eliminated from the statement of Theorem 3. On the other hand, it is necessary to require that (see Corollary 16).(4)If is abelian, then it is easy to show that every connected Cayley digraph on has a hamiltonian path. However, some abelian Cayley digraphs do not have a hamiltonian cycle. See Section 5 for more discussion of this.(5)The proof of Theorem 3 appears in Section 3, after some preliminaries in Section 2.
We recall some standard notation, terminology, and basic facts.
Notation. Let be a group, and let be a subgroup of . (All groups in this paper are assumed to be finite.)(i) is the identity element of ;(ii), for ;(iii)we write to say that is a normal subgroup of ;(iv) is the normal closure of in , so .
Definition 5. Let be a subset of the group .(i) is the arc-forcing subgroup, where .(ii)For any , is called the terminal coset. (This is independent of the choice of .)(iii)Any left coset of that is not the terminal coset is called a regular coset.(iv)For and , we use to denote the walk in that visits (in order) the vertices We usually omit the prefix when . Also, we often abuse notation when sequences are to be concatenated. For example,
Remark 6. Here are two observations about the arc-forcing subgroup.(1)It is important to note that , for every . Furthermore, we have , for every .(2)It is sometimes more convenient to define the arc-forcing subgroup to be , instead of (e.g., this is the convention used in [3, p. 42]). The difference is minor, because the two subgroups are conjugate: for any , we have
Definition 7. Suppose is a hamiltonian path in a Cayley digraph and .(i)A vertex travels by if contains the directed edge .(ii)A subset of travels by if every element of travels by .
Lemma 8 (see Housman [4, p. 82]). Suppose is a hamiltonian path in , with initial vertex , and let be the arc-forcing subgroup. Then,(1)the terminal vertex of belongs to the terminal coset ,(2)each regular coset either travels by or travels by .
3. Proof of Theorem 3
Let(i) be an even number that is relatively prime to , with ;(ii) a multiple of that is relatively prime to , with ;(iii) a generator of ;(iv) a generator of ;(v) a generator of ;(vi) a primitive root modulo ;(vii), where and ;(viii), so , and inverts ;(ix), so , and acts on via an automorphism of order ;(x). Suppose is a hamiltonian path in . This will lead to a contradiction.
It is well known (and easy to see) that Cayley digraphs are vertex-transitive, so there is no harm in assuming that the initial vertex of is . Note that(i)the terminal coset is ;(ii)since , we have .
Case 1. Assume at most one regular coset travels by in . Choose , such that is a regular coset, and assume it is the coset that travels by , if such exists.
For , let Letting , we have so Therefore , so we can choose two cosets and that do not belong to .
Recall that, by definition, is not the terminal coset , so is a nontrivial element of . Then, since , we can choose some , such that . Now, since we may multiply on the left by to see that Therefore, no element of is either the terminal coset or the regular coset that travels by . This means that every coset in travels by , so contains the cycle , which contradicts the fact that is a (hamiltonian) path.
Case 2. Assume at least two regular cosets travel by in . Let and be two regular cosets that both travel by . Since , we can choose some , such that .
Note that travels by , for every .(i)If is even, then so travels by .(ii)If is odd, then so travels by . Therefore contains the cycle , which contradicts the fact that is a (hamiltonian) path.
4. Cyclic Commutator Subgroups of Very Small Order
It is known that if , then every connected Cayley digraph on has a hamiltonian path. (Namely, we have , so is nilpotent, and the conclusion, therefore, follows from Theorem 14(2) below.) In this section, we prove the same conclusion when . We also provide counterexamples to show that the conclusion is not always true when or .
We begin with several lemmas. The first three each provide a way to convert a hamiltonian path in a Cayley digraph on an appropriate subgroup of to a hamiltonian path in a Cayley digraph on all of .
Lemma 9. Assume(i) is a finite group, such that , where is prime and ;(ii) is a generating set for ;(iii), such that ;(iv). If has a hamiltonian path, then has a hamiltonian path.
Proof. Since , we know that is an abelian group, so there is a hamiltonian path in (see Proposition 19 below). Also, by assumption, there is a hamiltonian path in . Then is a hamiltonian path in .
Definition 10. If is a subgroup of , then denotes the digraph whose vertices are the right cosets of in and with a directed edge for each and . Note that if .
Lemma 11 (“Skewed-Generators Argument,” cf. [3, Lem. 2.6], [5, Lem. 5.1]). Assume(i) is a generating set for the group ;(ii) is a subgroup of , such that every connected Cayley digraph on has a hamiltonian path;(iii) is a hamiltonian cycle in ;(iv). Then has a hamiltonian path.
Proof. Since , we know that is connected, so, by assumption, it has a hamiltonian path . Then is a hamiltonian path in .
Lemma 12. Assume(i) is a generating set of , with arc-forcing subgroup ;(ii)there is a hamiltonian path in every connected Cayley digraph on ;(iii)either , or is contained in a unique maximal subgroup of . Then has a hamiltonian path.
Proof. It suffices to show that
for then Lemma 11 provides the desired hamiltonian path in .
If , then every hamiltonian cycle in satisfies (see Remark 6 (1)). Thus, we may assume , so, by assumption, is contained in a unique maximal subgroup of . Since is generated by conjugates of (see Remark 6 (2)), there exist , such that .
We may also assume (since, by assumption, every Cayley digraph on has a hamiltonian path), so, letting , we have the two hamiltonian cycles and in . Since the two products and cannot both belong to . Hence, either or is a hamiltonian cycle in , such that . Since is the unique maximal subgroup of that contains , this implies as desired.
The final hypothesis of the preceding lemma is automatically satisfied when is cyclic of prime-power order.
Lemma 13. If is cyclic of order , where is prime, and is any subgroup of , then either or is contained in a unique maximal subgroup of .
Proof. Note that the normal closure is the (unique) smallest normal subgroup of that contains . Therefore, (since is normal in ). This implies that if is any proper subgroup of that contains , then Therefore, is the unique maximal subgroup of that contains .
The following known result handles the case where is nilpotent.
Theorem 14 (see Morris ). Assume is nilpotent, and generates . If either(1) or(2), where is prime and , then has a hamiltonian path.
We now state the main result of this section.
Theorem 15. Suppose (i) is cyclic of prime-power order,(ii)every element of either centralizes or inverts it. Then every connected Cayley digraph on has a hamiltonian path.
Proof. Let be a generating set for . Write for some and . Since every minimal generating set of has only one element, there exist , such that . Then, by Lemma 9, we may assume .
Let be the arc-forcing subgroup. We may assume , for otherwise we could assume, by induction on , that every connected Cayley digraph on has a hamiltonian path, and then Lemma 12 would apply (since Lemma 13 verifies the remaining hypothesis). So
If and both invert , then centralizes , so is nilpotent. Then Theorem 14 applies.
Therefore, we may now assume that does not invert . Then, by assumption, centralizes . Let , and write , where and . Then and . Since , this implies . Therefore, is a hamiltonian cycle in , so Lemma 11 applies.
Corollary 16. If or , then every connected Cayley digraph on has a hamiltonian path.
Proof. Theorem 15 applies, because the groups and have no nontrivial automorphisms, and inversion is the only nontrivial automorphism of or .
Remark 17 ([6, p. 266]). In the statement of Corollary 16, the assumption that cannot be replaced with the weaker assumption that . For a counterexample, let . Then , but it can be shown without much difficulty that does not have a hamiltonian path when and .
Here is a counterexample when .
Example 18. Let , where . Then , and the Cayley digraph is connected but does not have a hamiltonian path.
Proof. A computer search can confirm the nonexistence of a hamiltonian path very quickly, but, for completeness, we provide a human-readable proof.
Let and . The argument in Case 2 of the proof of Theorem 3 shows that no more than one regular coset travels by in any hamiltonian path. On the other hand, since a hamiltonian path cannot contain any cycle of the form , we know that at least vertices must travel by . Since , this implies that some regular coset travels by . So exactly one regular coset travels by in any hamiltonian path.
For and , let be the spanning subdigraph of in which(i)all vertices have outvalence , except , which has outvalence ;(ii)the vertices in the regular coset travel by ;(iii)a vertex in the terminal coset travels by if ;(iv)all other vertices travel by . An observation of D. Housman [7, Lem. 6.4(b)] tells us that if is a hamiltonian path from to , in which is the regular coset that travels by , then . Thus, from the conclusion of the preceding paragraph, we see that every hamiltonian path (with initial vertex ) must be equal to , for some and .
However, is not a (hamiltonian) path. More precisely, for each possible value of and , the following list displays a cycle that is contained in :(i)if and , (ii)if and , (iii)if and , (iv)if and , (v)if and , (vi)if and , (vii)if and : (viii)if and , Since is never a hamiltonian path, we conclude that does not have a hamiltonian path.
5. Nonhamiltonian Cayley Digraphs on Abelian Groups
When is abelian, it is easy to find a hamiltonian path in .
Proposition 19 (see [6, Thm. 3.1]). Every connected Cayley digraph on any abelian group has a hamiltonian path.
On the other hand, it follows from Lemma 8(2) that sometimes there is no hamiltonian cycle.
Proposition 20 (see Rankin [8, Thm. 4]). Assume is abelian. Then there is a hamiltonian cycle in if and only if there exist , such that , and .
Example 21. If and , then does not have a hamiltonian cycle.
All of the non-hamiltonian Cayley digraphs provided by Proposition 20 are -generated. However, a few -generated examples are also known. Specifically, the following result lists (up to isomorphism) the only known examples of connected, non-hamiltonian Cayley digraphs , such that (and ).
Remark 23. The precise conditions in (2) are (i) either or is odd, (ii) either is even or and are both even, (iii) , (iv) , and (v) .
It is interesting to note that, in the examples provided by Theorem 22, the group is cyclic (either or ), and either(1)one of the generators has order or(2)two of the generators differ by an element of order .S. J. Curran (personal communication) asked whether the constructions could be generalized by allowing to be an abelian group that is not cyclic. We provide a negative answer for case .
Proposition 24. Let be an abelian group (written additively), and let , such that is an element of order . (Also assume consists of three distinct, nontrivial elements of .) If the Cayley digraph is connected, but does not have a hamiltonian cycle, then is cyclic.
Proof. We prove the contrapositive: assume is not cyclic, and we will show that the Cayley digraph has a hamiltonian cycle (if it is connected). The argument is a modification of the proof of [9, Thm. 4.1].
Construct a subdigraph of as in [9, Defn. 4.2], but with in the place of , with in the place of , and with in the place of . (Case 1 is when ; Case 2 is when .) Every vertex of has both invalence and outvalence .
The argument in case 3 of the proof of [9, Thm. ] shows that the Cayley digraph has a hamiltonian cycle if . Therefore, we may assume . On the other hand, we know (because is not cyclic). Since , this implies . Since is not cyclic, this implies that has even order. Also, we may write and for some (unique) and . (Since , it is easy to see that , but we do not need this fact.)
Claim. has an odd number of connected components. Arguing as in the proof of [9, Lem. 4.1] (except that, as before, Case 1 is when and Case 2 is when ), we see that the number of connected components in is Since , we know that one of and is an even multiple of , and the other is an odd multiple. (Otherwise, the difference would be an even multiple of , so it would not generate .) Thus, one of and is even, and the other is odd. So is odd. This establishes the claim if .
We may now assume . This implies that the element has odd order (and must be nontrivial, but we do not need this fact). This means that is an even multiple of , so must be an odd multiple of (since ). Therefore, is odd, which means is odd. This completes the proof of the claim.
Now, if is odd, we can apply a very slight modification of the argument in case 4 of the proof of [9, Thm. ]. (Subcase 4.1 is when and subcase 4.2 is when .) We conclude that has a hamiltonian cycle, as desired.
Finally, if is even, then more substantial modifications to the argument in  are required. For convenience, let . Note that, since is even, the proof of the claim shows that is odd and .
Define as in subcase 4.1 of [9, Thm. ] (with in the place of and replacing with ). Let , and inductively construct, for , an element of , such that is a component of , and all other components are components of . The construction of from is the same as in subcase 4.1, but with replaced by .
We now let and inductively construct, for , an element of , such that is a single component of . Namely, [9, Lem. 4.2] implies there is an element , such that , , and are all in the same component of . Then, for , we see that is a hamiltonian cycle.
The author thanks Stephen J. Curran for asking the question that inspired Proposition 24. The other results in this paper were obtained during a visit to the School of Mathematics and Statistics at the University of Western Australia (partially supported by funds from Australian Research Council Federation Fellowship FF0770915). The author is grateful to colleagues there for making the visit so productive and enjoyable.
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