On automorphisms of direct products of Cayley graphs on abelian groups

Let X and Y be connected Cayley graphs on abelian groups, such that no two distinct vertices of X have exactly the same neighbours, and the same is true about Y . We show that if the number of vertices of X is relatively prime to the number of vertices of Y , then the direct product X × Y has only the obvious automorphisms (namely, the ones that come from automorphisms of its factors X and Y ). This was not previously known even in the special case where Y = K2 has only two vertices. The proof of this special case is short and elementary. The general case follows from the special case by standard arguments. Mathematics Subject Classifications: 05C25, 05C76


Introduction
The canonical bipartite double cover [13] of a graph X is the bipartite graph BX with V (BX) = V (X) × {0, 1}, where (v, 0) is adjacent to (w, 1) in BX ⇐⇒ v is adjacent to w in X.
Letting S 2 be the symmetric group on the 2-element set {0, 1}, it is clear that Aut X × S 2 is a subgroup of Aut BX. If this subgroup happens to be all of Aut BX, then it is easy to see (and well known) that X must be connected, and must also be "twin-free" (see Definition 4.1 below). B. Fernandez and A. Hujdurović [5] recently established that the converse is true when X is a circulant graph of odd order. This had been conjectured by Y.-L. Qin, B. Xia, and S. Zhou [11,Conj. 1.3], who proved the special case where X has prime order. See the introductions of [5] and [11] for additional history and motivation. All graphs in this paper are undirected, with no multiple edges, but loops are allowed Circulant graphs are examples of "Cayley graphs" (see Definition 2.1 below), and both sets of authors asked whether the converse can be generalized to all Cayley graphs on abelian groups of odd order ( [5,Problem 3.3] and [11, p. 157]). This note provides a short, elementary proof that the desired generalization is indeed true: Theorem 1.1. If X is a twin-free, connected Cayley graph on a finite abelian group of odd order, then Aut BX = Aut X × S 2 . Remark 1.2. A graph X is said to be stable if Aut BX = Aut X × S 2 [11, p. 155], so the theorem can be rephrased as the statement that: Every twin-free, connected Cayley graph on a finite abelian group of odd order is stable.
(However, the term "stable graph" is ambiguous, because it also has other meanings in graph theory [1,6].) is the nonabelian group of order 21, then X = Cay G; {a ±1 , x ±1 , (ax) ±1 } is twin-free and connected, but it can easily be verified by computer that |Aut X| = 42 and |Aut BX| = 252. (This was discovered with MAGMA and confirmed with sagemath.) Remark 1.4 ([5, Rem. 1.3]). For any Cayley graph X on an abelian group of odd order, the theorem makes it possible to obtain the automorphism group of BX from the automorphism group of X. For simplicity, let us assume that X is loopless. Then there exist integers c, d 1, and a twin-free, connected, abelian Cayley graph Y of odd order, such that X ∼ = K c (Y K d ). Then, by a well-known theorem of Sabidussi [12] on the automorphism group of a wreath product of graphs, the theorem implies that The canonical bipartite double cover BX can be realized as the direct product X × K 2 (see Definition 5.1), and the theory of direct products (see Section 5) implies that the theorem can be generalized by replacing BX = X × K 2 with X × Y , where Y is any graph in a much more general family: Corollary 1.5. Let X be a twin-free, connected Cayley graph on a finite abelian group of odd order, and let Y be any twin-free, connected graph, such that either: 1. Y is not bipartite, and |V (Y )| is relatively prime to |V (X)|, or 2. Y is bipartite, with bipartition V (Y ) = Y 0 ∪ Y 1 , such that (a) |Y 0 | and |Y 1 | are relatively prime to |V (X)|, and (b) either |Y 0 | = |Y 1 |, or Y has an automorphism that interchanges Y 0 and Y 1 .
Also assume that neither X nor Y is the one-vertex trivial graph. Then The statement of the corollary becomes much simpler if we assume that Y is also an abelian Cayley graph: Corollary 1.6. Let X and Y be twin-free, connected Cayley graphs on abelian groups, such that |V (X)| is relatively prime to |V (Y )|. Also assume that neither X nor Y is the one-vertex trivial graph. Then Remark 1.7. In Theorem 1.1, the Cayley graph X can be allowed to have an edge-colouring that is invariant under translation by elements of G (see Remark 3.4(2)). However, the proofs of Corollaries 1.5 and 1.6 do not allow colours on the edges.
Here is an outline of the paper: §1. Introduction §2. A crucial lemma §3. Comments on the lemma (optional) §4. Proof of the main theorem §5. Review of direct products (proof of Corollaries 1.5 and 1.6 )

A crucial lemma
As was already mentioned in the introduction, all graphs in this paper are undirected, with no multiple edges. Loops are allowed, but they are not necessary for any of the arguments, so readers are welcome to assume that all graphs are simple. Readers at the other extreme, who want to discuss multiple edges (or edge-colourings), are referred to Remark 3.4, but these complications are forbidden in this section. We now state a simple observation that is probably already in the literature somewhere. (Although the same proof also applies to Cayley digraphs, we state the result only for Cayley graphs, because they are the topic of this note.) Lemma 2.2. Let ϕ be an automorphism of a Cayley graph Cay(G; S), and let k ∈ Z + . If 1. G is abelian, and 2. ks = kt for all s, t ∈ S, such that s = t, Proof. Write k = p 1 p 2 · · · p r , where each p i is prime, and let k i = p 1 p 2 · · · p i for 0 i r. We will prove by induction on i that ϕ is an automorphism of Cay(G, k i S). The base case is true by assumption, since k 0 S = 1S = S.
For v, w ∈ G, let #(v, w) be the number of walks of length p i from v to w in the graph Cay(G, k i−1 S). These walks are in one-to-one correspondence with the p i -tuples (s 1 , s 2 , . . . , s p i ) of elements of k i−1 S, such that s 1 +s 2 +· · ·+s p i = w−v. Since G is abelian, any cyclic rotation of (s 1 , s 2 , . . . , s p i ) also corresponds to a walk from v to w. Therefore, the set of these walks can be partitioned into sets of cardinality (Also note that s is unique, if it exists, by assumption (2).) Hence, we see that Since p i k i−1 = k i , the desired conclusion that ϕ ∈ Aut Cay(G, k i S) now follows from the induction hypothesis that ϕ ∈ Aut Cay(G, k i−1 S) (and the observation that automorphisms preserve the value of the function #).

Comments on the lemma
This section is optional.
Remarks 3.1. Two comments on assumption (2) of Lemma 2.2: 1. This assumption holds for all S ⊆ G if and only if gcd k, |G| = 1.
2. This assumption can be weakened. For example, if k is prime, then it suffices to assume, for each s ∈ S, that |{ t ∈ S | ks = kt }| is not divisible by k.  Proof. (⇒) First, note that 0 / ∈ S, because a connected, edge-transitive graph with p vertices cannot have loops. Since p is prime, this implies S ⊆ Z × p . Now, let A 0 = { ϕ ∈ Aut X | ϕ(0) = 0 } be the stabilizer of the vertex 0 in Aut X. For every k ∈ Z × p , Lemma 2.2 tells us that kS is A 0 -invariant. Since S = 1S is also A 0invariant, this implies that kS ∩S is A 0 -invariant. However, A 0 is transitive on S (because S = N X (0) and X is edge-transitive), so this implies that either kS = S or S ∩ kS = ∅. Since this is true for all k (and S ⊆ Z × p ), this means that S is a block of imprimitivity for the regular representation of Z × p [3, p. 12]. So S is an orbit of some subgroup H of Z × p [3, Thm. 1.5A, pp. 13-14]: S = Hz for some z ∈ Z × p . Since we are dealing with the regular representation, this means that S is a coset of H.
(⇐) This is the easy direction (and does not require the assumption that p is prime). Assume S is a coset of the subgroup H of Z × p . Note that H acts by automorphisms on the group Z p (because multiplication by any k ∈ Z × p is an automorphism of Z p ). The set S is invariant under H; indeed, H is transitive on S (because S is a coset of H). The proof is now completed by a well-known, elementary argument [11,Lem. 2.8]: H is a group of automorphisms of the Cayley graph X (because it is a group of automorphisms of Z p that fixes S). Since H fixes the vertex 0, and acts transitively on the set S of neighbours of 0, this implies that X is edge-transitive.
Remarks 3.4. Unlike in [5,11], we do not need to assume that Cay(G; S) is a simple graph.
1. Graphs may have loops. Lemma 2.2 (as stated, without considering any edge-colourings) implies that ϕ is an automorphism of Cay(G; kS c ). Saying that this is true for every c is exactly the same as saying that ϕ is a (colour-preserving) automorphism of Cay(G; kS), if we let colour(t) = { colour(s) | t = ks, s ∈ S } for each t ∈ kS.
Also note that this proof only applies the original version of Lemma 2.2 to S c , not all of S, so hypothesis (2) can be replaced with the weaker assumption that: (2 ) for every colour c, we have ks = kt for all s, t ∈ S c , such that s = t. 4. Technically, we do not allow graphs to have multiple edges. However, since the statements of the results only consider automorphism groups, not other graphical properties, the multiplicity of an edge can be encoded as part of its colour (or "label"). For example, an edge coloured "2B, 3R, W" could be thought of as representing 2 blue edges, 3 red edges, and a white edge, all with the same endpoints. Remark 4.2. Synonyms for "twin-free" include "irreducible" [5], "R-thin" [8, p. 91], and "vertex-determining" [11].

Proof of the main theorem
Let S be a symmetric subset of a finite abelian group G of odd order, such that the Cayley graph X = Cay(G; S) is twin-free and connected. Given ϕ ∈ Aut BX, we wish to show that ϕ ∈ Aut X × S 2 .
Note that and that BX is connected and bipartite, with bipartition sets G × {0} and G × {1}. Since Aut X × S 2 contains an element that interchanges these two sets, we may assume 1. Each element x of N X (v) is labelled with the colour of the edge from v to x. Therefore, saying that N X (v) = N X (w) means that, for each x ∈ N X (v), the colour of the edge joining x to v is same as the colour of the edge joining x to w. Hence, an edge-coloured graph may be twin-free, even though its underlying uncoloured graph is not twin-free.
2. The edges of BX are coloured by colouring the edge from (v, 0) to (w, 1) with whatever colour appears on the edge from v to w in X.
It is now easy to complete the proof, by using the assumption that X is twin-free. For all g ∈ G, we have (definition of BX).
Since X is twin-free, this implies g = g , so ϕ(g, 1) = (g , 1) = (g, 1), which means ϕ(v) = v for all v ∈ G × {1}. Since this equality also holds for all v ∈ G × {0}, we conclude that ϕ is the identity element of Aut BX, and is therefore in the subgroup Aut X × S 2 .

Review of direct products
x 1 is adjacent to x 2 in X, and y 1 is adjacent to y 2 in Y .
Remark 5.2 ([8, p. 36]). The literature has numerous other names for the direct product, including "tensor product, " "Kronecker product, " "cardinal product, " and "conjunction. " The graph X in Theorem 1.1 is allowed to have edge-colours and multiple edges, but the theory of automorphisms of direct products does not seem to have been developed in this generality, so: Assumptions 5.3. In this section (and, therefore, in Corollaries 1.5 and 1.6), graphs do not have edge-colours or multiple edges (but they may have loops).
As was mentioned in Section 1, we have BX = X × K 2 . Generalizing the comments there about Aut BX, it is clear that Aut X ×Aut Y is a subgroup of Aut(X ×Y ), and that if this subgroup happens to be all of Aut(X × Y ), then X and Y must be connected, and must also be twin-free. (We ignore the situation where one of the graphs is the one-vertex trivial graph.) For direct products of non-bipartite graphs, the converse holds if and only if a certain "coprimality" condition holds. However, instead of stating the full strength of this classical theorem of W. Dörfler, we present only a simpler, weakened version of the result: Theorem 5.4 (Dörfler, cf. [4] or [8,Thm. 8.18,p. 103]). Let X and Y be twin-free, connected, non-bipartite graphs of relatively prime orders. Then The situation is more complicated (and not yet understood) when one of the factors of a direct product is bipartite. However, the following facts shed some light, as will be seen in the proposition that follows.
Facts 5.5. Let X and Y be connected graphs.
1. The Cartesian skeleton of X is a certain graph SX that is defined from X, such that V (SX) = V (X) [ The following straightforward consequence of these facts is presumably known to experts, but we do not have a reference. The gist is that, in order to understand the automorphism group of X × Y , where Y is bipartite, it often suffices to understand the special case where Y = K 2 .
Proposition 5.6. Let X and Y be twin-free, connected graphs that have at least one edge, such that: Then Aut(X × Y ) = Aut X × Aut Y .
Proof. Let ϕ ∈ Aut(X × Y ). From Fact 6, we know that SY has two connected components C 0 and C 1 , where V (C 0 ) = Y 0 and V (C 1 ) = Y 1 . By Facts 2 and 3, we have If |V (C 0 )| = |V (C 1 )|, then (by hypothesis (2b)) Y has an automorphism that interchanges C 1 and C 2 ; therefore, we may assume that ϕ fixes each connected component of SX SY . This means that ϕ restricts to an automorphism ϕ i of SX C i (for i = 0, 1). Since SX is connected (by Fact 5), and the connected component C i is obviously also connected, Fact 4 (and hypothesis (2a)) tells us that there exist permutations χ i of V (X) and η i of Y i , such that ϕ(x, y) = χ i (x), η i (y) for all x ∈ V (X) and y ∈ Y i .
Choose adjacent vertices y 0 and y 1 of Y , with y i ∈ Y i , and let y i = η i (y i ). Let B and B be the subgraphs of X × Y induced by V (X) × {y 0 , y 1 } and V (X) × {y 0 , y 1 }, respectively, so B = ϕ(B). By definition of the direct product, the maps (x, i) → (x, y i ) and (x, i) → (x, y i ) are isomorphisms from BX to B and B . Therefore, the map (x, i) → χ i (x), i is an automorphism of BX. So Assumption (3) tells us that χ 0 = χ 1 . This means that the V (X)-component of ϕ(x, y) depends only on x. (And we already knew that the V (Y )component of ϕ(x, y) depends only on y.) This easily implies ϕ ∈ Aut X × Aut Y .