Automorphisms of the double cover of a circulant graph of valency at most 7

A graph $X$ is said to be unstable if the direct product $X \times K_2$ (also called the canonical double cover of $X$) has automorphisms that do not come from automorphisms of its factors $X$ and $K_2$. It is nontrivially unstable if it is unstable, connected, and non-bipartite, and no two distinct vertices of X have exactly the same neighbors. We find all of the nontrivially unstable circulant graphs of valency at most $7$. (They come in several infinite families.) We also show that the instability of each of these graphs is explained by theorems of Steve Wilson. This is best possible, because there is a nontrivially unstable circulant graph of valency $8$ that does not satisfy the hypotheses of any of Wilson's four instability theorems for circulant graphs.


Introduction
Let X be a circulant graph. (All graphs in this paper are finite, simple, and undirected.) Theorem 1.4 (Wilson [17,Appendix A.1] (and [14, p. 156])). Let X = Cay(Z n , S) be a circulant graph, such that n is even. Let S e = S ∩ 2Z n and S o = S \ S e . If any of the following conditions is true, then X is unstable.
(C.1) There is a nonzero element h of 2Z n , such that h + S e = S e . (C.2 ) n is divisible by 4, and there exists h ∈ 1 + 2Z n , such that (a) 2h + S o = S o , and (b) for each s ∈ S, such that s ≡ 0 or −h (mod 4), we have s + h ∈ S. (C.3 ) There is a subgroup H of Z n , such that the set The main result of this paper establishes that Wilson's conjecture is true for graphs of valency at most 7: Theorem 1.9. Every nontrivially unstable circulant graph of valency at most 7 has Wilson type (C.1), (C.2 ), (C. 3 ), or (C.4).
We actually prove more precise (but more complicated) results, which show that all of the graphs in Theorem 1.9 belong to certain explicit families (and it is easy to see that the graphs in each family have a specific Wilson type). For example: Theorem 5.1. A circulant graph Cay(Z n , S) of valency 5 is unstable if and only if either it is trivially unstable, or it is one of the following: (1) Cay(Z 12k , {±s, ±2k, 6k}) with s odd, which has Wilson type (C.1).
(2) Cay(Z 8 , {±1, ±3, 4}), which has Wilson type (C.3 ), The following example shows that the constant 7 in Theorem 1.9 cannot be increased: Then the circulant graph X := Cay(Z n , S) has valency 8 and is nontrivially unstable, but does not have a Wilson type.
Here is an outline of the paper. After this introduction come two sections of preliminaries: Section 2 presents material from the theory of normal Cayley graphs and some other miscellaneous information that will be used; Section 3 lists some conditions that imply X is not unstable. The remaining sections each find the nontrivially unstable circulant graphs of a particular valency (or valencies). Namely, Section 4 considers valencies ≤ 4, whereas Sections 5 to 7 each consider a single valency (5, 6, or 7, respectively). The main results are Proposition 4.2 (valency ≤ 3), Theorem 4.3 (valency 4), Theorem 5.1 (valency 5), Theorem 6.1 and Corollary 6.8 (valency 6), and Theorem 7.1 (valency 7).
2A. Basic definitions and notation. For simplicity (and because it is the only case we need), the following definition is restricted to abelian groups, even though the notions easily generalize to nonabelian groups. Definition 2.2. Let S be a subset of an abelian group G, such that −s ∈ S for all s ∈ S.
(1) The Cayley graph Cay(G, S) is the graph whose vertices are the elements of G, and with an edge from v to w if and only if w = v + s for some s ∈ S (cf. [10, §1]). (2) For g ∈ G, we letg = (g, 1).
(4) For g ∈ G, we say that an edge {u, v} of the complete graph on G × Z 2 is a g-edge if v = u ±g. Note that {u, v} is an edge of BX if and only if it is an s-edge for some s ∈ S.
Notation 2.3. For convenience, proofs will sometimes use the following abbreviation: n = n/2.
We will employ the following fairly standard notation: Notation 2.4.
(1) K n is the complete graph on n vertices.
(2) K n is the complement of K n . It is a graph with n vertices and no edges. (3) K n,n is the complete bipartite graph with n vertices in each bipartition set. (4) C n is the cycle of length n. (5) For a ∈ Z n , we use |a| to denote the order of a as an element of the cyclic group Z n . So |a| = n gcd(n, a) .
It does not denote the absolute value of a. (6) φ denotes the Euler's totient function.
Throughout the paper various notions of graph products will be used. We now recall their definitions and notation. . Let X and Y be graphs.
(3) The wreath product X Y is the graph that is obtained by replacing each vertex of X with a copy of Y . (Vertices in two different copies of Y are adjacent in X Y if and only if the corresponding vertices of X are adjacent in X.) This is called the lexicographic product in [5, p. 43] (and denoted X • Y ).
Definition 2.6 (Kotlov-Lovász [8]). A graph X is twin-free if there do not exist two distinct vertices v and w, such that N X (v) = N X (w), where N X (v) denotes the set of neighbors of v in X.
The notion of a "block" (or "block of imprimitivity") is a fundamental concept in the theory of permutation groups, but we need only the following special case: Definition 2.7 (cf. [2, pp. 12-13]). Let G be a finite abelian group. Let X = Cay(G, S) be a Cayley graph. A nonempty subset B of V (X) is a block for the action of Aut X if, for every α ∈ Aut X, we have either α(B) = B or α(B) ∩ B = ∅.
It is easy to see that this implies B is a coset of some subgroup H of G. Then every coset of H is a block. Indeed, the action of Aut X permutes see note A.1 these cosets, so there is a natural action of Aut X on the set of cosets.
Remark 2.8. The most important instance of Definition 2.7 for us will be the case of canonical bipartite double covers. Indeed, if X = Cay(Z n , S) is a circulant graph then its canonical bipartite double cover BX = Cay(Z n × Z 2 , S), defined in Definition 2.2(3), is a Cayley graph. Therefore, every block for the action of Aut BX is a coset of some subgroup H of Z n × Z 2 .
2B. Normal Cayley graphs. Definition 2.9. (M.-Y. Xu [18,Defn. 1.4]) For each g ∈ G, it is easy to see that the translation g * , defined by g * (x) = g + x, is an automorphism of Cay(G, S). The set is a subgroup of Aut Cay(G, S). (It is often called the regular representation of G.) We say that Cay(G, S) is normal if the subgroup G * is normal in Aut Cay(G, S). This means that if ϕ is automorphism of the graph Cay(G, S), and ϕ(0) = 0, then ϕ is an automorphism of the group G.
Proof. Since X is unstable, we know that (0, 1) is not central in Aut BX. So it is conjugate to some other element of order 2. However, since BX is normal, we know that Z n × Z 2 is a normal subgroup of Aut BX. Therefore, (0, 1) cannot be the only element of order 2 in Z n × Z 2 ; so n is even. (This is also immediate from Theorem 3.1.) Also note that Z n × {0} is normal in Aut BX, because it consists of the elements of the normal subgroup Z n × Z 2 that preserve each bipartition set. Since (n, 0) is the unique element of order 2 in this normal subgroup, it must be central in Aut BX. Now, since (0, 1), (n, 0), and (n, 1) are the only elements of order 2 in Z n × Z 2 , and (n, 0) is not conjugate to any other elements, we see that (0, 1) must be conjugate to (n, 1), by some α ∈ Aut BX.
Corollary 2.11. If X = Cay(Z n , S) is a nontrivially unstable circulant graph of odd valency, then the Cayley graph BX is not normal.
Proof. If BX is normal, then the Lemma 2.10 implies n + gS = S, for some g ∈ Z × n . Also, since X has odd valency, we know that n ∈ S. Also, we know that g is odd (because n is even). Therefore 0 = n + n = n + gn ∈ n + gS = S. This contradicts our standing assumption that all graphs are simple (no loops). Lemma 2.12. Let X = Cay(Z n , S) be a nontrivially unstable circulant graph of odd valency, and let X 0 = Cay(Z n , S \ {n/2}), so BX 0 = Cay(Z n × Z 2 , (S \ {n/2}) × {1}).
If every automorphism of BX maps n/2-edges to n/2-edges, then BX 0 is not a normal Cayley graph. Moreover, if X 0 is bipartite, then X 0 is not a normal Cayley graph.
Proof. By the assumption on n-edges, it follows that every automorphism of BX induces an automorphism of BX 0 . If BX 0 is normal, it follows that BX is also normal, contradiction with Corollary 2.11. We conclude that BX 0 is non-normal.
Suppose now that X 0 is bipartite. It is not difficult to see that X 0 is connected, since X is connected. It follows that every element of S \ {n} is odd. Since X is nonbipartite, it follows that n is even. Suppose that X 0 is normal Cayley graph. Observe that BX 0 is isomorphic to the disjoint union of two copies of X 0 , and that the connected component containing the vertex (0, 0) is The map θ : Z n → H defined by θ(k) = (k, k mod 2) is an isomorphism between X 0 and X 1 . Then X 1 is a normal Cayley graph on H. Let ϕ be an automorphism of BX that fixes (0, 0). Then ϕ is also an automorphism of BX 0 , and consequently of its connected component X 1 . Since X 1 is normal, it follows that the action of ϕ on H is an automorphism of the group H, hence ϕ fixes the unique element of order 2 in H, which is (n, 0). Observe that the n-edge in BX incident with (n, 0), must also be fixed, hence (0, 1) is fixed. This shows that X is stable, a contradiction. The obtained contradiction shows that X 0 is non-normal.
Then the Cayley graph Cay(G, S) is normal.
Proof. For the reader's convenience, we essentially reproduce the proof that is given in [1], but add a few additional details.
The ⊇ containment is trivial. If y is an element of the left-hand side, then Using the assumption from the statement of the proposition and the fact that u = v, we obtain that u = t and v = s. In particular, y = x + u + v. This completes the proof of the claim.
Since σ is an automorphism fixing the identity, we also know that where in the last step we apply the hypothesis of (2.15). It now follows that: In particular, Case 2. Assume u = v. Since σ is a graph automorphism, we know it is a bijection from (x + u) + S onto (x σ + u σ ) + S. Case 1 shows it is also a bijection Hence, it must also hold that (x + 2u) σ = x σ + 2u σ . This finishes the proof.
2C. Miscellany. The following result is a very special case of the known results on automorphism groups of Cartesian products. (Note that C 4 is isomorphic to K 2 K 2 .) Proposition 2.16 (cf. [5, Thm. 6.10, p. 69]). Let X be a connected graph. If there does not exist a graph Y, such that X ∼ = K 2 Y, then Lemma 2.17. Let X = Cay(Z n , S) be a connected circulant graph of order n such that X is not twin-free, and let d be the valency of X.
(1) There is a connected circulant graph Y and some m ≥ 2, such that (3) If d = 4, then X is isomorphic either to K 4,4 or to C K 2 with = |V (X)|/2. Moreover, the unique twin of 0 in the second case is n/2.

Proof.
(1) Let ∼ be the relation of being twins on V (X) defined in Definition 2.6, i.e., write x ∼ y if and only if N X (x) = N X (y) for x, y ∈ V (X). Note that since X is assumed to have no loops, equivalence classes of ∼ are independent sets. Furthermore, they are clearly blocks for the action of Aut(X) since x ∼ y if and only if α(x) ∼ α(y) for all α ∈ Aut(X). Since X is a circulant (so in particular, a Cayley graph), by Definition 2.7 the blocks are cosets of some subgroup H. Clearly, each block is of size m := |H|. From the assumption that X is not twin-free, we obtain that m ≥ 2. It is easy to see that if x and y are adjacent, then x is adjacent to y for all x ∼ x and y ∼ y. It is now clear that X ∼ = Y K m with Y := Cay(Z n /H, S), where Z n /H is the quotient group and S := {s + H : s ∈ S}.
(2) By (1), we then represent X as Y K m , where Y is m-regular and connected, and m ≥ 2. As d = δm, and d is prime, it follows that m = d and δ = 1. In particular, Y = K 2 and X = K 2 K d ∼ = K d,d . (Conversely, it is clear that K d,d is a connected circulant graph, but is not twin-free.) (3) By (1), we then represent X as Y K m , where Y is m-regular and connected, and m ≥ 2. As 4 = δm and m ≥ 2, it follows that m ∈ {2, 4}. If m = 4, then δ = 1 and consequently X ∼ = K 2 K 4 ∼ = K 4,4 . If m = 2, then Y is connected and 2-regular, so it is isomorphic to the cycle C with = |V (X)|/2. It follows that X ∼ = C K 2 . Note that in this case, two vertices are twins in X if and only if they are in the same copy of K 2 (see Definition 2.5(3)). It is clear that these 2-element sets of twins form blocks for the action of Aut X by Definition 2.7. From Definition 2.7, it also follows that they are cosets of the subgroup of Z n of order 2. As this subgroup is {0, n/2}, the conclusion follows. Proposition 2.18 ([7,Cor. 4.6]). Let α be an automorphism of BX, where X is a circulant graph Cay(Z n , S), and let s, t ∈ S. If α maps some s-edge to a t-edge, and either gcd(|s|, |t| = 1, or S contains every element that generates s (e.g., if |s| ∈ {2, 3, 4, 6}), then S contains every element that generates t . Theorem 2.19 (Kovács [9], Li [11]). Let X be a connected, arc-transitive, circulant graph of order n. Then one of the following holds: (1) X = K n , (2) X is a normal circulant graph, and Y is a connected arctransitive circulant graph of order m, connected arc-transitive circulant graph of order m.
The statement of the following result in [7] requires the graph to have even order (because the statement refers to n/2), but the same proof applies to graphs of odd order. Although the proofs in this paper will only apply the lemma to graphs of even order, we omit this unnecessary hypothesis. Lemma 2.20 ([7,Cor. 4.3]). Let X = Cay(Z n , S) be a circulant graph, let ϕ be an automorphism of BX, and let S = { s ∈ S | 2t = 2s for all t ∈ S, such that t = s }.

Some conditions that imply stability
Theorem 3.1 (Fernandez-Hujdurović [3] (or [13])). There are no nontrivially unstable, circulant graphs of odd order. The complete graph on 2 vertices is bipartite, and therefore unstable. It is not difficult to see that all of the larger complete graphs are stable: Lemma 3.4. Let X = Cay(Z n , S) be a connected, nonbipartite circulant graph, and let S 0 be a nonempty subset of Z n \ {0} such that S 0 = −S 0 . If every automorphism of BX maps S 0 -edges to S 0 -edges, and some (or, equivalently, every) connected component of Cay(Z n , S 0 ) is a stable graph, then X is stable.
Proof. Let α be an automorphism of BX that fixes (0, 0), and let X 0 be the connected component of Cay(Z n , S 0 ) that contains 0. Since α maps S 0 -edges to S 0 -edges, it restricts to an automorphism of BX 0 . Since X 0 is a stable graph, this implies that α(0, 1) = (0, 1). Lemma 3.5. Let X = Cay(G, S) be a Cayley graph on an abelian group, and let k ∈ Z + , such that k is odd. Suppose there exists c ∈ S, such that 2c = s + t, for all s, t ∈ S \ {c}, and (3) for all a ∈ S of order 2k, there exist s, t ∈ S\{a}, such that 2a = s+t. Then X is stable.
Proof. Let us say that a cycle in BX is exceptional if, for every pair v i , v i+2 of vertices at distance 2 on the cycle, the unique path of length 2 from v i to v i+2 is v i , v i+1 , v i+2 . It is clear that every automorphism of BX must map each exceptional cycle of length k to an exceptional cycle of length k.
Let α be an automorphism of BX fixing (0, 0). If c is any element satisfying the conditions, then (c, 1) 2k is an exceptional cycle. Furthermore, every exceptional cycle of length 2k is of this form. Since k(c, 1) = (0, 1), for every such exceptional cycle, this implies that α fixes (0, 1). So X is stable by Lemma 3.2.
Lemma 3.6. Let X := Cay(Z n , S) be a circulant graph of even order and odd valency. Let S 0 ⊂ S be non-empty such that n/2 ∈ S 0 . Assume that the set of S 0 -edges is invariant under the elements of Aut BX (and S 0 = −S 0 ). If some (equivalently every) connected component X 0 of Cay(Z n , S 0 ) is not bipartite and has the property that BX 0 is normal, then X is stable.
Proof. We can take X 0 := Cay( S 0 , S 0 ). Because X is of odd valency, we know n ∈ S. By our assumptions, n is a vertex of X 0 . As X 0 is connected and assumed to be non-bipartite, BX 0 is connected. Let α ∈ Aut(BX) (0,0) . Then by our assumptions, α ∈ Aut(BX 0 ). Because α fixes (0, 0), it also fixes the connected component of BX 0 containing it, which is BX 0 . As BX 0 is normal, the restriction of α onto BX 0 is a group automorphism of S 0 × Z 2 . Note that as (0, 1), (n, 0), (n, 1) are the only elements of order 2 in S 0 × Z 2 , it follows that α must permute them among themselves. As α fixes the colors of BX 0 , it follows that α fixes (n, 0) because this is the unique element of order 2 in the set S 0 × 0. Because X is loopless, 0 / ∈ S and the only element of order 2 in the connection set of BX, which is S × 1, is (n, 1). Since α fixes S × 1 set-wise, it must hold that α fixes (n, 1) and consequently it also fixes (0, 1). It follows that X is stable.
Proofs in later sections assume that the following circulant graphs are known to be stable. Proof. This is easily verified by computer (in less than a second). For example, the MAGMA code in Figure 1 can be executed on http://magma. maths.usyd.edu.au/calc/, or the sagemath code in Figure 2 can be executed on https://cocalc.com . (Source code of these programs can also be found in the "ancillary files" directory of this paper on the arxiv.) . Let X be a twin-free, connected, vertex-transitive graph of diameter at least 4. If every edge of X is contained in a triangle, then X is stable.    Proof. It is clear that every edge of X k lies on a triangle. As the diameter of X k is equal to the diameter of C k , for k ≥ 8 it follows from Lemma 3.8 that X k is stable. It is easily verified by computer that X k is stable for all k ∈ {3, .., 7}, k = 4. (For example, we can note that X k ∼ = Cay(Z 2k , {±1, k ± 1}) for all k ≥ 3 and apply the code in Figure 1 or Figure 2.) The desired conclusion follows.
The union of the following two results provides a more precise formulation of the above theorem. Proof. We consider twin-free, connected, nonbipartite, circulant graphs of each valency ≤ 3.
(valency 1) K 2 is bipartite. (valency 2) If C n is a nonbipartite cycle, then n is odd, so BC n ∼ = C 2n , so |Aut BC n | = |Aut C 2n | = 2 · 2n = 2 · |Aut C n |, so C n is stable. (valency 3) A connected, nonbipartite, circulant graph X of valency 3 is either an odd prism or a nonbipartite Möbius ladder. In either case, the canonical double cover is the even prism K 2 C n , where n = |V (X)|. It is easy to check that K 4 is stable (see Example 3.3). And the following calculation (which uses Proposition 2.16) shows that X is also stable when n > 4: (1) n ≡ 2 (mod 4), gcd(a, n) = 1, and b = ma+(n/2), for some m ∈ Z × n , such that m 2 ≡ ±1 (mod n), or (2) n is divisible by 8 and gcd |a|, |b| = 4. In both of these cases, X has Wilson type (C.4).
(⇒) Assuming that X is nontrivially unstable, we will show that it satisfies the conditions of (1) or (2). Note that n must be even (see Theorem 3.1). Since X is connected, and not bipartite, the subgroup 2Z n must contain exactly one of the elements of {a, b}.
Case 2. Assume 2s = 2t, for all s, t ∈ S, such that s = t. We may assume α is not a group automorphism, for otherwise Case 1 applies. So BX is not normal. Therefore, Proposition 2.13 implies there exist s, t, u, v ∈ S such that s + t = u + v = 0 and {s, t} = {u, v}. From the assumption of this case, we see that this implies 3a = ±b (perhaps after interchanging a with b). This implies that a and b have the same parity, which contradicts the assumption that X is connected and nonbipartite.
Case 3. The remaining case. Since Case 2 does not apply, we have 2s = 2t, for some s, t ∈ S, such that s = t. We may assume s = a.
Subcase 3.1. Assume that t = −s = −a. Then |a| = 4. If n is divisible by 8, then condition (2) is satisfied. (If |a| = 4, and n is divisible by 8, then |b| must be divisible by 8, so gcd |a|, |b| = 4.) So we may assume n = 4k, where k is odd. Since X is nonbipartite, we know that |b| is not divisible by 4, so the fact that |a| = 4 implies | a ∩ b | ≤ 2. Hence, there is an automorphism of Z n that fixes a, but inverts b, so |Aut X| ≥ 4n.
Also, since k is odd and X is nonbipartite, we must have kb = ±a. Since This contradicts the assumption that X is unstable.
Subcase 3.2. Assume that t = −s. Therefore, we may assume s = a and t = b, so 2a = 2b. This means that a − b has order 2, and must therefore be equal to n, so S + n = S. This contradicts the fact that Cay(Z n , S) is twin-free.
The following observation can be verified by inspecting the list [1] of connected, non-normal Cayley graphs of valency 4 on abelian groups, and confirming that none of them are the canonical double cover of a nontrivially unstable circulant graph. (Recall that "normal" is defined in Definition 2.2(2.9).) For the reader's convenience, we provide a proof that avoids reliance on the entire classification, by extracting the relevant part of the proof in [1] (and by using Theorem 4.3 to reduce the number of cases). Proof. Write X = Cay(Z n , {±a, ±b}), and suppose BX is not normal. (This will lead to a contradiction.) By using Proposition 2.13 (and the fact that X is not bipartite), as in Case 2 of the proof of Theorem 4.3, we see that 2s = 2t, for some s, t ∈ S, such that s = t. Therefore, we may assume that either 2a = 2b or |a| = 4. However, we cannot have 2a = 2b, since X is twin-free. (If b = a + n, then S = S + n.) So we have |a| = 4. Then, since X is nontrivially unstable, we see from Theorem 4.3 that n is divisible by 8 and b = Z n . (We are now in the situation of [1, Lem. 3.4], but will briefly sketch the proof.) Let α ∈ Aut BX, such that α(0, 0) = (0, 0). The subgraph induced by the ball of radius 2 around (0, 0) has only two automorphisms, so the restriction of α to this ball is the same as the restriction of a group automorphism β (such that β(a) ∈ {±a} and β(b) ∈ {±b}). It is then easy to see that α(x) = β(x) for all x, so α is a group automorphism. This means that BX is normal.
The following technical result will be used in Sections 5 and 7.
Corollary 4.5. Let X = Cay(Z n , S) be a circulant graph of even order and odd valency, and let S 0 ⊂ S with |S 0 | = 4. Let X 0 := Cay(Z n , S 0 ) and let X 0 be a connected component of X 0 . Assume that the set of S 0 -edges is invariant under Aut(BX). If either (1) |V (X 0 )| is odd, or (2) X 0 is twin-free and nonbipartite, then X is stable.
Proof. Let us first assume that But also 4 = δm, so m is even, a contradiction.) Therefore X 0 is a connected, twin-free, circulant graph of odd order, so, by Theorem 3.1, it must be stable. It follows by Lemma 3.4 that X is stable.
Let us now suppose that | S 0 | = |V (X 0 )| is even. It then follows that n ∈ S 0 . By assumption (2), X 0 must be twin-free and nonbipartite. As all of its connected components are isomorphic to X 0 , it follows that X 0 is twin-free and nonbipartite. In particular, X 0 is not trivially unstable. If it is stable, we conclude that X is stable by Lemma 3.4. If it is not stable, it is nontrivially unstable so by Corollary 4.4 it follows that BX 0 is normal. Applying Lemma 3.6, we conclude that X is stable.

Unstable circulants of valency 5
Theorem 5.1. A circulant graph Cay(Z n , S) of valency 5 is unstable if and only if either it is trivially unstable, or it is one of the following: (1) Cay(Z 12k , {±s, ±2k, 6k}) with s odd, which has Wilson type (C.1), Remark 5.2. It is easy to see that each connection set listed in Theorem 5.1 contains both even elements and odd elements, so none of the graphs are bipartite. Then it follows from Lemma 2.17(2) that the graphs are also twin-free. Therefore, a graph in the list is nontrivially unstable if and only if it is connected. And this is easy to check: • the graph in (2) is connected (since 1 is in the connection set); • a graph in (1) is connected if and only if s is relatively prime to k.
The proof of Theorem 5.1 will use the following lemmas. The first lemma can be obtained by inspecting the list [1, Cor. 1.3] of connected, non-normal, circulant graphs of valency 4: K 5 , C m K 2 (with m ≥ 3) and K 2 K 5 − 5K 2 . For the reader's convenience, we reproduce the relevant parts of the proof in [1]. Proof (cf. [1, proof of Thm. 1.2]). From Proposition 2.13, we see that (perhaps after permuting the generators), we have either 2a = 2b or b = 3a or 4a = 0. However, we cannot have 2a = 2b, since X is twin-free. So there are two cases to consider.
For n = 8, we have 2b = 6a = −2a, which contradicts the assumption that X is twin-free.
For n = 10, we have the Cayley graph that is specified in the statement of the lemma.
For n ≥ 12, we have X ∼ = Cay(Z n , {±1, ±3}). This is the situation of [1, Lem. 3.5], but, for completeness, we sketch the proof. The only nontrivial automorphism of the ball of radius 2 centered at 0 is x → −x. From this, it easily follows that x → −x is the only nontrivial automorphism of X, so X is normal. This contradicts our hypothesis.
Case 2. Assume 4a = 0. This means |a| = 4. Since X is bipartite, we know that |b| is even, so b contains the unique element of order 2; this means 2a = b for some ∈ Z. Since X is bipartite, we know that is even. So |b| is divisible by 4. Therefore, b contains a. So X ∼ = Cay(n, {±1, ±n/4}). If we consider the subgraph induced by the ball of radius 2, we note that only the vertices ±1 have a pendant edge. In particular, every automorphism of X maps 1-edges to 1-edges and is therefore an automorphism of the graph X 0 := Cay(Z n , {±1}) ∼ = C n . Since this Cayley graph is normal, we can conclude the same about X, which is a contradiction. Proof. Recall that Notation 2.3 introduced n as an abbreviation for n/2. Since X has odd valency, we know that n is even; indeed, we may write X = Cay(Z n , {±a, ±b, n}).
Since every automorphism of BX maps n-edges to n-edges, we know that every automorphism of BX is an automorphism of BX 0 , where If X 0 is stable, then by Lemma 3.4 it follows that X is stable, a contradiction. So we may assume now that X 0 is unstable. Case 1. Assume X 0 is nontrivially unstable. As X 0 is 4-valent, by applying Corollary 4.4 we conclude that BX 0 is a normal Cayley graph. Because every automorphism of BX is an automorphism of BX 0 , it then follows that BX is normal as well. However, since X is nontrivially unstable and of valency 5, Corollary 2.11 implies that BX is not normal, a contradiction.
Case 2. Assume X 0 is trivially unstable. There are three possibilities to consider: Subcase 2.1. Assume X 0 is not connected. Then a and b generate a proper subgroup of Z n , while a, b and n generate the whole group. From here, n = 2k, where k is odd, and a, b = 2Z n has order k. The connected components of X 0 then have order k, and therefore have odd order. By applying Corollary 4.5(1) we conclude that X is stable, a contradiction. Subcase 2.2. Assume X 0 is connected, but is not twin-free. Then (by Lemma 2.17(1)) we can represent X 0 as a wreath product Y K m , where Y is a δ-regular connected graph and m > 1 is an integer such that δm = 4. Subsubcase 2.2.1. Assume m = 4. Then δ = 1, so we get that X 0 = K 2 K 4 = K 4,4 . Hence, X 0 is a connected, 4-valent Cayley graph on Z 8 and its connection set can only contain odd numbers, because it is also bipartite. This uniquely determines X 0 as Cay(Z 8 , {±1, ±3}). From here, X = Cay(Z 8 , {±1, ±3, 4}), so X is the graph in the statement of the lemma.
From Corollary 3.9, and the fact that X is not stable, we conclude that k = 4, i.e., X ∼ = C 4 K 2 . It is easy to see that this implies X = Cay(Z 8 , {±1, ±3, 4}). So X is again the graph in the statement of the lemma. Subcase 2.3. Assume X 0 is bipartite, connected and twin-free. By applying Lemma 2.12, we conclude that the Cayley graph X 0 is not normal. Then Lemma 5.3 tells us that X 0 = Cay(Z 10 , {±1, ±3}), meaning that X = Cay(Z 10 , {±1, ±3, 5}). But then X is bipartite, a contradiction.
The following simple observation provides a converse to Lemma 5.4.
. Since X 0 is bipartite, we know that BX 0 is isomorphic to the disjoint union of two copies of X 0 . But BX is connected, and the element (n, 1) of order 2 is the only element of its connection set that is not in the connection set of BX 0 . It follows that So we see from Proposition 2.16 that the set of n-edges is invariant under all automorphisms of BX. On the other hand, K 4,4 is edge-transitive, so the other edges are all in a single orbit.
Lemma 5.6. Let X = Cay(Z n , S) be a nontrivially unstable, circulant graph of valency 5. If the set of n/2-edges is not invariant under Aut BX, then: (1) X = Cay(Z 12k , {±s, ±2k, 6k}), for some s, k ∈ Z + , with s odd, and (2) Aut BX has exactly two orbits on the set of edges of BX.
Proof. Write X = Cay(Z n , {±a, ±b, n}). By assumption, some automorphism of BX maps an n-edge to an a-edge (perhaps after interchanging a and b). Then Proposition 2.18 shows that every generator of a is in S \ {n}. So the number of generators of a is ≤ 4 (and |a| > 2), so |a| ∈ {3, 4, 5, 6, 8, 12}.
Case 1. Assume |a| ∈ {5, 8, 12}. The four generators of a are in S, so they must coincide with ±a and ±b. Therefore, a, n = a, b, n = Z n , so n ∈ {10, 8, 12}. Therefore, X is one of the following Cayley graphs: Note that the first and third graphs appear in Lemma 3.7 under (3a) and (3b) respectively, and are therefore both stable. Lemma 5.5 (2) implies that the second graph is also not possible (since the statement of the lemma requires that the set of n-edges is not invariant under Aut BX). So this case cannot occur. This graph is stable by Lemma 3.7(3c).
Since (3, 1) ∩ (2, 1), (6, 1) = {0, 0}, this implies where M 6 is the Möbius ladder with 6 vertices. Then Proposition 2.16 implies that BX is not edge-transitive. Since n-edges are in the same orbit as a-edges, this establishes part (2) of the statement of the lemma for this graph.
Subcase 2.2. Assume |b| ∈ {3, 4, 6}. From here, we see from Proposition 2.18 that no automorphism of BX can map an n-edge to a b-edge (because S cannot contain more than 2 generators of b , in addition to ±a). Hence, the set of b-edges is invariant under all automorphisms of BX.
(Note that this establishes part (2) of the statement of the lemma for this subcase.) Now, we see that every automorphism of BX is also an automorphism of the graphs which are the canonical double covers of respectively. Note that BX 1 is arc-transitive (because a-edges and n-edges are in the same orbit of Aut BX). If |a| = 3, then every connected component of BX 1 is isomorphic to a 6-prism, which is not arc-transitive. If |a| = 4, then every connected component of X 1 is isomorphic to K 4 , which is a stable graph by Example 3.3, so it follows from Lemma 3.4 that X is stable, a contradiction. Therefore, we must have |a| = 6.
The connected components of X 2 are |b|-cycles. If |b| is odd, then these are stable (by Theorem 3.1, for example). By another application of Lemma 3.4, it follows that X is stable, a contradiction.
Note that if is odd, then a and b must both be odd (since |a| and |b| are even). Since n = 3 is also odd, this means that all elements of S are odd, so X is bipartite, a contradiction.
Therefore, we know that is even, so we may write = 2k. Then n = 12k, n = 6k and {±a} = {±2k}. In particular, ±a and n are all even. So c must be odd (since X is connected). This means that X appears in part (1) of the statement of the lemma with parameter s = c.
Proof of Theorem 5.1. (⇐) It suffices to show that each of the graphs in (1) and (2) has the specified Wilson type. For any member of (1), it holds that S e = {2k, 6k, 10k}; therefore 4k + S e = S e , so the graph has Wilson type (C.1). For the graph in (2), see Lemma 5.5(1).
(⇒) Assume X = Cay(Z n , S) has valency 5, and is nontrivially unstable. Then either Lemma 5.4 or Lemma 5.6 must apply (depending on whether the set of n-edges is invariant or not). So X is one of the graphs listed in these two lemmas, and is therefore listed in the statement of the theorem.
Combining Lemmas 5.4 to 5.6 also yields the following observation that will be used in Section 7: Corollary 5.7. If X is a nontrivially unstable, 5-valent, circulant graph, then Aut BX has exactly two orbits on the edges of BX.
6. Unstable circulants of valency 6 See Corollary 6.8 for a more explicit formulation of the following theorem. Proof. Let X = Cay(Z n , S) be a nontrivially unstable, circulant graph of valency 6, and write S = {±a, ±b, ±c}. The proof is by contradiction, so assume that X does not have any of the four listed Wilson types. As usual, we let n = n/2, for convenience. The proof considers several cases. Case 1. Assume 2a = 2b. This means b = a + n (and −b = −a + n). Since X does not have Wilson type (C.1), we also know that Therefore, we must have S ∩ 2Z n = {±a, ±b}. Since S ⊆ 2Z n , this implies We claim that |c| is not divisible by 4. To see this, note that otherwise n/gcd(c, n) = |c| is even, so c/gcd(c, n) is odd, and we also know that |2c| is even, so n ∈ 2c . This contradicts the fact that X does not have Wilson type (C.3 ) (with H = n , R = {±c}, and d = gcd(c, n)). This completes the proof of the claim.
If |c| is odd, then letting S 0 = {±c} in Lemma 3.4 implies that X is stable (because cycles of odd length are stable), which is a contradiction.
Also note that, since α preserves the set of c-edges in BX, it must also preserve the complement, which consists of the a-edges and b-edges. Hence, α is an automorphism of the canonical double cover of the 5-valent circulant graph X = Cay(Z n , S ), where S = {±a, ±b, n}.
Let X 0 be the connected component of X that contains 0. Note that X 0 is not stable (since α does not fix (0, 1)). Also note that X 0 is connected, by definition. Furthermore, it is not bipartite, because X is not bipartite and n = kc where k is odd. We therefore see from Lemma 2.17 that it is also twin-free. So X 0 is nontrivially unstable, and must therefore be one of the graphs listed in Theorem 5.1 (after identifying the cyclic group V (X 0 ) with some Z m by a group isomorphism). Since 2a = 2b, it follows that X 0 is the graph in part (2)  (perhaps after interchanging a with b and/or replacing both of them with their negatives). However, we know from (6.3) that 2c = −2c and 2c = 2a. Also, if 2c = a − b = n, then |c| = 4, which contradicts the fact that |c| is not divisible by 4. Thus, only two possibilities need to be considered. Subsubcase 1.2.1. Assume 2c = a + b = 2a + n. Since 2a and 2c are even, this implies n is even, which means n ≡ 0 (mod 4). It also implies that a and b have the same parity, so we conclude from (6.2) that a and b are odd. Since X is not bipartite, then c must be even. Therefore, 0 ≡ 2c = 2a + n ≡ 2 + n (mod 4), so n ≡ 2 (mod 4), which means n/4 is odd. Also note that 4c = 4a = 4b. Therefore, setting k := n/4, we get that If c ≡ 0 (mod 4), this graph has Wilson type (C.2 ) (with h = k).
So we may assume c ≡ 2 (mod 4). We will show that this implies X is stable (which is a contradiction). Any two vertices of BX that are in the same coset of the subgroup (k, 0) have 4 common neighbors, but no see note A.2 two vertices of BX that are in different cosets of (k, 0) have more than 3 common neighbors. Therefore, each coset of (k, 0) is a block for Aut BX. see note A.3 Also note that (c + 2k, 1) is the only element of the coset (c, 1) + (k, 0) that is not adjacent to (0, 0). So every automorphism of BX must preserve the set of (c + 2k)-edges. Since c + 2k is an element of odd order (because c + 2k ≡ 2 + 2 ≡ 0 (mod 4) and n = 4k is not divisible by 8), we conclude from Lemma 3.4 that X is stable.
We may now assume that |c| is odd. Since 2a = 2b and we may assume that Subsubcase 1.2.1 does not apply, it is easy to see that X is stable by Lemma 3.5, which is a contradiction. • If |b| = 4, then a and b are generators of the same cyclic subgroup of order 4, but then {±a} = {±b}, a contradiction.
• We may now assume |b| ∈ {3, 6}. We consider two subsubcases. Then we can assume that no isomorphism of BX maps an a-edge to a c-edge (for otherwise an earlier argument would apply after interchanging b and c). Since a-edges can be mapped to b-edges, this implies that no b-edge can be mapped to a c-edge. Therefore, every automorphism of BX is an automorphism of BX 0 , where X 0 = Cay(Z n , {±a, ±b}). Let X 0 be the connected component of X 0 that contains 0. Recalling that |a| = 4, we see that if |b| = 3 then X 0 is isomorphic to Cay(Z 12 , {±3, ±4}), and if |b| = 6, then X 0 is isomorphic to Cay(Z 12 , {±2, ±3}). Both of these graphs are stable (by Lemma 3.7(2) or Theorem 4.3). So Lemma 3.4 tells us that X is stable as well, a contradiction.
Subcase 2.2. Assume every automorphism of BX maps a-edges to a-edges. Then every automorphism of BX is also an automorphism of BX 0 , where X 0 = Cay(Z n , {±b, ±c}). As usual, let X 0 be the connected component of X 0 that contains 0. Subsubcase 2.2.1. Assume X 0 is not twin-free. Then b = c + n (perhaps after replacing c with −c). Since a + n = −a, this implies S + n = S, which contradicts the fact that X is twin-free. Subsubcase 2.2.2. Assume X 0 is not connected (but is twin-free). We know that X 0 is not stable (for otherwise Lemma 3.4 contradicts the fact that X is not stable). Since X 0 is connected (by definition) and twin-free (by assumption), this implies that it has even order (see Theorem 3.1). Since a, b, c = Z n and |a| = 4, we conclude that b, c = 2Z n and n ≡ 4 (mod 8).
• If X 0 is bipartite, then BX 0 is isomorphic to the union of four disjoint copies of X 0 . Since |a| = 4 and BX is connected, this implies that BX ∼ = C 4 X 0 . Since X 0 has even valency, and |V (X 0 )|/2 is odd, it it is easy to see that X 0 does not have K 2 as a Cartesian factor. (If X 0 ∼ = K 2 Y , then Y is a regular graph of odd valency and odd order, which is impossible.) This implies (by Proposition 2.16) that |Aut BX| = |Aut(C 4 X 0 )| = |Aut C 4 | · |Aut X 0 | = 8 |Aut X 0 |. Also note that, since X 0 is a bipartite circulant graph whose order is congruent to 2 modulo 4, it is (isomorphic to) the canonical double cover of a circulant graph of odd order. Since connected, twin-free, circulant graphs of odd order are stable (see Theorem 3.1) and 2a is the element of order 2 in Z n , we conclude that if β is any automorphism of X 0 , then β(v +2a) = β(v)+2a for every v ∈ V (X 0 ). This implies that we can extend β to an automorphism β of X by defining β (a + v) = a + β(v) for v ∈ V (X 0 ); so Aut X contains a copy of Aut X 0 . Since Aut X also contains the translation v → v + a and the negation automorphism v → −v, we conclude that |Aut X| ≥ 4 |Aut X 0 |. Combining this with the above calculation of |Aut BX| contradicts the fact that X is not stable.
Subsubcase 2.2.4. Assume X 0 is nontrivially unstable. Then Theorem 4.3 tells us that X 0 has Wilson type (C.4). Then X also has Wilson type (C.4), with the same value of m.
Case 3. Assume neither of the previous cases apply (even after permuting and/or negating some of the generators). Let α be an automorphism of BX that fixes (0, 0). The assumption of this case implies that 2s = 2t for all s, t ∈ S, such that s = t. Therefore, Lemma 2.20 implies that the cosets of 2Z n × {0} are blocks for the action of Aut BX. So α must fix the two cosets that are in Z n × {0}, and either fixes or interchanges the other two. However, also note that S is the disjoint union of S e := S ∩ 2Z n and S o := S ∩ (1 + 2Z n ).
Each of these two sets has even cardinality (since it is closed under inverses), and |S e | + |S o | = 6, so it is easy to see that |S e | = |S o |. Therefore, α cannot interchange S e and S o , which means that α must fix all four cosets of 2Z n × {0}. So α maps S e -edges to S e -edges, and maps S o -edges to S o -edges. (6.5) Hence, by Lemma 3.4, we know that the connected components of Cay(Z n , S e ) are unstable.
(The connected components of Cay(Z n , S o ) are always unstable, since they are bipartite.) We also know that Cay(Z n , S e ) is twin-free (since X does not have Wilson type (C.1)). Therefore, we see from Theorem 3.1 that | S e | is even, so n ≡ 0 (mod 4). From Lemma 2.10, we see that BX is not normal. So applying Proposition 2.13 to BX implies that either |a| = 4 or 3a = b or 2a = 2b or 2a = b+c (perhaps after permuting and/or negating some of the generators). Since neither of the previous cases apply, this implies that We will consider each of these two possibilities as a separate subcase.
Therefore, we must have |a| = 10. (Then |b| = 10 and they generate the same cyclic subgroup.) Since n ≡ 0 (mod 4), we may write n = 20k for some k ∈ Z.
Also note that the vertices (0, 1) and (10k, 1) lie in the same connected component of Y 1 , which is a cycle (of length 20k or 4k), and that these two vertices are diametrically opposite on this cycle. Since we already know that α fixes (10k, 1), it must also fix (0, 1). We conclude that X is stable, a contradiction.   Figure 3, under the assumption of this subsubcase that |ã| ≥ 14. From this drawing, it can be seen that ±b are the only vertices in B 2 that have a pendant edge. (These edges are colored white in the figure.) So {±b} is α-invariant. This means that α maps b-edges to b-edges. Since we already know that α maps c-edges to c-edges, it must also map a-edges to a-edges.
In the terminology of [6], this means that α is a color-preserving graph automorphism. We will now use a simple argument from [6, §4] to establish that α is a group automorphism (so BX is normal, and then it follows from Lemma 2.10 that X has Wilson type (C.4)). We provide only a sketch of the proof. By composing with negation, if necessary, we may assume that α(ã) =ã. This implies α(kã) = kã for all k ∈ Z. Let m ∈ Z + , such that mc ∈ ã . Then α(kmc) = kmc for all k ∈ Z. If 2mc = 0, this implies that α( c) = c for all ∈ Z; in fact, α(kã + c) = kã + c, for all k, ∈ Z, which means that α is the identity map, contradicting the fact that α / ∈ Aut X × Z 2 . Therefore, we may assume 2mc = 0, for all m ∈ Z, such that mc ∈ ã . This means that | c ∩ ã | ≤ 2, so there is a group automorphism of Z n that fixesã and negatesc. So we may assume that α(c) =c. Sinceã +c is the only common neighbor ofã andc, we must have α(ã +c) =ã +c. Similarly, we must have α(2ã +c) = 2ã +c and α(ã + 2c) =ã + 2c. Repeating the argument shows that α(kã + c) = kã + c, for all k, ∈ Z, so, once again, α is the identity map. (Note that this implies 2ã =b +c.) We will show that α(kb + ã) = kα(b) + α(ã), for all k, ∈ Z ≥0 . (This implies that α is a group automorphism of Z n , so BX is normal, so Lemma 2.10 implies that X has Wilson type (C.4).) Since b and c have the same parity, we see from (6.5) that α maps {b, c}edges to {b, c}-edges, and maps a-edges to a-edges. In particular, we may assume (by composing with negation if necessary) that α fixes every element of ã . (6.6) Since α maps {b, c}-edges to {b, c}-edges, α is an automorphism of the graph Cay(Z n ×Z 2 , {±b, ±c}), which has at most two connected components. Let X 0 be its connected component containing (0, 0) with the vertex set b ,c . Since α fixes (0, 0), it restricts to an automorphism of X 0 . Because we are assuming Case 1 and Case 2 do not hold and we can additionally assume Subcase 3.1 does not hold either, Proposition 2.13 applies to X 0 . It follows that the restriction of α to V (X 0 ) is a group automorphism of b ,c . We let b , c ∈ {±b, ±c}, such that α(b) =b and α(c) =c . It follows that: =b +c (6.7).
As we have already established that the hypothesis of Proposition 2.13 holds, we obtain that {b , c } = {b, c}.
To complete the proof of this subcase, we now prove by induction on k that, for all k, ∈ Z ≥0 , we have α(kb + ã) = kb + ã.
The base case is provided by (6.6), so assume k > 0. Since α maps a-edges to a-edges, there exists ∈ {±1}, such that α(kb + ã) = kb + ã for all ∈ Z. We wish to show = 1, so suppose = −1. (This will lead to a contradiction.) Letting = −2 tells us This implies −2b = 2c , which contradicts the fact that Case 1 does not apply.
The following result provides a more explicit version of Theorem 6.1. Remark 6.9 explains which of these graphs are nontrivially unstable. Corollary 6.8. A circulant graph X = Cay(Z n , {±a, ±b, ±c}) of valency 6 is unstable if and only if either it is trivially unstable, or it is one of the following: (1) Cay(Z 8k , {±a, ±b, ±2k}), where a and b are odd, which is of Wilson type (C.1). (2) Cay(Z 4k , {±a, ±b, ±b + 2k}), where a is odd and b is even, which is of Wilson type (C.1). (3) Cay Z 4k , ±a, ±(a + k), ±(a − k) , where a ≡ 0 (mod 4) and k is odd, which is of Wilson type (C.2 ).
(5) Cay(Z 8k , {±a, ±k, ±3k}), where a ≡ 0 (mod 4) and k is odd, which is of Wilson type (C.3 ).  Proof. (⇐) It is easy to see that each graph has the specified Wilson type, and is therefore unstable. (The arguments in the other direction of the proof can provide some hints, if required.) (⇒) If X is unstable, then we know from Theorem 6.1 that X has Wilson type (C.1), (C.2 ), (C.3 ), or (C.4). We treat each of these possibilities as a separate case. Case 1. Assume X has Wilson type (C.1). Subcase 1.1. Assume |S e | = 2. Then we may assume S e = {±c}. Since X has Wilson type (C.1), we must have −c = c + n, so |c| = 4. (Since c ∈ S e , this implies that n is divisible by 8.) Therefore X is as described in part (1) of the corollary. Subcase 1.2. Assume |S e | = 4. Then we may assume S e = {±b, ±c}. Since X has Wilson type (C.1) (and |S e | is a power of 2), we must have S e + n = S e . Therefore, we may assume c = b + n (because we cannot have b + n = −b and c + n = −c). Since b and c are even, this implies n is even, so n is divisible by 4. Hence, X is as described in part (2) of the corollary.  We see from part (b) of condition (C.2 ) that S contains an element that is divisible by 4, so we must have a ≡ 0 (mod 4). Also, since a + (n/4) = b is odd, we know that n/4 is odd. Hence, X is as described in part (3) of the corollary.  . We may assume that X does not have Wilson type (C.1), so R contains at least one element of S e ; for definiteness, let us say that a is in R ∩ S e . Since r/d is odd for every r ∈ R, we know that all elements of R have the same parity, so this implies R ⊆ S e .  Assume |h| = 2 (so h = n). Since n/d is even, we know that n ∈ dZ n . Therefore, the last condition in (C.3 ) implies that |a| is divisible by 4. (Since a is even, this implies that n is divisible by 8.) Hence, X is as described in part (4) of the corollary. shows {±b, ±c} = {±n/8, ±3n/8}. Since the elements of S cannot all be even, we know that n/8 is odd. Then, since a is even, we know that 2aZ n does not contain an element of order 4, so we conclude from the last sentence of condition (C.3 ) that aZ n does not contain H. This means that a is divisible by 4. Hence, X is as described in part (5) of the corollary. Subcase 3.2. Assume |R| = 4. This means we may assume R = {±a, ±b}, H = n , and |c| = 4. Since n/d is even, we know that the elements of R have even order, so n ∈ R = dZ n . Therefore, the last sentence of condition (C.3 ) implies n ∈ 2dZ n , which means n/d is divisible by 4. Since r/d is odd for every r ∈ R, this implies |a| ≡ |b| ≡ 0 (mod 4). Since a ∈ R ⊆ S e , we conclude that n is divisible by 8. So c = ±n/4 is even. This contradicts the fact that at least one element of S must be odd (since S generates Z n ). Since |S e | = |S o |, we know that α(S e ) = S o . Therefore, we must have α(S e ) = S e and α(S o ) = S o . This means that n is even (so n is divisible by 4).
Assume, without loss of generality, that b and c have the same parity (and a has the opposite parity).
We must have n + ma ∈ {±a}. So we may assume n + ma = a (by replacing m with −m if necessary), so (m − 1)a = n.
Proof. For convenience, let S = {±a, ±b, ±c}. It is clear that X is connected if and only if gcd(S ∪ {n}) = 1. Therefore, the first condition in each part of the remark is precisely the condition for X to be connected.
Knowing that gcd(S ∪ {n}) = 1 implies that at least one element of S is odd (since n is even). Therefore, X is nonbipartite if and only if at least one element of S is even. It is obvious that S has an even element in all parts of Corollary 6.8 other than (6) and (7), so the statement of the remark only adds this as an explicit condition for parts (6) and (7). (In part (7), the fact that (m + 1)b ≡ (m + 1)c ≡ 4k (mod 8k) implies that b and c have the same parity, so there is no need to mention the possibility that c is even.) Now, let us suppose that X is not twin-free (but is connected and nonbipartite). Then by Lemma 2.17(1), it follows that X ∼ = Y K m with Y being a connected circulant of valency δ and m ≥ 2 an integer. Clearly, 6 = δm, so m ∈ {2, 3, 6}.
• If m = 6, then δ = 1 and Y ∼ = K 2 . Consequently, X ∼ = K 2 K 6 ∼ = K 6,6 , which contradicts the fact that X is not bipartite. • If m = 3, then δ = 2 and Y is a cycle of even length, which again contradicts the fact that X is not bipartite. • If m = 2, then from the proof of Lemma 2.17(3), it can be concluded that the unique twin of 0 is n. Therefore, it must hold that n+S = S. Thus, we see that X is twin-free if and only if n + S = S.
In parts (3) and (5), it is clear that if n + S = S, then a + n = −a, which means a = ±n/4. Since a ≡ 0 (mod 4), this implies n is divisible by 16, which contradicts the fact that k is odd. So all of the graphs of these two types are twin-free. All other parts of the remark add a final condition that specifically rules out the possibility that n + S = S.

Remark 7.2.
It is easy to see that each connection set listed in Theorem 7.1 contains both even elements and odd elements, so none of the graphs are bipartite. Then it follows from Lemma 2.17(2) that the graphs are also twin-free. Therefore, a graph in the list is nontrivially unstable if and only if it is connected. And this is easy to check: a graph listed in Theorem 7.1 is connected if and only if gcd(t, k) = 1 (except that the condition for part (2) is gcd(b, c, k) = 1).

Remark 7.3.
In the statement of Theorem 7.1, it is implicitly assumed that k, t ∈ Z + . In order for the graphs to have valency 7, the parameters k and t (or k, b, and c) must be chosen so that all of the listed elements of the connection set are distinct in the cyclic group. (Note that this implies k > 1 in parts (4), (1), and (5).) Proof of Theorem 7.1. (⇐) This is the easy direction. For each family of graphs in the statement of the theorem, we briefly justify the specified Wilson type (which implies that the graphs are unstable): (1) Type (C.1) with S e = {±2t, ±2(k − t), ±2(k + t)} and 2k + S e = S e .
We consider the following cases.
Subcase 1.1. Assume |a| ∈ {7, 9}. Then S must contain the 6 generators of a . Since |a| is odd, we know n / ∈ a , so we conclude that Z n = a × n . In particular, X is one of the following: Note that the first graph appears in Lemma 3.7(5b), which contradicts the assumption that X is (nontrivially) unstable. The second graph is listed in part (1) of the statement of the theorem (with parameters k = 3 and t = 1). Subcase 1.2. Assume |a| ∈ {14, 18}. Then S again contains the 6 generators of a . Additionally, since |a| is even, we have n ∈ a . We conclude that a generates Z n and it follows that X is one of the following: It is clear that both of these graphs are bipartite, which contradicts the assumption that X is nontrivially unstable. Subcase 1.3. Assume |a| = 5. Then S contains the 4 generators of a . We may suppose without loss of generality that besides ±a, the remaining two generators of a are ±b.   By the assumption of Case 1, there is an automorphism α of X that maps an n-edge to an a-edge. By composing with translations on the left and right, we may assume that α fixes the vertex (0, 0), and maps an n-edge that is adjacent to (0, 0) to an a-edge that is adjacent to (0, 0). We see from Proposition 2.18 (by the same argument as in Subsubcase 1.3.2) that the set of c-edges is invariant under all automorphisms of BX. So α restricts to an automorphism of BX 0 , where X 0 is the connected component of Cay(Z n , {±a, ±b, n}) that contains 0. However, since X 0 ∼ = Cay(Z 8 , {±1, ±3, 4}), we know from Lemma 5.5(2) that the set of n-edges is invariant under all automorphisms of BX 0 . This contradicts the choice of α. Subcase 1.5. Assume |a| = 10. Once again, we may assume that ±a and ±b are the four generators of a (which contains n, because |a| is even). Subsubcase 1.5.1. Assume |c| ∈ {3, 4, 6}. Then X is one of the following graphs: (1) |c| = 3 =⇒ X = Cay(Z 30 , {±3, ±9, ±10, 15}). By Lemma 3.7(5f), this graph is stable.
As before, we see from Proposition 2.18 that the set of c-edges is invariant under all automorphisms of BX. Then Lemma 3.4 implies that |c| is even (since cycles of odd length are stable).
If m is odd, then c must also be odd (since |c| is odd and n = 10m is not divisible by 4), so all elements of S are odd. Then X is bipartite, which contradicts the assumption that X is nontrivially unstable.
Subsubcase 1.6.2. Assume |c| / ∈ {3, 4, 6}. As usual, we see from Proposition 2.18 (by the same argument as in Subsubcase 1.3.2) that the set of c-edges is invariant under all automorphisms of BX. Therefore, if we let S 0 := {±a, ±b, n}, then the set of S 0 -edges is also invariant under every automorphism of BX. Since the connected components of Cay(Z n , S 0 ) are isomorphic to Cay(Z 12 , {±1, ±5, 6}), we see from Lemma 3.7(3b) that these connected components are stable. It therefore follows from Lemma 3.4 that X is stable.  By Lemma 3.7(5a), this graph is stable. So we must have |c| / ∈ {3, 4, 6}. Then, yet again, Proposition 2.18 implies that the set of c-edges is invariant under every automorphism of BX. Therefore, if we let S 0 := {±a, ±b, n}, then the set of S 0 -edges is also invariant.
In this situation, a connected component X 0 of Cay(Z n , S 0 ) is isomorphic to Cay(Z 12 , {±2, ±3, 6}). This graph is nontrivially unstable, as it is listed in Theorem 5.1(1) with parameters k = 1 and s = 3 (and Remark 5.2 tells us that it is nontrivially unstable, not merely unstable). So Corollary 5.7 tells us that the group Aut BX 0 has precisely two orbits on the edges of BX 0 . We know from the assumption of Case 1 that there exists an automorphism of BX that maps an n-edge to an a-edge. Since the set of S 0 -edges is invariant, this implies there is an automorphism of BX 0 that maps an n-edge to an a-edge. (So the n-edges are in the same orbit as the a-edges.) It follows that the set of b-edges is invariant.
If |a| = 4, then the invariant subgraph Cay(Z n , {±a, n}) of X has connected components isomorphic to K 4 , which is stable by Example 3.3. It then follows from Lemma 3.4 that X is stable.
So we must have |a| = 6. Then |b| = 4. Write n = 12k. Then, since |a| is not divisible by 4, we see that a must be even.
If b and c are of opposite parity, then the connected components of Cay(Z n , {±b, ±c}) are nonbipartite. It is not difficult to see that they are also twin-free (since |b| = 4 and the valency of the graph is so small). Since see note A. 4 the set of b-edges is invariant under Aut BX and the set of c-edges is also invariant, we know that the set of {b, c}-edges is invariant. It therefore follows from Corollary 4.5(2) that X is stable. So b and c must have the same parity. However, they cannot both be even, since a is known to be even, and X is not connected if every element of S is even. So b and c are odd. Since |a| = 6, we now see that X is listed in part (2) of the theorem. If |a| = 3, then the connected components of Cay(Z n , S 0 ) are isomorphic to Cay(Z 6 , {±2, 3}), which is stable by Lemma 3.7(1a). It follows by Lemma 3.4 that X is also stable.
If |a| = 4, then the connected components of Cay(Z n , S 0 ) are isomorphic to Cay(Z 4 , {1, 2, 3}), which is further isomorphic to K 4 , which is stable by Example 3.3. It follows again by Lemma 3.4 that X is stable.
Let X 0 be the connected component of X 0 that contains 0. As X is assumed to be unstable and X 0 is 4-valent, we can conclude from Corollary 4.5 that |V (X 0 )| is even and that either X 0 is bipartite or X 0 is not twin-free.
Suppose, first, that X 0 is bipartite. This implies that b and c are of the same parity.
(1) If b and c are even, then m must be odd. (Otherwise, S would contain only even integers, so X would not be connected.) Consequently, since n = 6m, it follows that b and c are both of odd order, so X 0 contains an odd cycle. This contradicts the assumption that X 0 is bipartite.
(2) If b and c are odd, then m must be even. (Otherwise, every element of S is odd, so X is bipartite, which contradicts the fact that X is nontrivially unstable.). Hence, we may write m = 2k, and then we see that X is listed in part (2) of the theorem.
We can now assume that X 0 is not bipartite and is not twin-free. By Lemma 2.17 (3), it follows that X 0 is isomorphic to K 4,4 or C K 2 with = |V (X 0 )|/2. As X 0 is assumed to be nonbipartite, the first case is not possible, and in the second case, we see that |V (X 0 )|/2 is odd.
From the fact that X 0 is not twin-free (and the valency of X is smallonly 4) we obtain that c = b + n (perhaps after replacing b with −b). For see note A.5 every v ∈ Z n × Z 2 , we deduce that v + (n, 0) is the unique twin of v in the graph BX 0 . Since automorphisms must map twin vertices to twin vertices, and the set of {b, c}-edges is invariant under Aut BX, we conclude that the cosets of the subgroup (n, 0) are blocks for the action of Aut BX (see Definition 2.7). Note that quotient graph of BX 0 with respect to the partition induced by cosets of (n, 0) is a cycle. Since |V (X 0 )|/2 is odd, the length of this cycle is |V (X 0 )|. Therefore, in this cycle, the vertices corresponding to the cosets {(0, 0), (n, 0)} and {(0, 1), (n, 1)} are at maximum distance.
This completes the proof of Case 1. For the remaining cases, we may assume that every automorphism of BX maps n-edges to n-edges, and is therefore an automorphism of the canonical double cover of the following subgraph X 0 of X: Notation 7.4. For the remainder of the proof, we let X 0 := Cay(Z n , {±a, ±b, ±c}) be the graph that is obtained from X by removing all of the n-edges.
Case 2. Assume X 0 is not connected. (We also assume that every automorphism of BX maps n-edges to n-edges, for otherwise Case 1 applies.) Then a, b, c is a proper subgroup of Z n , but a, b, c, n is the whole group Z n . It follows that n = 2m, where m is odd, and | a, b, c | = m. Let X 0 be the connected component of X 0 that contains 0. Note that X 0 is connected by definition and also that it is not bipartite (since it is vertex-transitive and of odd order).
If X 0 is twin-free, then it follows by Theorem 3.1 that X 0 is stable. By Lemma 3.4, we conclude that X is stable, which is a contradiction.
Therefore, we know that X 0 is not twin-free. Then by Lemma 2.17(1), where Y is a vertex-transitive, connected graph and d ≥ 2. Let δ be the valency of Y . Since X 0 is 6-valent, it follows that 6 = δd and therefore, d ∈ {2, 3, 6}. Because d |V (Y )| = |V (X 0 )| is odd, it cannot happen that d is even. Hence, we conclude that d = 3. It follows that δ = 2, so Y must be a cycle. Letting k := |V (Y )| = m/3 = n/6, we conclude that X 0 ∼ = C k K 3 . Since X 0 is the disjoint union of two copies of X 0 , we now see that for some s ∈ Z n , such that | s, 2k | = n/2. (This final condition means gcd(s, 2k, 6k) = 2, so s must be even; write s = 2t. Then This graph is listed in part (1) of the theorem.
Case 3. Assume that X 0 is bipartite and that BX 0 is arc-transitive. (We also assume that every automorphism of BX maps n-edges to n-edges, and that X 0 is connected, for otherwise a previous case applies.) Since X 0 is bipartite, it follows that BX 0 is isomorphic to a disjoint union of two copies of X 0 . Since BX 0 is assumed to be arc-transitive, it follows that X 0 is a connected arc-transitive circulant graph. Consequently, it is one of the four types that are listed in Theorem 2.19. Type 2.19(1) is impossible because X 0 is bipartite (and has valency 6). By Lemma 2.12, it follows that X 0 is not a normal Cayley graph, so it does not have type 2.19(2) either.
Then Y is a cycle of even length 2m. We obtain that Then X = Cay(Z 6m , {±t, 2m ± t, 4m ± t, 3m}). Since X 0 is connected, we have gcd(t, 2m) = 1. In particular, t is odd. Since X is not bipartite, this implies m is even. Writing m = 2k, we get that So X is listed in part (6) of the theorem.
Then Y is a connected, cubic, arctransitive, circulant graph, so Corollary 2.23 tells us that Y is either K 4 or K 3,3 . Since X 0 is bipartite, it follows that Y is isomorphic to K 3,3 . We obtain that X 0 ∼ = K 6,6 , a case that has already been considered in Subsubcase 3.1.1.
Assume, now, that δ = 2 and d = 4. Since δ = 2, we have Y = C m . Then m must be even, because X 0 is bipartite. This contradicts the fact that gcd(d, m) = 1.
Case 4. Assume that none of the preceding cases apply. This means that: (1) every automorphism of BX maps n-edges to n-edges, (2) X 0 is connected, and (3) either X 0 is not bipartite or BX 0 is not arc-transitive. is not bipartite. Let X 1 := Cay(Z n , {±s, n}), and let X 1 be a connected component of X 1 . Then X 1 is connected, cubic, and nonbipartite. The graph X 1 must also be twin-free, because otherwise Lemma 2.17(2) would imply that X 1 ∼ = K 3,3 , which contradicts the fact that X 1 is not bipartite. We therefore conclude from Proposition 4.2 that X 1 is stable. By our assumptions, the set of s-edges is invariant under the action of Aut(BX), and the set of n-edges is also invariant. So it follows by Lemma 3.4 that X is stable, which is a contradiction. Subcase 4.2. Assume there exists s ∈ {a, b, c}, such that the set of s-edges is invariant under Aut BX. Since X is not bipartite, we know that S contains two elements of opposite parity. Therefore, we may assume without loss of generality that a + n is odd. We may assume that the set of a-edges is not invariant under Aut BX, for otherwise Subcase 4.1 applies (with s = a). So s = a. Hence, we may assume, without loss of generality, that s = c, which means the set of c-edges is invariant under Aut BX. This implies that the a-edges and the b-edges are in the same orbit of Aut BX.
Denote their connected components containing 0 by X 1 and X 2 , respectively. We may assume that X 1 is bipartite. (Otherwise, Subcase 4.1 applies with s = c.) This implies that |c| is even, so n ∈ c . More precisely, we have n = (|c|/2) c. Since X 1 is bipartite, this implies that |c|/2 is odd.
Also, since X is unstable, it follows from Corollary 4.5 that |V (X 2 )| is even and either X 2 is not twin-free or X 2 is bipartite. Subsubcase 4.2.1. Assume X 2 is not twin-free. We claim that X 2 ∼ = K 4,4 .
From Lemma 2.17 (3), we obtain that X 2 is isomorphic to K 4,4 or C K 2 with := | a, b |/2. Let X * 2 be the connected component of Cay(Z n , {±a, ±b, n}) containing 0. Note that X * 2 is obtained by adding n-edges to X 2 . Since X 2 ∼ = C K 2 , this implies X * 2 ∼ = C K 2 . However, since n-edges are also invariant, X * 2 cannot be stable, since in this case Lemma 3.4 would imply that X is stable. By Corollary 3.9, we therefore conclude that = 4. Finally, note that C 4 K 2 ∼ = K 4,4 . Hence, we may assume that X 2 ∼ = K 4,4 . This completes the proof of the claim.
Therefore, (the vertex sets of) the connected components of this graph are blocks for the action of Aut BX 0 . These blocks are the four cosets of the subgroup 2Z n ×{0}. As BX 0 is 6-valent and arc-transitive (and all neighbors of (0, 0) are in Z n ×{1}), it follows that either all 6 neighbors of (0, 0) in BX 0 lie in the same coset of 2Z n ×{0} or three of its neighbors lie in 2Z n ×{1} and the other three lie in (1 + 2Z n ) × {1}. In the first case, it follows that a, b, c are all of the same parity, which contradicts the fact that X 0 is connected and nonbipartite. In the second case, it follows that exactly three elements of the set {±a, ±b, ±c} are odd, which is impossible, since −s has the same parity as s. (1) S contains no element of order 4, and (2) we may assume, without loss of generality, that 2a = 2b. By Lemma 2.20, every automorphism of BX 0 is also an automorphism of Cay(Z n × Z 2 , {(±2c, 0)}). This implies that the cosets of 2c × {0} are blocks for the action of Aut BX 0 . The two c-neighbors of (0, 0) are both in the coset (c, 1) + 2c × {0} . Therefore, by arc-transitivity, either all neighbors of (0, 0) in BX 0 are in this coset or there are three different cosets, each containing two neighbors of (0, 0).
However, if all neighbors of (0, 0) are in (c, 1) + 2c × {0} , then a and b have the same parity as c. This contradicts the fact that X 0 is connected and nonbipartite.
So there are three different cosets that each contain two neighbors of (0, 0). Consider the quotient graph of BX 0 with respect to the coset partition induced by 2c × {0}. This is a cubic, connected, bipartite, arc-transitive graph Q, which is a Cayley graph on Z m × Z 2 , where m is the index of 2c in Z n . It follows from Theorem 2.19 that there are only three cubic, connected Cayley graphs on abelian groups (up to isomorphism): K 4 , K 3,3 , and the cube Q 3 .
• K 4 is not bipartite.
• If Q ∼ = K 3,3 , then 2c is of index 3 in Z n . But this means 2c cannot be of index 2 in c , so it follows that 2c = c , so c is of odd order. Then n = 3|c| is also odd. By Theorem 3.1, X is stable. Therefore, Q must be the cube (and therefore has exactly 8 vertices).
It follows that 2c × 0 is of index 8 in Z n × Z 2 , so the order of 2c is n/4. This means n/gcd(n, 2c) = n/4, so gcd(n, 2c) = 4. Therefore, c is an even integer, and n is also even. Since X is connected, then a and b = a + n must be odd.
There is more than one path of length 2 from (0, 0) to each of these vertices. Due to arc-transitivity, this implies that the the path (0, 0), (c, 1), (2c, 0) is not the only path of length 2 from (0, 0) to (2c, 0). Because the first coordinate of (2c, 0) is an even integer, we conclude that 2c is a sum of two integers from {±a, ±a + n, ±c} of the same parity, besides c + c. As we have assumed there is no element of order 4 in S, the case 2c = (−c) + (−c) is not possible. So 2c must be a sum of two (odd) integers from {±a, ±a + n}. Recalling that 2c = 2s for s = c, we see that (up to relabeling a, −a and a + n), we must have either 2c = a + (a + n) = 2a + n or 2c = a − (a + n) = n.
However, the case 2c = n is not possible, because no element of S has order 4. Therefore, we have 2c = 2a + n. This means 2(c − a) = n, so either c = a + (n/4) or c = a − (n/4). However, if c = a + (n/4), then −c = −a − (n/4); therefore, we may assume c = a − (n/4), by replacing a, b, and c with their negatives, if necessary.
As a is an odd integer, we know that n/|a| = gcd(a, n) is odd, so a contains every element of Z n whose order is a power of 2. In particular, it contains n and ±n/4. Since b = a + n and c = a + (n/4), we conclude that a = Z n . Then, writing n = 4k, we have: X = Cay(Z 4k , {±a, ±(a + 2k), ±(a − k), 2k}).
If a ≡ k (mod 4), then this is listed in part (4) of the theorem (with parameter t = a).
To complete the proof, we will show that if a ≡ k (mod 4), then the subgraph X 0 is stable. This then implies by Lemma 3.4 that X is stable as well, which is a contradiction.
This implies that no vertex in u − c + (k, 0) is adjacent to v, so all of the common neighbors are in u + c + (k, 0) . Since the only neighbors of u that are in this coset are u + c, u + (c + k), and u + (c − k), this obviously implies that u and v have no more than 3 common neighbors.
A.4. Details missing in the proof of Subsubcase 1.7.1 Proof. We are assuming that b and c are of opposite parity, so the graph Γ := Cay(Z n , {±b, ±c}) is nonbiparite. Assume that it is not twin-free. Then by Lemma 2.17(3) and the fact that Γ is nonbipartite, it follows that Γ ∼ = C K 2 with = |V (Γ)|/2. Furthermore, the unique twin of 0 is n i.e. n + {±b, ±c} = {±b, ±c}. Since |b| = 4, it follows that b + n = −b and −b + n = b. Hence, c + n = −c and −c + n = c, which implies |c| is a divisor of 4, a contradiction.
A.5. Details missing in the proof of Subsubcase 1.7.2.