On Cayley Digraphs That Do Not Have Hamiltonian Paths

Abstract

We construct an infinite family {Cay(𝐺_𝑖;𝑎_𝑖;𝑏_𝑖)} of connected, 2-generated Cayley digraphs that do not have hamiltonian paths, such that the orders of the generators 𝑎_𝑖 and 𝑏_𝑖 are unbounded. We also prove that if 𝐺 is any finite group with |[𝐺,𝐺]|≤3, then every connected Cayley digraph on 𝐺 has a hamiltonian path (but the conclusion does not always hold when |[𝐺,𝐺]|=4 or 5).

1. Introduction

Definition 1. For a subset 𝑆 of a finite group 𝐺, the Cayley digraph  Cay(𝐺;𝑆) is the directed graph whose vertices are the elements of 𝐺 and with a directed edge 𝑔→𝑔𝑠 for every 𝑔∈𝐺 and 𝑠∈𝑆. 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 [1] 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 𝐺 is generated by two elements 𝑎 and 𝑏, such that 𝑎²=𝑏³=𝑒. If |𝐺|≥9|𝑎𝑏²|, then the Cayley digraph 𝐶𝑎𝑦(𝐺;𝑎,𝑏) 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, 𝑎 and 𝑏 can both be of arbitrarily large order.

Theorem 3. For any 𝑛∈ℕ, there is a connected Cayley digraph 𝐶𝑎𝑦(𝐺;𝑎,𝑏), such that(1)𝐶𝑎𝑦(𝐺;𝑎,𝑏) does not have a hamiltonian path,(2)𝑎 and 𝑏 both have order greater than 𝑛.Furthermore, if 𝑝 is any prime number such that 𝑝>3 and 𝑝≡3 (mod4), then we may construct the example so that the commutator subgroup of 𝐺 has order 𝑝. More precisely, 𝐺=ℤ_𝑚⋉ℤ_𝑝 is a semidirect product of two cyclic groups, so 𝐺 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 [3] 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 “𝑝≡3 (mod 4)” can be eliminated from the statement of Theorem 3. On the other hand, it is necessary to require that 𝑝>3 (see Corollary 16).(4)If 𝐺 is abelian, then it is easy to show that every connected Cayley digraph on 𝐺 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.

2. Preliminaries

We recall some standard notation, terminology, and basic facts.

Notation. Let 𝐺 be a group, and let 𝐻 be a subgroup of 𝐺. (All groups in this paper are assumed to be finite.)(i)𝑒 is the identity element of 𝐺;(ii)𝑥^𝑔=𝑔⁻¹𝑥𝑔, for 𝑥,𝑔∈𝐺;(iii)we write 𝐻⊴𝐺 to say that 𝐻 is a normal subgroup of 𝐺;(iv)𝐻^𝐺=⟨ℎ^𝑔∣ℎ∈𝐻,𝑔∈𝐺⟩ is the normal closure of 𝐻 in 𝐺, so 𝐻^𝐺⊴𝐺.

Definition 5. Let 𝑆 be a subset of the group 𝐺.(i)𝐻=⟨𝑆𝑆⁻¹⟩ is the arc-forcing subgroup, where 𝑆𝑆⁻¹={𝑠𝑡⁻¹∣𝑠,𝑡∈𝑆}.(ii)For any 𝑎∈𝑆, 𝑎⁻¹𝐻 is called the terminal coset. (This is independent of the choice of 𝑎.)(iii)Any left coset of 𝐻 that is not the terminal coset is called a regular coset.(iv)For 𝑔∈𝐺 and 𝑠₁,…,𝑠_𝑛∈𝑆, we use [𝑔](𝑠_𝑖)^𝑛_𝑖=1 to denote the walk in Cay(𝐺;𝑆) that visits (in order) the vertices 𝑔,𝑔𝑠₁,𝑔𝑠₁𝑠₂,…,𝑔𝑠₁𝑠₂⋯𝑠_𝑛.(1) We usually omit the prefix [𝑔] when 𝑔=𝑒. Also, we often abuse notation when sequences are to be concatenated. For example, 𝑎⁴,𝑠_𝑖³_𝑖=1,𝑡_𝑗²_𝑗=1=𝑎,𝑎,𝑎,𝑎,𝑠₁,𝑠₂,𝑠₃,𝑡₁,𝑎,𝑎,𝑎,𝑎,𝑠₁,𝑠₂,𝑠₃,𝑡₂.(2)

Remark 6. Here are two observations about the arc-forcing subgroup.(1)It is important to note that ⟨𝑆𝑆⁻¹⟩⊆⟨𝑆𝑔⟩, for every 𝑔∈𝐺. Furthermore, we have ⟨𝑆𝑆⁻¹⟩=⟨𝑆𝑎⁻¹⟩, for every 𝑎∈𝑆.(2)It is sometimes more convenient to define the arc-forcing subgroup to be ⟨𝑆⁻¹𝑆⟩, instead of ⟨𝑆𝑆⁻¹⟩ (e.g., this is the convention used in [3, p. 42]). The difference is minor, because the two subgroups are conjugate: for any 𝑎∈𝑆, we have 𝑆⁻¹𝑆=𝑎⁻¹𝑆=𝑆𝑎^−1𝑎=𝑆𝑆^−1𝑎.(3)

Definition 7. Suppose 𝐿 is a hamiltonian path in a Cayley digraph Cay(𝐺;𝑆) and 𝑠∈𝑆.(i)A vertex 𝑔∈𝐺  travels by  𝑠 if 𝐿 contains the directed edge 𝑔→𝑔𝑠.(ii)A subset 𝑋 of 𝐺  travels by  𝑠 if every element of 𝑋 travels by 𝑠.

Lemma 8 (see Housman [4, p. 82]). Suppose 𝐿 is a hamiltonian path in 𝐶𝑎𝑦(𝐺;𝑎,𝑏), with initial vertex 𝑒, and let 𝐻=⟨𝑎𝑏⁻¹⟩ be the arc-forcing subgroup. Then,(1)the terminal vertex of 𝐿 belongs to the terminal coset 𝑎⁻¹𝐻,(2)each regular coset either travels by 𝑎 or travels by 𝑏.

3. Proof of Theorem 3

Let(i)𝛼 be an even number that is relatively prime to (𝑝−1)/2, with 𝛼>𝑛;(ii)𝛽 a multiple of (𝑝−1)/2 that is relatively prime to 𝛼, with 𝛽>𝑛;(iii)𝑎 a generator of ℤ_𝛼;(iv)𝑏 a generator of ℤ_𝛽;(v)𝑧 a generator of ℤ_𝑝;(vi)𝑟 a primitive root modulo 𝑝;(vii)𝐺=(ℤ_𝛼×ℤ_𝛽)⋉ℤ_𝑝, where 𝑧^𝑎=𝑧⁻¹ and 𝑧^𝑏=𝑧^𝑟2;(viii)𝑎=𝑎𝑧, so |𝑎|=𝛼, and 𝑎 inverts ℤ_𝑝;(ix)𝑏=𝑏𝑧, so |𝑏|=𝛽, and 𝑏 acts on ℤ_𝑝 via an automorphism of order (𝑝−1)/2;(x)𝐻=⟨𝑎𝑏⁻¹⟩=⟨𝑎𝑏⁻¹⟩=ℤ_𝛼×ℤ_𝛽. Suppose 𝐿 is a hamiltonian path in Cay(𝐺;𝑎,𝑏). 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 𝐿 is 𝑒. Note that(i)the terminal coset is 𝑎⁻¹𝐻=𝑧⁻¹𝐻;(ii)since 𝑝≡3(mod4), we have ℤ^×_𝑝=⟨−1,𝑟²⟩.

Case 1. Assume at most one regular coset travels by 𝑎 in 𝐿. Choose 𝑧^′∈ℤ_𝑝, such that 𝑧^′𝐻 is a regular coset, and assume it is the coset that travels by 𝑎, if such exists.
For 𝑔∈𝐺, let ℬ_𝑔=𝑔𝑏^𝑘𝐻∣𝑘∈ℤ.(4) Letting 𝑝^′=(𝑝−1)/2, we have 𝑟^2𝑝′−1+𝑟^2𝑝′−2+⋯+𝑟²¹=𝑟+1_2𝑝′−1𝑟²=𝑟−1^𝑝−1−1𝑟²,−1≡0mod𝑝(5) so 𝑏^(𝑝−1)/2=𝑏𝑧^𝑝′=𝑏^𝑝′𝑧^{(𝑟2)𝑝′−1+(𝑟2)𝑝′−2+⋯+(𝑟2)1+1}=𝑏^𝑝′∈ℤ_𝛼×ℤ_𝛽=𝐻.(6) Therefore #ℬ_𝑒≤(𝑝−1)/2≤𝑝−2, so we can choose two cosets 𝑧^𝑖𝐻 and 𝑧^𝑗𝐻 that do not belong to ℬ_𝑒.
Recall that, by definition, 𝑧^′𝐻 is not the terminal coset 𝑧⁻¹𝐻, so 𝑧^′𝑧 is a nontrivial element of ℤ_𝑝. Then, since ℤ^×_𝑝=⟨−1,𝑟²⟩, we can choose some ℎ∈⟨𝑎,𝑏⟩=𝐻, such that (𝑧^𝑗−𝑖)^ℎ=𝑧^′𝑧. Now, since 𝑧^𝑖𝐻,𝑧^𝑗𝐻∉ℬ_𝑒,𝑧⁻¹ℎ⁻¹𝑧^𝑗−𝑖∈𝑧⁻¹𝑧^{𝑗−𝑖ℎ}𝐻=𝑧⁻¹𝑧^′𝑧𝐻=𝑧^′𝐻,(7) we may multiply on the left by 𝑔=𝑧⁻¹ℎ⁻¹𝑧^−𝑖 to see that 𝑧⁻¹𝐻,𝑧^′𝐻∉ℬ_𝑔.(8) Therefore, no element of ℬ_𝑔 is either the terminal coset or the regular coset that travels by 𝑎. This means that every coset in ℬ_𝑔 travels by 𝑏, so 𝐿 contains the cycle [𝑔](𝑏^𝛽), which contradicts the fact that 𝐿 is a (hamiltonian) path.

Case 2. Assume at least two regular cosets travel by 𝑎 in 𝐿. Let 𝑧^𝑖𝐻 and 𝑧^𝑗𝐻 be two regular cosets that both travel by 𝑎. Since ℤ^×_𝑝=⟨−1,𝑟²⟩, we can choose some ℎ∈⟨𝑎,𝑏⟩=𝐻, such that (𝑧⁻¹)^ℎ=𝑧^𝑗−𝑖.
Note that 𝑧^𝑖ℎ⁻¹𝑎^𝑘 travels by 𝑎, for every 𝑘∈ℤ.(i)If 𝑘=2ℓ is even, then 𝑎^𝑘=(𝑎𝑧)^2ℓ=(𝑎𝑧𝑎𝑧)^ℓ=𝑎²𝑧^𝑎𝑧^ℓ=𝑎²𝑧⁻¹𝑧^ℓ=𝑎^2ℓ∈𝐻,(9) so 𝑧^𝑖ℎ⁻¹𝑎^𝑘∈𝑧^𝑖𝐻 travels by 𝑎.(ii)If 𝑘=2ℓ+1 is odd, then 𝑎^𝑘=(𝑎𝑧)^2ℓ+1=(𝑎𝑧)^2ℓ(𝑎𝑧)=𝑎^2ℓ(𝑎𝑧)=𝑎^𝑘𝑧,(10) so 𝑧^𝑖ℎ⁻¹𝑎^𝑘=𝑧^𝑖ℎ⁻¹𝑎^𝑘𝑧=𝑧^𝑖ℎ⁻¹𝑧⁻¹𝑎^𝑘=𝑧^𝑖𝑧^−1ℎℎ⁻¹𝑎^𝑘∈𝑧^𝑖𝑧^𝑗−𝑖𝐻=𝑧^𝑗𝐻(11) travels by 𝑎. Therefore 𝐿 contains the cycle [𝑧^𝑖ℎ⁻¹](𝑎^𝛼), which contradicts the fact that 𝐿 is a (hamiltonian) path.

4. Cyclic Commutator Subgroups of Very Small Order

It is known that if |[𝐺,𝐺]|=2, then every connected Cayley digraph on 𝐺 has a hamiltonian path. (Namely, we have [𝐺,𝐺]⊆𝑍(𝐺), so 𝐺 is nilpotent, and the conclusion, therefore, follows from Theorem 14(2) below.) In this section, we prove the same conclusion when |[𝐺,𝐺]|=3. We also provide counterexamples to show that the conclusion is not always true when |[𝐺,𝐺]|=4 or |[𝐺,𝐺]|=5.

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 𝐺 to a hamiltonian path in a Cayley digraph on all of 𝐺.

Lemma 9. Assume(i)𝐺 is a finite group, such that [𝐺,𝐺]≅ℤ_𝑝𝑘, where 𝑝 is prime and 𝑘∈ℕ;(ii)𝑆 is a generating set for 𝐺;(iii)𝑎,𝑏∈𝑆, such that ⟨[𝑎,𝑏]⟩=[𝐺,𝐺];(iv)𝑁=⟨𝑎,𝑏⟩. If 𝐶𝑎𝑦(𝑁;𝑎,𝑏) has a hamiltonian path, then 𝐶𝑎𝑦(𝐺;𝑆) has a hamiltonian path.

Proof. Since [𝐺,𝐺]⊆𝑁, we know that 𝐺/𝑁 is an abelian group, so there is a hamiltonian path (𝑠_𝑖)^𝑚_𝑖=1 in Cay(𝐺/𝑁;𝑆) (see Proposition 19 below). Also, by assumption, there is a hamiltonian path (𝑡_𝑗)^𝑛_𝑗=1 in Cay(𝑁;𝑎,𝑏). Then 𝑡_𝑗^𝑛_𝑗=1,𝑠_𝑖^𝑚_𝑖=1,𝑡_𝑗^𝑛_𝑗=1(12) is a hamiltonian path in Cay(𝐺;𝑆).

Definition 10. If 𝐾 is a subgroup of 𝐺, then 𝐾⧵Cay(𝐺;𝑆) denotes the digraph whose vertices are the right cosets of 𝐾 in 𝐺 and with a directed edge 𝐾𝑔→𝐾𝑔𝑠 for each 𝑔∈𝐺 and 𝑠∈𝑆. Note that 𝐾⧵Cay(𝐺;𝑆)=Cay(𝐺/𝐾;𝑆) if 𝐾⊴𝐺.

Lemma 11 (“Skewed-Generators Argument,” cf. [3, Lem. 2.6], [5, Lem. 5.1]). Assume(i)𝑆 is a generating set for the group 𝐺;(ii)𝐾 is a subgroup of 𝐺, such that every connected Cayley digraph on 𝐾 has a hamiltonian path;(iii)(𝑠_𝑖)^𝑛_𝑖=1 is a hamiltonian cycle in 𝐾⧵𝐶𝑎𝑦(𝐺;𝑆);(iv)⟨𝑆𝑠₂𝑠₃⋯𝑠_𝑛⟩=𝐾. Then 𝐶𝑎𝑦(𝐺;𝑆) has a hamiltonian path.

Proof. Since ⟨𝑆𝑠₂𝑠₃⋯𝑠_𝑛⟩=𝐾, we know that Cay(𝐾;𝑆𝑠₂𝑠₃⋯𝑠_𝑛) is connected, so, by assumption, it has a hamiltonian path (𝑡_𝑗𝑠₂𝑠₃⋯𝑠_𝑛)^𝑚_𝑗=1. Then 𝑡_𝑗,𝑠_𝑖^𝑛_𝑖=2^𝑚−1_𝑗=1,𝑡_𝑚,𝑠_𝑖^𝑛−1_𝑖=2(13) is a hamiltonian path in Cay(𝐺;𝑆).

Lemma 12. Assume(i)𝑆 is a generating set of 𝐺, with arc-forcing subgroup 𝐻=⟨𝑆𝑆⁻¹⟩;(ii)there is a hamiltonian path in every connected Cayley digraph on 𝐻^𝐺;(iii)either 𝐻=𝐻^𝐺, or 𝐻 is contained in a unique maximal subgroup of 𝐻^𝐺. Then 𝐶𝑎𝑦(𝐺;𝑆) has a hamiltonian path.

Proof. It suffices to show that thereexistsahamiltoniancycle𝑠_𝑖^𝑛_𝑖=1𝐺inCay𝐻^𝐺;𝑆,suchthat𝐻^𝐺=𝑆𝑠₂⋯𝑠_𝑛,(∗) for then Lemma 11 provides the desired hamiltonian path in Cay(𝐺;𝑆).
If 𝐻^𝐺=𝐻, then every hamiltonian cycle in Cay(𝐺/𝐻^𝐺;𝑆) satisfies (∗) (see Remark 6 (1)). Thus, we may assume 𝐻^𝐺≠𝐻, so, by assumption, 𝐻 is contained in a unique maximal subgroup 𝑀 of 𝐻^𝐺. Since 𝐻^𝐺 is generated by conjugates of 𝑆⁻¹𝑆 (see Remark 6 (2)), there exist 𝑎,𝑏,𝑐∈𝑆, such that (𝑎⁻¹𝑏)^𝑐∉𝑀.
We may also assume 𝐻^𝐺≠𝐺 (since, by assumption, every Cayley digraph on 𝐻^𝐺 has a hamiltonian path), so, letting 𝑛=|𝐺∶𝐻^𝐺|≥2, we have the two hamiltonian cycles (𝑎^𝑛−1,𝑐) and (𝑎^𝑛−2,𝑏,𝑐) in Cay(𝐺/𝐻^𝐺;𝑆). Since 𝑎^𝑛−1𝑐⁻¹𝑎^𝑛−2𝑏𝑐=𝑎⁻¹𝑏^𝑐∉𝑀,(14) the two products 𝑎^𝑛−1𝑐 and 𝑎^𝑛−2𝑏𝑐 cannot both belong to 𝑀. Hence, either (𝑎^𝑛−1,𝑐) or (𝑎^𝑛−2,𝑏,𝑐) is a hamiltonian cycle (𝑠_𝑖)^𝑛_𝑖=1 in Cay(𝐺/𝐻^𝐺;𝑆), such that 𝑠₁𝑠₂⋯𝑠_𝑛∉𝑀. Since 𝑀 is the unique maximal subgroup of 𝐻^𝐺 that contains 𝐻, this implies 𝐻^𝐺=𝐻,𝑠₁𝑠₂⋯𝑠_𝑛=𝑆𝑠₂𝑠₃⋯𝑠_𝑛,(15) as desired.

The final hypothesis of the preceding lemma is automatically satisfied when [𝐺,𝐺] is cyclic of prime-power order.

Lemma 13. If [𝐺,𝐺] is cyclic of order 𝑝^𝑘, where 𝑝 is prime, and 𝐻 is any subgroup of 𝐺, then either 𝐻=𝐻^𝐺 or 𝐻 is contained in a unique maximal subgroup of 𝐻^𝐺.

Proof. Note that the normal closure 𝐻^𝐺 is the (unique) smallest normal subgroup of 𝐺 that contains 𝐻. Therefore, 𝐻^𝐺⊆𝐻[𝐺,𝐺] (since 𝐻[𝐺,𝐺] is normal in 𝐺). This implies that if 𝑀 is any proper subgroup of 𝐻^𝐺 that contains 𝐻, then 𝑀=𝐻⋅(𝑀∩[𝐺,𝐺])⊆𝐻⋅𝐻^𝐺∩[𝐺,𝐺]^𝑝.(16) Therefore, 𝐻⋅(𝐻^𝐺∩[𝐺,𝐺])^𝑝 is the unique maximal subgroup of 𝐻^𝐺 that contains 𝑀.

The following known result handles the case where 𝐺 is nilpotent.

Theorem 14 (see Morris [3]). Assume 𝐺 is nilpotent, and 𝑆 generates 𝐺. If either(1)#𝑆≤2 or(2)|[𝐺,𝐺]|=𝑝^𝑘, where 𝑝 is prime and 𝑘∈ℕ, then 𝐶𝑎𝑦(𝐺;𝑆) has a hamiltonian path.

We now state the main result of this section.

Theorem 15. Suppose (i)[𝐺,𝐺] is cyclic of prime-power order,(ii)every element of 𝐺 either centralizes [𝐺,𝐺] or inverts it. Then every connected Cayley digraph on 𝐺 has a hamiltonian path.

Proof. Let 𝑆 be a generating set for 𝐺. Write [𝐺,𝐺]=ℤ_𝑝𝑘 for some 𝑝 and 𝑘. Since every minimal generating set of ℤ_𝑝𝑘 has only one element, there exist 𝑎,𝑏∈𝑆, such that ⟨[𝑎,𝑏]⟩=[𝐺,𝐺]. Then, by Lemma 9, we may assume 𝑆={𝑎,𝑏}.
Let 𝐻=⟨𝑏𝑎⁻¹⟩ be the arc-forcing subgroup. We may assume 𝐻^𝐺=𝐺, for otherwise we could assume, by induction on |𝐺|, that every connected Cayley digraph on 𝐻^𝐺 has a hamiltonian path, and then Lemma 12 would apply (since Lemma 13 verifies the remaining hypothesis). So 𝐻ℤ_𝑝𝑘=𝐻[𝐺,𝐺]⊃𝐻^𝐺=𝐺.(17)
If 𝑎 and 𝑏 both invert ℤ_𝑝𝑘, then 𝐻=⟨𝑏𝑎⁻¹⟩ centralizes ℤ_𝑝𝑘=[𝐺,𝐺], so 𝐺 is nilpotent. Then Theorem 14 applies.
Therefore, we may now assume that 𝑎 does not invert ℤ_𝑝𝑘. Then, by assumption, 𝑎 centralizes ℤ_𝑝𝑘. Let 𝑛=|𝐺∶𝐻|, and write 𝑎=𝑎𝑧, where 𝑎∈𝐻 and 𝑧∈ℤ_𝑝𝑘. Then 𝑎=𝑎𝑧∈𝐻𝑧 and 𝑏=(𝑏𝑎⁻¹)(𝑎𝑧)∈𝐻𝑧. Since ⟨𝑎,𝑏⟩=𝐺, this implies 𝐻⟨𝑧⟩=𝐺. Therefore, [𝐻]𝑎^𝑛=𝐻,𝐻𝑧,𝐻𝑧²,…,𝐻𝑧^𝑛−1,𝐻(18) is a hamiltonian cycle in 𝐻⧵Cay(𝐺;𝑆), so Lemma 11 applies.

Corollary 16. If |[𝐺,𝐺]|≤3 or [𝐺,𝐺]≅ℤ₄, then every connected Cayley digraph on 𝐺 has a hamiltonian path.

Proof. Theorem 15 applies, because the groups {𝑒} and ℤ₂ have no nontrivial automorphisms, and inversion is the only nontrivial automorphism of ℤ₃ or ℤ₄.

Remark 17 ([6, p. 266]). In the statement of Corollary 16, the assumption that [𝐺,𝐺]≅ℤ₄ cannot be replaced with the weaker assumption that |[𝐺,𝐺]|=4. For a counterexample, let 𝐺=𝐴₄×ℤ₂. Then |[𝐺,𝐺]|=4, but it can be shown without much difficulty that Cay(𝐺;𝑎,𝑏) does not have a hamiltonian path when 𝑎=((12)(34),1) and 𝑏=((123),0).

Here is a counterexample when |[𝐺,𝐺]|=5.

Example 18. Let 𝐺=ℤ₁₂⋉ℤ₅=⟨ℎ⟩⋉⟨𝑧⟩, where 𝑧^ℎ=𝑧³. Then |[𝐺,𝐺]|=5, and the Cayley digraph Cay(𝐺;ℎ²𝑧,ℎ³𝑧) 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 𝑎=ℎ²𝑧=𝑧⁴ℎ² and 𝑏=ℎ³𝑧=𝑧³ℎ³. The argument in Case 2 of the proof of Theorem 3 shows that no more than one regular coset travels by 𝑎 in any hamiltonian path. On the other hand, since a hamiltonian path cannot contain any cycle of the form [𝑔](𝑏⁴), we know that at least ⌊(|𝐺|−1)/4⌋=14 vertices must travel by 𝑎. Since |𝑎𝑏⁻¹|=12<14, this implies that some regular coset travels by 𝑎. So exactly one regular coset travels by 𝑎 in any hamiltonian path.
For 0≤𝑖≤3 and 0≤𝑚≤11, let 𝐿_𝑖,𝑚 be the spanning subdigraph of Cay(𝐺;𝑎,𝑏) in which(i)all vertices have outvalence 1, except 𝑏⁻¹(𝑎𝑏⁻¹)^𝑚=𝑧⁴ℎ^9−𝑚, which has outvalence 0;(ii)the vertices in the regular coset 𝑧^𝑖𝐻 travel by 𝑎;(iii)a vertex 𝑏⁻¹ℎ^−𝑗=𝑧⁴ℎ^9−𝑗 in the terminal coset travels by 𝑎 if 0≤𝑗<𝑚;(iv)all other vertices travel by 𝑏. An observation of D. Housman [7, Lem. 6.4(b)] tells us that if 𝐿 is a hamiltonian path from 𝑒 to 𝑏⁻¹(𝑎𝑏⁻¹)^𝑚, in which 𝑧^𝑖𝐻 is the regular coset that travels by 𝑎, then 𝐿=𝐿_𝑖,𝑚. Thus, from the conclusion of the preceding paragraph, we see that every hamiltonian path (with initial vertex 𝑒) must be equal to 𝐿_𝑖,𝑚, for some 𝑖 and 𝑚.
However, 𝐿_𝑖,𝑚 is not a (hamiltonian) path. More precisely, for each possible value of 𝑖 and 𝑚, the following list displays a cycle that is contained in 𝐿_𝑖,𝑚:(i)if 𝑖=0 and 0≤𝑚≤8, 𝑧²ℎ⟶^3𝑏𝑧ℎ⟶^6𝑏𝑧³ℎ⟶^9𝑏𝑧⟶^4𝑏𝑧²ℎ³;(19)(ii)if 𝑖=0 and 9≤𝑚≤11, ℎ⟶^2𝑎𝑧ℎ⟶^4𝑏𝑧⁴ℎ⟶^7𝑎𝑧ℎ⟶^9𝑏𝑧²_𝑏⟶ℎ⟶^3𝑎𝑧²ℎ⟶^5𝑏𝑧³ℎ⟶^8𝑏𝑧ℎ^𝑏11⟶ℎ²;(20)(iii)if 𝑖=1 and 0≤𝑚≤7, ℎ⟶^4𝑏𝑧³ℎ⟶^7𝑏𝑧²ℎ^𝑏10⟶𝑧⁴ℎ⟶^𝑏ℎ⁴;(21)(iv)if 𝑖=1 and 8≤𝑚≤11, ℎ⟶^𝑏𝑧ℎ⟶^4𝑎ℎ⟶^6𝑏𝑧²ℎ⟶^9𝑏𝑧⟶^3𝑏𝑧ℎ³_𝑎⟶𝑧³ℎ⟶^5𝑏𝑧⁴ℎ⟶^8𝑎𝑧³ℎ^𝑏10⟶ℎ;(22)(v)if 𝑖=2 and 0≤𝑚≤9, ℎ⟶^5𝑏𝑧ℎ⟶^8𝑏𝑧⁴ℎ^𝑏11⟶𝑧³ℎ⟶^2𝑏ℎ⁵;(23)(vi)if 𝑖=2 and 10≤𝑚≤11, 𝑧²ℎ⟶^3𝑎𝑧⁴ℎ⟶^5𝑎𝑧²ℎ⟶^7𝑎𝑧⁴ℎ⁹_𝑎⟶𝑧²ℎ^𝑎11⟶𝑧⁴ℎ⟶^𝑎𝑧²ℎ³;(24)(vii)if 𝑖=3 and 0≤𝑚≤10: ℎ⟶^7𝑏𝑧⁴ℎ^𝑏10⟶𝑧ℎ⟶^𝑏𝑧²ℎ⟶^4𝑏ℎ⁷;(25)(viii)if 𝑖=3 and 𝑚=11, 𝑧³ℎ⟶^2𝑎𝑧⁴ℎ⟶^4𝑎𝑧³ℎ⟶^6𝑎𝑧⁴ℎ⁸_𝑎⟶𝑧³ℎ^𝑎10⟶𝑧⟶^4𝑎𝑧³ℎ².(26)Since 𝐿_𝑖,𝑚 is never a hamiltonian path, we conclude that Cay(𝐺;𝑎,𝑏) does not have a hamiltonian path.

5. Nonhamiltonian Cayley Digraphs on Abelian Groups

When 𝐺 is abelian, it is easy to find a hamiltonian path in Cay(𝐺;𝑆).

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 𝐺=⟨𝑎,𝑏⟩ is abelian. Then there is a hamiltonian cycle in 𝐶𝑎𝑦(𝐺;𝑎,𝑏) if and only if there exist 𝑘,ℓ≥0, such that ⟨𝑎^𝑘𝑏^ℓ⟩=⟨𝑎𝑏⁻¹⟩, and 𝑘+ℓ=|𝐺∶⟨𝑎𝑏⁻¹⟩|.

Example 21. If gcd(𝑎,𝑛)>1 and gcd(𝑎+1,𝑛)>1, then Cay(ℤ_𝑛;𝑎,𝑎+1) does not have a hamiltonian cycle.

All of the non-hamiltonian Cayley digraphs provided by Proposition 20 are 2-generated. However, a few 3-generated examples are also known. Specifically, the following result lists (up to isomorphism) the only known examples of connected, non-hamiltonian Cayley digraphs Cay(𝐺;𝑆), such that #𝑆>2 (and 𝑒∉𝑆).

Theorem 22 (see Locke and Witte [9]). The following Cayley digraphs do not have hamiltonian cycles: (1)𝐶𝑎𝑦(ℤ_12𝑘;6𝑘,6𝑘+2,6𝑘+3), for any 𝑘∈ℤ⁺;(2)𝐶𝑎𝑦(ℤ_2𝑘;𝑎,𝑏,𝑏+𝑘), for 𝑎,𝑏,𝑘∈ℤ⁺, such that certain technical conditions (Remark 23) are satisfied.

Remark 23. The precise conditions in (2) are (i) either 𝑎 or 𝑘 is odd, (ii) either 𝑎 is even or 𝑏 and 𝑘 are both even, (iii) gcd(𝑎−𝑏,𝑘)=1, (iv) gcd(𝑎,2𝑘)≠1, and (v) gcd(𝑏,𝑘)≠1.

It is interesting to note that, in the examples provided by Theorem 22, the group 𝐺 is cyclic (either ℤ_12𝑘 or ℤ_2𝑘), and either(1)one of the generators has order 2 or(2)two of the generators differ by an element of order 2.S. J. Curran (personal communication) asked whether the constructions could be generalized by allowing 𝐺 to be an abelian group that is not cyclic. We provide a negative answer for case (2).

Proposition 24. Let 𝐺 be an abelian group (written additively), and let 𝑎,𝑏,𝑘∈𝐺, such that 𝑘 is an element of order 2. (Also assume {𝑎,𝑏,𝑏+𝑘} consists of three distinct, nontrivial elements of 𝐺.) If the Cayley digraph 𝐶𝑎𝑦(𝐺;𝑎,𝑏,𝑏+𝑘) is connected, but does not have a hamiltonian cycle, then 𝐺 is cyclic.

Proof. We prove the contrapositive: assume 𝐺 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 𝐻₀ of 𝐺 as in [9, Defn. 4.2], but with 𝐺 in the place of ℤ_2𝑘, with |𝐺| in the place of 2𝑘, and with |𝑎| in the place of 𝑑. (Case 1 is when 𝑘∉⟨𝑎⟩; Case 2 is when 𝑘∈⟨𝑎⟩.) Every vertex of 𝐻₀ has both invalence 1 and outvalence 1.
The argument in case 3 of the proof of [9, Thm. 4.1(⇐)] shows that the Cayley digraph Cay(𝐺;𝑎,𝑏,𝑏+𝑘) has a hamiltonian cycle if ⟨𝑎−𝑏,𝑘⟩≠𝐺. Therefore, we may assume ⟨𝑎−𝑏,𝑘⟩=𝐺. On the other hand, we know ⟨𝑎−𝑏⟩≠𝐺 (because 𝐺 is not cyclic). Since |𝑘|=2, this implies 𝐺=⟨𝑎−𝑏⟩⊕⟨𝑘⟩. Since 𝐺 is not cyclic, this implies that 𝑎−𝑏 has even order. Also, we may write 𝑎=𝑎^′+𝑘^′ and 𝑏=𝑏^′+𝑘^′′ for some (unique) 𝑎^′,𝑏^′∈⟨𝑎−𝑏⟩ and 𝑘^′,𝑘^′′∈⟨𝑘⟩. (Since 𝑎^′−𝑏^′∈⟨𝑎−𝑏⟩, it is easy to see that 𝑘^′=𝑘^′′, but we do not need this fact.)
Claim. 𝐻₀ 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 𝑘∉⟨𝑎⟩ and Case 2 is when 𝑘∈⟨𝑎⟩), we see that the number of connected components in 𝐻₀ is |⟨+|⟨⟨𝑎⟩,|⟨⟨𝑎⟩.𝐺∶𝑎,𝑘⟩|𝐺∶𝑏,𝑘⟩|if𝑘∉𝐺∶𝑏,𝑘⟩|if𝑘∈(27) Since ⟨𝑎^′−𝑏^′⟩=⟨𝑎−𝑏⟩, we know that one of 𝑎^′ and 𝑏^′ is an even multiple of 𝑎−𝑏, and the other is an odd multiple. (Otherwise, the difference would be an even multiple of 𝑎−𝑏, so it would not generate ⟨𝑎−𝑏⟩.) Thus, one of |𝐺∶⟨𝑎,𝑘⟩| and |𝐺∶⟨𝑏,𝑘⟩| is even, and the other is odd. So |𝐺∶⟨𝑎,𝑘⟩|+|𝐺∶⟨𝑏,𝑘⟩| is odd. This establishes the claim if 𝑘∉⟨𝑎⟩.
We may now assume 𝑘∈⟨𝑎⟩. This implies that the element 𝑎^′ has odd order (and 𝑘^′ must be nontrivial, but we do not need this fact). This means that 𝑎^′ is an even multiple of 𝑎−𝑏, so 𝑏^′ must be an odd multiple of 𝑎−𝑏 (since ⟨𝑎^′−𝑏^′⟩=⟨𝑎−𝑏⟩). Therefore, |⟨𝑎−𝑏⟩∶⟨𝑏^′⟩| is odd, which means |𝐺∶⟨𝑏,𝑘⟩| is odd. This completes the proof of the claim.
Now, if |𝐺∶⟨𝑏,𝑘⟩| is odd, we can apply a very slight modification of the argument in case 4 of the proof of [9, Thm. 4.1(⇐)]. (Subcase 4.1 is when 𝑘∉⟨𝑎⟩ and subcase 4.2 is when 𝑘∈⟨𝑎⟩.) We conclude that Cay(𝐺;𝑎,𝑏,𝑏+𝑘) has a hamiltonian cycle, as desired.
Finally, if |𝐺∶⟨𝑏,𝑘⟩| is even, then more substantial modifications to the argument in [9] are required. For convenience, let 𝑚=|𝐺∶⟨𝑎,𝑘⟩|. Note that, since |𝐺∶⟨𝑏,𝑘⟩| is even, the proof of the claim shows that 𝑚 is odd and 𝑘∉⟨𝑎⟩.
Define 𝐻^′₀ as in subcase 4.1 of [9, Thm. 4.1(⇐)] (with 𝐺 in the place of ℤ_2𝑘 and replacing gcd(𝑏,𝑘) with |𝐺∶⟨𝑏,𝑘⟩|). Let 𝐻₁=𝐻^′₀, and inductively construct, for 1≤𝑖≤(𝑚+1)/2, an element 𝐻_𝑖 of ℰ, such that 𝑣∣𝑧_𝑣=0,0≤𝑦_𝑣∪≤2𝑖−2𝑣∣𝑧_𝑣=1,𝑥_𝑣=0or1(mod|𝐺∶⟨𝑏,𝑘⟩|)(28) is a component of 𝐻_𝑖, and all other components are components of 𝐻₀. The construction of 𝐻_𝑖 from 𝐻_𝑖−1 is the same as in subcase 4.1, but with 2𝑖 replaced by 2𝑖−1.
We now let 𝐾₁=𝐻_(𝑚+1)/2 and inductively construct, for 1≤𝑖≤|𝐺∶⟨𝑏,𝑘⟩|/2, an element 𝐾_𝑖 of ℰ, such that 𝑣∣𝑧_𝑣∪=0𝑣∣𝑧_𝑣=1,𝑥_𝑣≡0,1,…,or2𝑖−1(mod|𝐺∶⟨𝑏,𝑘⟩|)(29) is a single component of 𝐾_𝑖. Namely, [9, Lem. 4.2] implies there is an element 𝐾_𝑖=𝐾^′_𝑖−1, such that (2𝑖−2)𝑎, (2𝑖−2)𝑎+𝑘, and (2𝑖−1)𝑎+𝑘 are all in the same component of 𝐾_𝑖. Then, for 𝑖=|𝐺∶⟨𝑏,𝑘⟩|/2, we see that 𝐾_𝑖 is a hamiltonian cycle.

Acknowledgments

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.