On the asymptotic enumeration of Cayley graphs

In this paper, we are interested in the asymptotic enumeration of Cayley graphs. It has previously been shown that almost every Cayley digraph has the smallest possible automorphism group: that is, it is a digraphical regular representation (DRR). In this paper, we approach the corresponding question for undirected Cayley graphs. The situation is complicated by the fact that there are two infinite families of groups that do not admit any graphical regular representation (GRR). The strategy for digraphs involved analysing separately the cases where the regular group R has a nontrivial proper normal subgroup N with the property that the automorphism group of the digraph fixes each N-coset setwise, and the cases where it does not. In this paper, we deal with undirected graphs in the case where the regular group has such a nontrivial proper normal subgroup.


Introduction
We consider only finite groups and finite (di)graphs in this paper. A digraph Γ is an ordered pair (V, E) with V a finite non-empty set of vertices and with E a subset of the Cartesian product V × V . In particular, Γ is a binary relation on V. We say that Γ = (V, E) is a graph if E = {(w, v) | (v, w) ∈ E} , that is, if Γ is a symmetric binary relation. An automorphism of a digraph or of a graph is a permutation on V that preserves the set E. Definition 1.1 Let R be a group and let S be a subset of R. The Cayley digraph Γ(R, S) is the graph with V = R and with (r, t) ∈ E if and only if tr −1 ∈ S.
When the set S is inverse-closed (that is, S = S −1 ∶= {s −1 | s ∈ S} ), the digraph Γ(R, S) is actually a graph, which we refer to as the Cayley graph on R with connection set S.
The problem of finding digraphical and graphical regular representations (DRRs and GRRs) for groups has a long history. Mathematicians have studied graphs with specified automorphism groups at least as far back as the 1930s, and in the 1970s, there were many papers devoted to the topic of finding GRRs (see for example [2, 10-13, 19-21, 24]), although the "DRR" and "GRR" terminology was coined somewhat later. Definition 1.2 A digraphical regular representation (DRR) for a group R is a digraph whose automorphism group is the group R acting regularly on the vertices of the graph.
A graphical regular representation (GRR) for a group R is a digraphical regular representation which is a graph.
It is an easy observation that when Γ(R, S) is a Cayley (di)graph, the group R acts regularly on the vertices as a group of graph automorphisms. A DRR (respectively, GRR) for R is therefore a Cayley digraph (respectively, Cayley graph) on R that admits no other automorphisms.
The main thrust of much of the work through the 1970s was to determine which groups admit GRRs. This question was ultimately answered by Godsil in [8].

Theorem 1.3 (Godsil, [8]) A group has a graphical regular representation if and only if it is not one of:
• a generalised dicyclic group (see Definition 1.9); • an abelian group of exponent greater than 2; or • one of 13 small groups (of order at most 32).
A corresponding result for DRRs by Babai [2] was much simpler, requiring no excluded families and finding only 5 exceptional small groups.
Babai and Godsil made the following conjecture.

Conjecture 1.4 ( [3]
; Conjecture 3.13, [9]) If R is not generalised dicyclic or abelian of exponent greater than 2, then for almost all inverse-closed subsets S of R, Γ(R, S) is a GRR.
The details of this conjecture are somewhat imprecise; we are interested in the following more specific formulation: Given a finite group R, we let 2 (R) denote the number of inverse-closed subsets of R, see also Definition 1.8. From Godsil's theorem, as r → ∞ , the condition "R admits a GRR" is equivalent to "R is neither a generalised dicyclic group, nor abelian of exponent greater than 2." The corresponding result for Cayley digraphs (which does not require any families of groups to be excluded) was proved by the first and third authors in [17].
The strategy used in [17] (which was based on previous work in [3] by Babai and Godsil) to prove that almost every Cayley digraph is a DRR, involved three major pieces. One piece was to show that there are not many Cayley digraphs admitting digraph automorphisms that are also group automorphisms. A second piece of the proof involved considering the possibility that the group R has a proper nontrivial normal subgroup N, and there is a digraph automorphism that fixes every orbit of N setwise. This piece itself naturally divides into two parts. If |N| is relatively small in comparison with |R|, then showing that roughly 2 |R|∕|N| digraphs do not admit a particular type of automorphism is significant, while if |N| is relatively large (for example if |N| = |R|∕c for some constant c) this sort of bound is not useful for our purposes. Conversely, if |N| is relatively large then showing that roughly 2 |N| digraphs do not admit a particular type of automorphism is significant, but such a bound is not useful if |N| is relatively small. So, we need to combine bounds of each type to come up with an overall bound. The third and final piece of the proof involved considering the possible existence of digraph automorphisms that do not fix all orbits of any normal subgroup N of R.
While the second piece may not seem entirely natural, it is important to consider because it covers a possibility that does not readily succumb to induction. If a graph only admits automorphisms that fix every orbit of N setwise, then the quotient graph on the orbits of N may be in fact a GRR. The induced subgraph on a single orbit may very well also be a GRR, so an inductive argument will reduce a non-GRR to two smaller GRRs, making induction virtually impossible to use effectively.
Similarly to the results about existence of GRRs and DRRs, the requirement that a connection set for a graph must be inverse-closed creates complications that make the proof of the Babai-Godsil conjecture more difficult for graphs than for digraphs. Rather than trying to accomplish the full result in a single paper, it makes sense to divide the work into the main pieces that were used to prove the DRR result and attempt to show each of these pieces for GRRs.
The first piece, showing that there are not many Cayley graphs admitting graph automorphisms that are also group automorphisms (unless the group is generalised dicyclic or abelian of exponent greater than 2) was accomplished by the third author in [22]. Some of the main results from that work are also used in this paper, and we have included them as Theorem 1.13 and Proposition 1.14.
The goal of this paper is to complete the second piece of the proof: that is, to show that the number of Cayley graphs on R that admit nontrivial graph automorphisms that fix the vertex 1 and normalise some proper nontrivial normal subgroup N of R, is vanishingly small as a proportion of all Cayley graphs on R.
As in the work on DRRs, this problem naturally divides into the cases where the normal subgroup N is "large" or "small" relative to |R|. Our main results are Theorem 1.5 and Theorem 1.6, which we prove in Sects. 3 and 4, respectively. In the case of graphs, lim r→∞ min |{S ⊆ R ∶ Aut(Γ(R, S)) = R}| 2 (R) ∶ R admits a GRR and |R| = r = 1.
be the set of elements of R having order at most 2. Given a subset X of R, we write (X) ∶= X ∩ (R) . Given an inverse-closed subset X of R, we let Definition 1.9 Let A be an abelian group of even order and of exponent greater than 2, and let y be an involution of A. The generalised dicyclic group Dic(A, y, x) is the group ⟨A, x | x 2 = y, a x = a −1 , ∀a ∈ A⟩ . A group is called generalised dicyclic if it is isomorphic to some Dic(A, y, x) . When A is cyclic, Dic(A, y, x) is called a dicyclic or generalised quaternion group. We let ̄A ∶ Dic(A, y, x) → Dic(A, y, x) be the mapping defined by (ax)̄A = ax −1 and āA = a , for every a ∈ A . In particular, ̄A is an automorphism of Dic(A, y, x) . The role of the label "A" in ̄A seems unnecessary; however, we use this label to stress one important fact. An abstract group R might be isomorphic to Dic(A, y, x) , for various choices of A. Therefore, since the automorphism ̄A depends on A and since we might have more than one choice of A, we prefer a notation that emphasizes this fact.
It follows from [18, Section 2.1 and 4] that if D = Dic(A, x, y) is generalized dicyclic over A, then either A is characteristic in D, or D ≅ Q 8 × C 2 for some ∈ ℕ . In particular, when D is not isomorphic to Q 8 × C 2 , the automorphism ̄A is uniquely determined by D.
When D = Q 8 × C 2 , the group D is generalized dicyclic over three distinct abelian subgroups, namely if Q 8 = ⟨i, j⟩ , then D is generalized dicyclic over ⟨i⟩ × C 2 , ⟨j⟩ × C 2 and ⟨ij⟩ × C 2 . In particular, we have three distinct options for the automorphism ̄A : one for each of these abelian subgroups. For simplicity, we denote by ̄i,̄j and ̄k the corresponding automorphisms. It is not hard to check that ̄k =̄īj and hence ⟨̄i,̄j⟩ is elementary abelian of order 4. Definition 1.10 Let A be an abelian group. We let A ∶ A → A denote the automorphism of A defined by x A = x −1 ∀x ∈ A . Very often, we drop the label A from A because this should cause no confusion.
In what follows we use the following facts repeatedly.

Remark 1.11
Let X be a finite group. Since a chain of subgroups of X has length at most log 2 (|X|) , X has a generating set of cardinality at most ⌊log 2 (�X�)⌋ ≤ log 2 (�X�).

Lemma 1.12
Let R be a finite group and let X be an inverse-closed subset of X. The number of inverse-closed subsets S of X is 2 (X) . In particular, R has 2 (R) inverse-closed subsets.
Proof Given an arbitrary inverse-closed subset S of X, S ∩ (X) is an arbitrary subset of (X) whereas in S ∩ (X ⧵ (X)) the elements come in pairs, where each element is paired up to its inverse. Thus, the number of inverse-closed subsets of X is The last statement follows using X = R . ◻ The following important results by the third author deal with the case where there is a graph automorphism that is also a group automorphism of R. Theorem 1.13 ([22], Lemma 2.7) Let R be a finite group and let be a non-identity automorphism of R. Then, one of the following holds R ( ) is abelian of exponent greater than 2 and has index 2 in R, R is a generalized dicyclic group over R ( ) and =̄ R ( ) , (3) R is abelian of exponent greater than 2 and is the automorphism of R mapping each element to its inverse.

Notation 1.15
With R a finite group that is neither abelian of exponent greater than 2 nor generalised dicyclic, we define so that |S N | is a value we aim to bound to prove Theorem 1.5. We divide S N into three subsets: so Observe that Proposition 1.14 already provides us with a bound for |S 1 N | . In the next section, we will show that |U N | = 0.

Graph automorphisms that fix or invert every group element
The bulk of this section consists of a long lemma in which we show that if a nontrivial permutation that fixes or inverts every element of a group exists, then the normaliser of R in the appropriate group is in fact larger than R. This means that any connection sets that could arise in U N have actually already arisen in S 1 N , and therefore do not appear in U N . Lemma 2.1 Let G be a subgroup of Sym(R) with R < G and with the property that r g ∈ {r, r −1 } , for every r ∈ R and for every g ∈ G 1 . Then, G (R) > R.
Proof We argue by contradiction and, among all groups satisfying the hypothesis of this lemma, we choose G with |R||G| as small as possible and with In this proof, we denote by r g the image of the point r ∈ R via the permutation g and we denote by r g ∶= g −1 rg the conjugation of r via g.
Let M be a subgroup of G with R < M . For every r ∈ R and for every x ∈ M 1 = M ∩ G 1 , r x ∈ {r, r −1 } , and, from the modular law, Therefore, by the minimality of our counterexample, we get M = G . As M was an arbitrary subgroup of G with R < M , we deduce Let K be the core of R in G, that is, K ∶= ⋂ g∈G R g . We claim that To prove this claim, we argue by contradiction and we suppose that K ≠ 1 . Let Ḡ be the permutation group induced by G on the action on K-orbits. Moreover, we let ̄∶ G →Ḡ denote the natural projection.
Let H be the kernel of ̄ . Thus, H is the largest subgroup of G fixing each K-orbit setwise and H ≤ G 1 K . Since R is a maximal subgroup of G and R ≤ RH ≤ G , we have that either R = RH or G = RH.
In the first case, H ≤ R and, since H ≤ G 1 K , from the modular law we obtain and, for every r ∈R and for every ḡ ∈Ḡ 1 , we have r̄g ∈ {r,r −1 } . Using our assumption that K ≠ 1 , we get that |R| < |R| , and by the minimality of our couterexample we have that Ḡ = G∕K = R∕K =R . That is, G = R contradicting the fact that R is a proper subgroup of G.
So the second case holds, and G = RH , so G 1 acts trivially on K-orbits. In other words, G 1 fixes each K-orbit setwise. Thus, H = KG 1 , and consequently (2.1) R is a maximal subgroup of G.
(2.2) the core of R in G is 1.
Suppose there exist x ∈ G 1 and r ∈ R such that r x = r −1 and o(rK) ≥ 3 . Then r x = r −1 ∈ r −1 K = (rK) −1 ≠ rK , contradicting the fact that G 1 fixes each K-orbit. This shows that Let L be the subgroup of R fixed pointwise by G 1 , that is, L ∶= {r ∈ R | G r = G 1 } . (The set L is indeed a subgroup of R, because it is a block of imprimitivity for the action of G on R containing the point 1.) Clearly, L < R , because G 1 ≠ 1 . Now, from (2.4), we deduce that for every r ∈ R ⧵ L , o(rK) ≤ 2 . Hence, Now, by (2.5), we must have ⟨xK ∈ R∕K | x 2 ∉ K⟩ ≤ L∕K . Since either �R∕K ∶ ⟨xK ∈ R∕K | x 2 ∉ K⟩� = 2 or R/K is a 2-group, we deduce that one of the following holds In what follows, we analyze these three alternatives.
Case (1) Since R/K and G 1 are elementary abelian 2-groups, we deduce that G/K is a 2-group. From R∕K < G∕K , it follows that G∕K (R∕K) > R∕K . So G (R) > R , but this contradicts our choice of G and R.
Case (2) Let f ∈ G 1 with f ≠ 1 . Now, as G 1 normalizes K, the action of f on the points in K coincides with the action of f by conjugation on K. Thus, k f = k f ∈ {k, k −1 } , for every k ∈ K .
In particular, f is a non-trivial automorphism of K with the property that it maps each element to itself or to its inverse (so every inverse-closed subset of K is invariant under f ). Therefore using Theorem 1.13 only one of the following holds true: • K is abelian of exponent greater than 2 and f = is the automorphism inverting each element of K, • K is generalised dicyclic over an abelian subgroup A of exponent greater than 2 and f =̄A, Since R = KL and since G 1 fixes L pointwise, the action of g ∈ G 1 on R is uniquely determined once the action of g on K is determined. Since we have at most four choices for the action of g ∈ G 1 on K, we deduce that |G 1 | divides 4. If |G 1 | = 2 , then |G ∶ R| = 2 and hence R ⊴ G , which contradicts R = G (R) . Thus 4 = |G 1 | = |G ∶ R| and K ≅ Q 8 × C 2 , for some ≥ 0.
Since |G ∶ R| = 4 , the transitive action of G on the right cosets of R gives rise to a permutation group of degree 4 and hence G/K is isomorphic to a transitive subgroup of Sym(4) . As R∕K = G∕K (R∕K) , we deduce that G/K is isomorphic to either Sym(4) or Alt(4).

(2.4)
for every x ∈ G 1 and for every r ∈ R either r x = r or o(rK) ≤ 2.
is an involution.
If R/K were a 2-group, we reach a contradiction using the same argument as in Case (1). So R/K is a maximal subgroup of G/K which is not a 2-group, hence R/K isomorphic to either Sym(3) or Alt (3).
Let C be a Sylow 3-subgroup of R. Thus C = ⟨c⟩ is a cyclic group of order 3. Since K is a 2-group and R = KL , replacing C by a suitable R-conjugate, from Sylow's theorem, we can assume that C ≤ L . Let k ∈ K with k ∉ L . As k is not fixed by each element of G 1 , there exists x ∈ G 1 such that k x = k −1 ≠ k . Now, as c x −1 = c , we obtain On the other hand, (ck) x ∈ {ck, (ck) −1 } . If (ck) x = ck , then we deduce k = k −1 , contradicting the fact that k x ≠ k . If (ck) x = (ck) −1 , we deduce k −1 c −1 = ck −1 and hence k −1 = ck −1 c = c 2 (k −1 ) c . Again we obtain a contradiction because k and k c belong to K but c 2 ∉ K.
Case (3) Before proceeding with this case, we collect some information on G/K. Observe that in this case, R/K is a generalized dihedral group over the abelian group KL/K. Consider the set Ω of the right cosets of R/K in G/K. By (2.1) R/K is a maximal subgroup of G/K. So G/K is a primitive permutation with generalised dihedral point stabilisers.
These groups were classified in [7,Lemma 2.2]. Using this and the fact that G 1 is 2-elementary abelian group, the only possibility that can occur is that G/K is a primitive group of affine type of degree |R ∶ K| = |G 1 | . Since G = G 1 R and R ∩ G 1 = 1 , G 1 K∕K acts regularly on Ω . Moreover, as KG 1 ⊴ G by (2.3), G 1 K∕K is the socle of G/K. Since every element of G 1 is an involution (it fixes or inverts each element of R), then G 1 K∕K is an elementary abelian 2-group. Now, R/K acts by conjugation irreducibly as a linear group over the elementary abelian 2-group G 1 K∕K. Let K ∈ LK∕K ⧵ {K}. Since LK/K is abelian, then G 1 K∕K ( K) = {aK ∈ G 1 K∕K | −1 a K = aK} is stable under the conjugation by uK, for every uK ∈ LK∕K. Further, since R∕K = ⟨rK, LK∕K⟩ , where rK = r −1 K , and r −1 rK = −1 K, for every K ∈ LK∕K, then G 1 K∕K ( K) is stable under the conjugation by xK. In other words, we proved that G 1 K∕K ( K) is a proper R-submodule of the irruducible R-module G 1 K∕K, and consequently G 1 K∕K ( K) is trivial. Summing up, KL/K is abelian and G 1 K∕K ( K) is trivial for every K ∈ LK∕K ⧵ {K}. Thus, KL/K is a cyclic group of odd order. Moreover, as the socle G 1 K∕K has even order, |KL/K| must be odd. We let t ∶= |KL∕K| . At this point, the reader might find it useful to consider Fig. 1. Since KL/K is cyclic, there exists c ∈ L with ⟨c⟩K = KL and with o(cK) = t. Suppose now that K ≰ L and let k ∈ K ⧵ L . As k is not fixed by each element of G 1 , there exists x ∈ G 1 with k x = k −1 ≠ k . Now, since x fixes c, we are in position to use the same argument as in Case (2). That is (2.6) holds, and consequently either k = k −1 or c 2 ∈ K. Since k ≠ k −1 and o(cK) = t is odd, in both cases we get a contradiction.
We conclude that K ≤ L . (For the proof here, it might be useful again considering Fig. 1.) In particular, KL = L . Fix r ∈ R ⧵ L . As |R ∶ L| = 2 , we have R = L ∪ rL . Now, LG 1 fixes L and rL setwise. The action induced by LG 1 on L is the regular action of L because G 1 fixes L pointwise. As LG 1 ⊴ G , we must also have that the action of LG 1 on rL is simply the regular action of L. In particular, for every x ∈ G 1 , there exists x ∈ L with the property that The set { x | x ∈ G 1 } forms a subgroup of L, which we denote by T. As G 1 is elementary abelian, so is T.
Summing up, we have Using this and the fact that T is a group we see that if x ∈ G 1 fixes some point in rL, then x = 1 and consequently x fixes all points in rL. Further, x fixes all points in L, hence x = 1 . Therefore, each element in We have shown that none of the three alternatives is possible. Therefore, we obtain a contradiction, and the contradiction has arisen from assuming K ≠ 1 . Hence, K = 1 , which is our original claim (2.2). Now, as R is maximal in G and as R is core-free in G, we may view G as a primitive permutation group on the set Ω = G�R of right cosets of R in G. Observe that in this action G 1 acts as a regular subgroup and it is an elementary abelian 2-group which itself is core-free in G.
The primitive permutation groups containing an abelian regular subgroup have been classified by Cai Heng Li in [14]. Applying this classification [14, Theorem 1.1] to our group G in its action on Ω and to its elementary abelian regular subgroup G 1 , we deduce that one of the following holds: (1) G is an affine primitive permutation group, (2) the set Ω admits a Cartesian decomposition Ω = Δ (for some ≥ 1 ) and the primitive group G preserves this cartesian decomposition; moreover, T ≤ G ≤Twr Sym( ) , where the action of T wr Sym( ) on Δ is the natural primitive product action. The group T is either Alt(Δ) or Sym(Δ) , G 1 = G 1,1 × G 1,2 × ⋯ × G 1, with G 1,i ≤T and with G 1,i acting regularly on Δ , for each i. Now, we shall see that neither of these two alternatives is possible.

Case (1)
Let V be socle of G. Thus V ⊴ G and V is an elementary abelian 2-group. Observe that where the first equality follows from the fact that V acts transitively on Ω with point stabiliser R and the second equality follows because G acts also transitively on R with point stabilizer G 1 . Moreover, where the first equality follows because V acts regularly on Ω with point stabilizer R and the second equality follows because R acts regularly on itself with point stabilizer G 1 .
Since G 1 is a regular subgroup of the affine group G, from [5, Corollary 5 (1)], we deduce Let Since G 1 is abelian, we have G 1 ≤ N and hence Similarly, since V is abelian, we have V ≤ N and hence Thus, Let r ∈ R and let v ∈ V ∩ G 1 . We recall that r v ∈ {r, r −1 }.
If r v = r −1 , then 1 r −1 = r −1 = r v = 1 rv and hence rvr = r 2 (r −1 vr) ∈ G 1 . As V ⊴ G , we have r −1 vr ∈ V and hence r 2 V ∈ G 1 V∕V . Since all the elements of G 1 V∕V have order at most 2, it follows that r 4 V = V , that is r 4 ∈ V ∩ R = 1 . This shows that if o(r) ≠ 4 , then r −1 vr ∈ V ∩ G 1 . Therefore, all elements of R of order different from 4 normalise V ∩ G 1 and hence they all lie in Q.
This shows that R ⧵ Q is either empty, or contains only elements of order 4. In the first case (2.8) yields However, this contradicts the fact that G 1 is corefree in G. Thus Q < R and every element in R ⧵ Q has order 4. For every r ∈ R ⧵ Q , r 2 does not have order 4, so r 2 ∈ Q . This shows that Q contains the square of each element of R, hence and R/Q is an elementary abelian 2-group.
Let x ∈ G 1 and let r ∈ R . If r x = r , then rxr −1 ∈ G 1 ≤ G 1 Q = N . If r x = r −1 , then rxr ∈ G 1 and hence rxr = r 2 (r −1 xr) ∈ G 1 ≤ G 1 Q = N . Since r 2 ∈ Q , we deduce that r −2 ⋅ r 2 (r −1 xr) = r −1 xr ∈ N . We have shown that From (2.9) and (2.10), we deduce that R normalises G 1 Q = N . Since G 1 also normalizes N, we have that RG 1 = G normalises N, that is, Since Q ⊴ R and since R is a maximal subgroup of G by (2.1), we deduce that either G (Q) = G or G (Q) = R . If G (Q) = G , then Q is a normal subgroup of G contained in the core-free subgroup R. Therefore, Q = 1.
From (2.8), we have G 1 = QG 1 = N = QV = V , contradicting the fact that G 1 is corefree in G. Thus, When G is viewed as a permutation group on R, QG 1 is the setwise stabilizer in G of Q ⊆ R ; hence, we can consider the permutation group induced by N = QG 1 in its action on Q.
and, for every rH ∈ QH∕H and for every gH ∈ G 1 ∕H , we have r g H ∈ {rH, r −1 H} . If N∕H (QH∕H) = QH∕H , from the minimality of our counterexample, we deduce that either N = G or G 1 acts trivially on Q.
In the first case, Moreover, for every g ∈ G , from (2.13), we have G g 1 ⊴ N g = N . Since G 1 is an elementary abelian 2-group, then each G g 1 is a normal 2-subgroup of N, for every g ∈ G . Consequently U is a normal 2-subgroup of G. In particular, U ∩ R is a normal 2-subgroup of R.
Since V is an irreducible 2 R-module and U ∩ R ⊴ R , we deduce that V is completely reducible 2 (U ∩ R)-module by Clifford's theorem. Since V has characteristic 2 and since U ∩ R is a 2-group, this can happen only when Therefore, we can assume that N∕H (QH∕H) > QH∕H. That is, there exists a non-identity element g ∈ G 1 normalizing QH/H. Hence, for every r ∈ Q , g −1 rg = uh , for some u ∈ Q and for some h ∈ H . Since g ∈ G 1 , and r g ∈ {r, r −1 } , we get u = u h = 1 uh = 1 g −1 rg = r g . This means that g −1 rgH ∈ {rH, (rH) −1 } for every r ∈ Q , and consequently g is a nonidentity automorphism of QH/H with the property that (rH) g ∈ {rH, (rH) −1 } , for every rH ∈ QH∕H . Thus from Theorem 1.13, Q ≅ QH∕H is either an abelian group of exponent greater than 2 or a generalized dicyclic group.
Since V is an irreducibly 2 R-module and 2 (Q) ⊴ R , we deduce that V is completely reducible 2 (Q)-module by Clifford's theorem. Since V has characteristic 2 and since 2 (Q) is a 2-group, this can happen only when If Q is a generalised dicyclic group, that is, Q = Dic(A, y, x) , with A an abelian group of even order and of exponent greater than 2, and y an involution in A, then ⟨y⟩ is a characteristic subgroup of order 2, which contradicts (2.14). Thus, Q is an abelian group, and Q has odd order by (2.14). Since N = QV = QG 1 by (2.11), and since V ⊴ N , then V is the unique Sylow 2-subgroup of N. As |G 1 | = |V| and G 1 ≤ N , we get G 1 = V . This contradicts the fact that G 1 is core-free in G.
Case (2) We identify Ω with Δ , and we recall that and, since G 1,1 is a 2-group, rearranging the points from 3 onwards if necessary, we can assume (Observe that |Δ| ≥ 8 because |Δ| is a power of 2 larger than 5.) Let consider the 3-cycle r = ( 2 3 4 ) and observe that it lies in R because it fixes the point 1 and R = G .
In this new setting, to look at the original action of G on R, we have to identify the set R with the set of right cosets of G 1 in G. In particular, is such a point. We have Since neither rgr −1 ∈ G 1 nor rgr ∈ G 1 , then G 1 rg ∉ {G 1 r, G 1 r −1 } . This contradicts our hypotheses.
We have shown that neither of the alternatives is possible. Therefore, we have contradicted the existence of such G and R. ◻ During the refereeing process of this paper, we found out that a short and elementary proof of Lemma 2.1 can be easily deduced from a classical result of Bergman and Lenstra [4, Theorem 1]. We have decided to keep our more elaborate proof hoping that it can play some role in possible generalizations. Lemma 2.1 is sufficient to show that U N is empty.

Corollary 2.2
When R is neither abelian of exponent greater than 2 nor generalised dicyclic, U N = �.
Proof Recall from Notation 1.15 that when R is neither abelian of exponent greater than 2 nor generalised dicyclic (2.14) 2 (Q) = 1.
while and Notice that the set of all elements of Aut(Γ(R, S)) that fix the vertex 1 and fix or invert every other element of R is a subgroup of Aut(Γ(R, S)) . By Lemma 2.1 with G being generated by R and the set of all such elements, we have U N = � . This is because every set that could lie in U N must appear in S 1 N . ◻

Groups with a "large" normal subgroup
We begin this section with a lovely little general result showing that in a non-abelian group, there cannot be a group automorphism such that the result of computing nn is constant for more than 3/4 of the group elements (and in fact in an abelian group, this can only happen if is the automorphism that inverts every group element). For the special case where is trivial and the constant is 1, our proof relies on (so does not replace) classical work by Liebeck and MacHale [15]. Lemma 3.1 Let N be a group, let be an automorphism of N and let t ∈ N . Then one of the following holds: Then nn = t because n ∈ S , and n (n ) = t because n ∈ S . Therefore, From this we obtain mt = t −1 m , that is, t m = t −1 . As n = mt , we also have t n = t −1 . We have shown that for every n ∈ S ⋅ t ∩ S , we have t n = t −1 . For every two elements n 1 , n 2 ∈ N with t n 1 = t −1 = t n 2 , we have n 1 n −1 2 ∈ N (t) . Therefore, we deduce that |N|∕2 < |S ⋅ t ∩ S| ≤ | N (t)| . Thus, N = N (t) and t ∈ (N) . Moreover, for every n ∈ St ∩ S , we have t n = t −1 and, as t ∈ (N) , we have t n = t . Thus, t 2 = 1 . Summing up, t is a central element of N of order at most 2. Suppose that t = 1 . Then S = {n ∈ N | n = n −1 } . In particular, is an automorphism inverting more than 3|N|/4 of the elements of N. From a classical result of Liebeck and MacHale [15], we deduce that N is abelian and is the automorphism inverting each element of N, that is, n = n −1 ∀n ∈ N.
Suppose that t ≠ 1 . Since t ∈ (N) and since t = t , we may consider the group N ∶= N∕⟨t⟩ and the induced automorphism ̄∶N →N . In particular, in N , the set S projects to the set S = {n ∈N |n̄=n −1 } . Since this set has cardinality larger than 3|N|∕4 , applying again the theorem of Liebeck and MacHale, we deduce that N is abelian and n̄=n −1 ∀n ∈N . It follows that for every n ∈ N , contradicting the fact that |S � | < |N|∕4 . This contradiction has arisen assuming that N is not abelian and hence N is abelian. Now, for every n, m ∈ S , we have (nm) = n −1 t ⋅ m −1 t = n −1 m −1 t 2 = (nm) −1 and hence nm ∈ S � . Therefore, S ⋅ S ⊆ S ′ , but this is impossible because |S ′ | < |S| . This contradiction has arisen from assuming t ≠ 1 and hence t = 1 and the proof is now complete. ◻ We will also require a similar result that considers when inversion is applied after the automorphism. Lemma 3.2 Let N be a group, let be an automorphism of N and let t ∈ N . Then one of the following holds: Proof The proof of this is very similar to the proof of Lemma 3.1, so we omit some of the repeated details.
As before, by taking some n ∈ S −1 ∩ S , we can conclude that t = t .
As |S| > 3|N|∕4 , we can argue as before that |S −1 t ∩ S| > |N|∕2 . Let n ∈ S −1 t ∩ S . Then n = mt , for some m ∈ S −1 ; that is, m −1 ∈ S . Notice that this means (m −1 ) = t −1 m −1 , so m = mt . Therefore From this we obtain mt = t −1 m , that is, t m = t −1 . As n = mt , we also have t n = t −1 . We have shown that for every n ∈ S −1 t ∩ S , we have t n = t −1 . As before, this implies that |N|∕2 < |S −1 t ∩ S| ≤ | N (t)| . Thus, N = N (t) and t ∈ (N) . As before, this implies that In particular, is an automorphism fixing more than half of the elements of N. Since the set of fixed points of an automorphism is a subgroup of N, we deduce that = 1 ; that is, n = n ∀n ∈ N.
Suppose that t ≠ 1 . Since t ∈ (N) and since t = t , we may consider the group N ∶= N∕⟨t⟩ and the induced automorphism ̄∶N →N . In particular, in N , the set S projects to the set S = {n ∈N |n̄=n} . Since this set has cardinality larger than |N|∕2 , again we see that n̄=n ∀n ∈N . It follows that for every n ∈ N , n ∈ ⟨t⟩n = {n, tn}.
Now, for every n, m ∈ S , we have (nm) = (tn)(tm) = (nm)t 2 = nm since t is central of order 2, and hence nm ∈ S � . Therefore, S ⋅ S ⊆ S ′ , but this is impossible because |S ′ | < |S| . Again this contradiction completes our proof. ◻ Our next few results show that except in some very special cases, if we have a group T with an index-2 subgroup N and a permutation of T that has a very specific sort of action on every element of the nontrivial coset of N in T, then the number of subsets of T that are closed under both inversion and this permutation is vanishingly small relative to the number of Cayley graphs on T. Lemma 3.3 Let T be a finite group, let N be a subgroup of T having index 2, let ∈ T ⧵ N , let t ∈ N and let t ∶ T → T be any permutation defined by Then one of the following holds:

is the only non-identity square in T and N is an elemen
In parts (2), (3) and (4), if n t ∈ {n, n −1 } for every n ∈ N , then we have Proof If t = 1 , then we obtain part (4). Thus, for the rest of the argument, we assume t ≠ 1.
Observe that t fixes N setwise and induces on T ⧵ N a permutation which is the product of disjoint cycles each of whose lengths is o(t). For simplicity, we let n t ∈ N and ( n) t = tn, ∀n ∈ N. and hence part (1) follows. The only remaining possibility is The orbits of H on T ⧵ N have even cardinality because o( t ) = o(t) = 2 and t has no fixed points on T ⧵ N . There are only two possibilities for H having an orbit of cardinality 2 on T ⧵ N: • this orbit is { n, tn} where both n and tn are involutions (in this case fixes both n and tn), , the elements of the first type are in the set Let n 1 be an element in N with ( n 1 ) −1 = tn 1 . Let n ∈ N and suppose that n 1 n ∈ T ⧵ N also satisfies ( n 1 n) −1 = tn 1 n . This means n −1 tn 1 = tn 1 n , that is, n ( tn 1 ) −1 = n −1 . Therefore, the elements of the second type are in the set Observe that A or B might be the empty set: A = � when there is no involution in Since X ∈ S if and only X is a union of orbits of H, we get If |B| ≤ 3|N|∕4 , then and part (1) follows. Suppose now that |B| > 3|N|∕4 , that is, |{n ∈ N | n tn 1 = n −1 }| > 3|N|∕4 . This means that the action of tn 1 by conjugation on N inverts more than 3/4 of the elements of N. From [15], N is abelian and the action of tn 1 by conjugation on N inverts each element of N. Therefore B ⊃ N and hence ∈ B . Therefore −1 = t , that is, t = 2 (since o(t) = 2 ). When N is an elementary abelian 2-group, we deduce T ≅ C 4 × C 2 for some ∈ ℕ and hence part (2) holds. When N has exponent greater than 2, we deduce T = Dic(N, 2 , ) and hence part (3) The hypotheses of the next lemma look much like the previous one, with the additional assumption that N is abelian (of exponent greater than 2), and a different action on the nontrivial coset of N. The exceptional cases and the proof are quite different, though.
Lemma 3.4 Let T be a finite group, let N be an abelian subgroup of T having index 2 and exponent greater than 2, let t ∈ N , let ∈ T ⧵ N , let t ∶ T → T be any permutation defined by Further suppose that either o( ) = 2 , or ( n) t = n whenever o( n) = 2 . Then one of the following holds: In parts (2), (3) and (4), if n t ∈ {n, n −1 } for every n ∈ N , then we have Proof We let ∶ T → T the permutation defined by x = x −1 ∀x ∈ T . Since N is abelian, for every n ∈ N , we have Thus, t is a permutation having order 2. Clearly, has also order 2. For simplicity, we let S ∶= {X ⊆ T | X = X −1 , X t = X} . In particular, X ∈ S if and only if X is ⟨ t , ⟩-invariant, that is, X is a union of ⟨ t , ⟩-orbits.
Given n ∈ N , the ⟨ ⟩-orbit containing n is { n, n −1 −1 } . Now, there are only two possibilities for t not fusing this ⟨ ⟩-orbit with another ⟨ ⟩-orbit. The first possibility is when t fixes both n and n −1 −1 ; the second possibility is when ( n) t = ( n) , that is, Given n ∈ A , we have tn −1 = ( n) t = n and, from (3.1), t n = (n −1 −1 ) t = n −1 −1 . The first equality yields n 2 = t . The second equality yields where in the second equality we have used that 2 ∈ N and that N is abelian. Therefore, if n ∈ A , then n 2 = t and t = −1 t −1 −3 . Observe that the second condition does not depend on n any longer. This means that we have two possibilities for A ; either A = � , or  Since n 2 1 = 2 is the unique involution that is a square in N, we get part (3). In the second possibility, −2 = t = n 2 0 . If we also have ( n 0 ) 2 = t , then T = Dic(N, 2 , ) and we obtain again part (3). If ( n 0 ) 2 ≠ t , then ⟨ , n 0 ⟩ has order 16 and is isomorphic to the group with presentation ⟨x, y | x 4 = y 4 = (xy) 4 = 1, x 2 = y 2 ⟩ and we obtain part (4).
Case o( ) = 2 . For every n ∈ N , from (3.1) (and using o( ) = 2 ), we have Moreover, n t t ∈ N ∀n ∈ N . Define z ∶= (tt � ) −1 and ∶ T → T by If this group has order 2, then |T⧵N must be either |T⧵N or the identity permutation. Suppose that |T⧵N = |T⧵N . Then, for every n ∈ N , we have n −1 = zn , so n = zn −1 and hence nn = z . But since z ≠ 1 , Lemma 3.1 implies that we cannot have z = nn for every n ∈ N.
⟨ ⟩ on T ⧵ N have all length o(z) = 2 , we have |S| ≤ 2 (N)+|N|∕2 = 2 (T)−|N|∕2 and part (1) follows. It remains to consider the case that ⟨ �T⧵N , �T⧵N ⟩ has order 4. By the orbit counting lemma, the number of orbits of ⟨ ⟩ on T ⧵ N is Also, by the orbit counting lemma, the number of orbits of ⟨ �T⧵N , �T⧵N ⟩ on T ⧵ N is where in the first equality we have used (3.3) and in the second equality we have used the fact that has no fixed points on T ⧵ N . Now, n ∈ Fix T⧵N ( ) if and only if n = ( n) = z(n −1 ) , that is, z = nn . From Lemma 3.1, we deduce |Fix T⧵N ( )| ≤ 3|N|∕4 because z ≠ 1 . Thus and part (1) follows.
In this case, tt = z = 1 and t = t −1 . In this case, for every n ∈ N , we have This shows that t = t on T ⧵ N , and hence (in particular) ⟨ �T⧵N , ( t ) �T⧵N ⟩ is an elementary abelian 2-group of order 1, 2 or 4. If ( t ) |T⧵N is the identity mapping, then n = ( n) t = tn −1 , for every n ∈ N . In particular, t = tt −1 which implies t = 1 . This means that for every n ∈ N , n = ( n) t = n −1 , so that N is an elementary abelian 2-group, contradicting our hypothesis that N has exponent greater than 2.
It only remains to consider the case that ⟨ �T⧵N , ( t ) �T⧵N ⟩ has order 4. By the orbit counting lemma, the number of orbits of ⟨ , t ⟩ on T ⧵ N is where the equality between the two members follows by (3.3). If |Fix T⧵N ( t )| ≤ |N|∕3 and |Fix T⧵N ( t )| ≤ |N|∕2 , or |Fix T⧵N ( t )| ≤ |N|∕2 and |Fix T⧵N ( t )| ≤ |N|∕3 , then we immediately obtain part (1). Therefore, we suppose that this does not hold. An easy computation reveals that As The next lemma again has a similar flavour. This time we are assuming that the index-2 subgroup N of T is generalised dicyclic, and we need to assume that our permutation fixes each of the cosets of the abelian subgroup A of N setwise.

Lemma 3.5 Let T be a finite group, let N = Dic(A, y, x) be a generalised dicyclic subgroup of T having index 2, let t ∈ N , let ∈ T ⧵ N , let t ∶ T → T be any permutation defined by
Recall that ̄A is given in Definition 1.9. Then one of the following holds: 2 = y = t and a = a −1 ∀a ∈ A, (3) t = 1 , ⟨ , A⟩ is abelian, and T = Dic(⟨ , A⟩, y, x).
In parts (2) and (3), if n t ∈ {n, n −1 } for every n ∈ N , then we have z t ∈ {z, z −1 } ∀x ∈ T. Proof We let ∶ T → T the permutation defined by z = z −1 ∀z ∈ T . For simplicity, we let S ∶= {X ⊆ T | X = X −1 , X t = X} . Observe that for every a ∈ A , we have a t ∈ A and Suppose o(t) ≥ 3 . Then, the orbits of ⟨ t ⟩ on A all have length o(t) ≥ 3 and hence and part (1) follows in this case. In particular, for the rest of the proof we may suppose that o(t) ≤ 2 . Since N is generalised dicyclic and t ∈ N , we obtain t ∈ A . Now, for every a ∈ A , we have ( a) t = ta ∈ A and hence A is t -invariant. Therefore, t has |A|/o(t) cycles on A . This also means that xA is t -invariant.
Suppose that 2 ∉ A , that is, A ≠ −1 A . Then, T/A is a cyclic group and N = ⟨ 2 , A⟩ . If o(t) ≠ 1 , then and part (1) follows in this case. Suppose then t = 1 . In this case, t fixes A pointwise. For every a ∈ A , we have As ⟨ 2 , A⟩ = N = Dic(A, y, x) and as all elements in N ⧵ A have order 4, we deduce o( 2 ) = 4 and o( ) = 8 . In particular, 3 ≠ −1 and from (3.6) we deduce that t has no fixed points on −1 A . Hence, t has at most |A|/2 cycles on −1 A . Therefore and part (1) follows in this case.
Henceforth, we may assume that 2 ∈ A . Then, ⟨ , A⟩ is a group having a subgroup A of index 2. Furthermore, since both N = ⟨x, A⟩ and ⟨ , A⟩ are index-2 subgroups of T, we must have ( x) 2 ∈ N ∩ ⟨ , A⟩ = A . Also, since and x both normalise A, so does x . So ⟨ x, A⟩ is a group having a subgroup of index 2 and t restricts to a permutation of ⟨ x, A⟩ . Since t ∈ A and o(t) ≤ 2 we see that x and t commute, so for every a ∈ A we have a t ∈ A, (xa) t ∈ xA, ∀a ∈ A, and ( n) t = tn̄A , ∀n ∈ N.
So, we can apply Lemma 3.3 to the group ⟨ x, A⟩ and the permutation ( t ) �⟨ x,A⟩ with x taking the role of the " " in that lemma, and x 2 t taking the role of "t." If part (1) in Lemma 3.3 holds, then and conclusion (1) holds. If part (2) in Lemma 3.3 holds, then A is an elementary abelian 2-group, but this contradicts our definition of a generalised dicyclic group together with our hypothesis that N is such a group.
We can also apply Lemma 3.3 to the group ⟨ , A⟩ and the permutation t . In this case takes the role of " " in the lemma, and t takes the role of "t".
If part (1)  We have now applied Lemma 3.3 to two different subgroups of T and have completed the proof except in the cases where parts (3) or (4) arise from both applications. We now consider these final four possible outcomes individually.
It is not possible that part (4) holds in both applications, since this would imply that t = 1 and x 2 = t , contradicting o(x) = 4 from the definition of a generalised dicyclic group.
If part (3) holds in both applications, then ⟨ x, A⟩ = Dic(A, ( x) 2 , x) implies that a x = a x = a −1 , so a = a for every a ∈ A . But ⟨ , A⟩ = Dic(A, t, ) implies that a = a −1 for every a ∈ A . Taken together, these imply that A is an elementary abelian 2-group, again a contradiction.
If part (3) holds in the first application and part (4) holds in the second, then we have t = 1 , ( o(x 2 t) = 2 ), x 2 t = ( x) 2 , and ⟨ x, A⟩ = Dic(A, ( x) 2 , x) . Since ⟨ x, A⟩ = Dic(A, ( x) 2 , x) , we see that a x = a x = a −1 , so a = a for every a ∈ A , and ⟨ , A⟩ is abelian. Since A⟩, y, x) . This is Finally, if part (4) holds in the first application and part (3) holds in the second, then we have y = x 2 = t , o(t) = 2 , t = 2 , and ⟨ , A⟩ = Dic (A, t, ) . This is conclusion (2). ◻ With these preliminary results in hand, we are ready to prove bounds on the number of connection sets that admit various types of graph automorphisms. Recall Notation 1.15. We already have bounds on |S 1 N | and on |U N | . Our goal in this section is to bound |T N | when |N| is relatively large. In order to do this, we need to further subdivide T N .

Notation 3.6
For what follows, R is a group that is neither generalised dicyclic, nor abelian of exponent greater than 2. We let N be normal subgroup of R and we let It should be clear from this definition that We will bound the cardinality of each of these sets. Most of the bounds we find will only be vanishingly small relative to 2 (R) if |N| is relatively large compared to |R|. Specifically, they will all work if |N| ≥ 9 log 2 |R| . In order to create the best possible bound, however, we will want to balance |N| against |R/N|, so we will use these bounds only when �N� ≥ √ �R�.
The first bound is only useful if |N|/2 dominates 2 log 2 |R| . In particular, it will be useful if |N| ≥ 5 log 2 |R|.

Proposition 3.7 We have |T
. Proof Let S ∈ T 1 N and set G S ∶= Aut(Γ(R,S)) (N) . Say, (xN) f = yN , for some xN, yN ∈ R∕N with yN ∉ {xN, x −1 N} and for some f ∈ G S with 1 f = 1 . Now, x f = yt , for some t ∈ N . Observe that where we are denoting by f ∶ N → N the automorphism induced by the conjugation via f on N. Observe that we have at most | Aut(N)| ≤ 2 (log 2 |N|) 2 choices for the automorphism f . Therefore, as t ∈ N , given xN and yN, we deduce from (3.8) that we have at most |N|2 (log 2 |N|) 2 choices for the permutation f |xN ∶ xN → yN restricted to xN.
We consider various possibilities: We consider these cases in turn: we let B i , B ii , B iii , B iv be the subsets of S 2 N satisfying, respectively, (i), (ii), (iii) or (iv). In the first case, the number of inverse-closed subsets of ∃f ∈ Aut(Γ(R,S)) (N)⧵ Aut(Γ(R,S)) (N) with 1 f = 1 and N is neither abelian of exponent greater than 2 nor generalised dicyclic, or N is abelian of exponent greater than 2 and n f ≠ n −1 for some n ∈ N, or N = Dic(A, y, x) ≇ Q 8 × C 2 and n f ≠ n̄A for some n ∈ N, or N ≅ Q 8 × C 2 and n f ∉ {n̄i , n̄j , n̄k } for some n ∈ N}, T N |; ∃x ∈ R and ∃f ∈ Aut(Γ(R,S)) (N) with 1 f = 1, (xN) f ≠ xN and either N is non -abelian or there exists n ∈ N with (xn) f ≠ (xn) −1 }, T N |; ∃x ∈ R and ∃f ∈ Aut(Γ(R,S)) (N) with 1 f = 1 and xN ∪ yN is at most 2 (xN) , because once T ∩ xN has been chosen the set T ∩ yN must equal (T ∩ xN) f . Therefore

xN)− (yN) and the number of inverse-closed f-invariant subsets T of
In the second case, the number of inverse-closed subsets of In the third case, the number of inverse-closed subsets of R ⧵ (xN ∪ yN ∪ y −1 N) is 2 (R)− (xN)−|N| and the number of inverse-closed f-invariant subsets of xN ∪ yN ∪ y −1 N is at most 2 |N| , because once we choose a subset of xN all the others are uniquely determined. Therefore The fourth case is similar to the third case and we have The proof now follows by adding the contribution of the four sets B i , B ii , B iii and B iv . ◻ Our second bound is useful whenever |N| grows with |R|.

Proposition 3.8 We have |T
Proof Given S ∈ T 2 N , we let G S ∶= Aut(Γ(R,S)) (N) . Given f ∈ (G S ) 1 , we let f ∶ N → N denote the automorphism induced by the action of conjugation of f on N. Let f ∈ (G S ) 1 ⧵ (G S ) 1 (N) witnessing that S ∈ T 2 N , that is, • N is neither an abelian group of exponent greater than 2 nor a generalised dicyclic group, or • N is an abelian group of exponent greater than 2 and f ≠ (where ∶ N → N is defined by x = x −1 , for every x ∈ N ), or • N = Dic(A, x, y) ≇ Q 8 × C 2 and f ≠̄A (where ̄A is given in Definition 1.9), or • N ≅ Q 8 × C 2 and f ∉ {̄i,̄j,̄k} (where ̄i,̄j,̄k are given in Definition 1.9).
In each of these cases, by Theorem 1.13 applied to N, we deduce that the number of f-invariant inverse-closed subsets of N is at most 2 (N)−|N|∕96 . In particular, where the first factor accounts for the number of inverse-closed subsets of R ⧵ N , the second factor accounts for the number of inverse-closed f-invariant subsets of N and the third factor accounts for the number of choices of f . ◻ For our third bound to be useful, we need |N|/8 to dominate log 2 |R| . In particular, it will be useful if |N| ≥ 9 log 2 |R|.

Proposition 3.9 We have |T
Proof Given S ∈ T 3 N , we let G S ∶= Aut(Γ(R,S)) (N) . Given any element ∈ G S , we let ∶ N → N denote the automorphism induced by the action of conjugation of on N. Let x ∈ R and let f ∈ (G S ) 1 ⧵ (G S ) 1 (N) with o(xN) > 2 and assume either • N is non-abelian, or • N is abelian and there exists n ∈ N with (xn) f ≠ (xn) −1 .
Observe that From (3.9), we deduce that we have at most | Aut(N)||N| ≤ 2 (log 2 |N|) 2 +log 2 |N| choices for the restriction f |xN ∶ xN → x −1 N of f to xN. Let ∶ xN → xN be the permutation obtained by composing first f |xN and then ∶ Let � ∶ N → N the permutation defined by n � = (n −1 ) fx (t −1 ) x ∀n ∈ N . An easy computation reveals that n ∈ Fix N ( � ) if and only if n −1 (n −1 ) fx = t x . In particular, we are in the position to apply Lemma 3.1 (with = fx and with the element t there replaced by t x here). From Lemma 3.1, we have two possibilities: • |Fix N ( � )| ≤ 3|N|∕4 , or • N is abelian, t = 1 and n fx = n −1 ∀n ∈ N.
If the second possibility holds, then N is abelian, f = x −1 and from (3.9) we get (xn) f = x −1 (n x −1 ) −1 = x −1 xn −1 x −1 = (xn) −1 for every n ∈ N ; however, this contradicts the fact that S ∈ T 3 N . Therefore, |Fix N ( � )| ≤ 3|N|∕4. The definition of ′ and the previous paragraph yield that has at most orbits. Since S ∩ xN is -invariant, the number of choices for S ∩ xN is at most 2 7|N|∕8 . By taking in account the contributions of f , xN and t, we obtain ◻ Our fifth bound is again useful whenever |N| grows with |R|.

Proposition 3.10 We have |T
Proof Given S ∈ T 4 N , we let G S ∶= Aut(Γ(R,S)) (N) . Given any element ∈ G S , we let ∶ N → N denote the automorphism induced by the action of conjugation of on N. Let ∈ R and let f ∈ (G S ) 1 with f ∉ { , −1 } . Furthermore, if possible we will choose so that o( ) = 2 . Therefore, we may assume that if o( ) ≠ 2 , then ( � ) f = � for every � ∈ R with o( � ) = 2 . (This will be important when we apply Lemma 3.4.) We now consider various possibilities depending on the behaviour of N , but first, we state the fact that the set S does not lie in T 2 N in a manner tailored to our current needs: Case a (G S ) 1 = (G S ) 1 (N) , or Case b N is abelian of exponent greater than 2 and, for every , the automorphism f induced by f on N is one of ̄i , ̄j , ̄k.
In particular, in cases B, C, and D, n f ∈ {n, n −1 } ∀n ∈ N. Suppose that ∈ N . Since 1 f = 1 and since f normalises N, we have f = f ∈ { , −1 } . For the rest of the proof, we may suppose that ∉ N . For the remainder of the proof, we may assume that N = −1 N , meaning that N is an index-2 subgroup of ⟨ , N⟩.
(where |N| is the number of choices for the restriction f N ∶ N → N of f to N , and |R/N| is the number of choices for N ∈ R∕N ). Therefore, for the rest of the proof we may suppose that f ∉ G S (N) . In particular, only Case B, C or D may arise. Suppose that Case B holds. Then, (3.10) becomes n f = n −1 and ( n) f = tn −1 , ∀n ∈ N , so n f = n −1 for every n ∈ N . As already observed at the beginning, if cannot be chosen with o( ) = 2, then for every n ∈ N with o( n) = 2, we have ( n) f = n. So we may apply Lemma 3.4 with f �⟨ ,N⟩ taking the role of t .
When f | N is given, from Lemma 3.4, we deduce that the number of choices for S ∩ ⟨ , N⟩ is at most 2 (⟨ ,N⟩)− �N� 24 (again, the other cases cannot arise since f ∉ { , −1 } ). Therefore (again, |N| is the number of choices for the restriction f N ∶ N → N of f to N , and |R/N| is the number of choices for N ∈ R∕N).
Cases C and D can be dealt with simultaneously. Here, (3.10) becomes n f = n̄A and ( n) f = tn̄A , ∀n ∈ N . When f | N is given, from Lemma 3.5, we deduce that the number of choices for S ∩ ⟨ , N⟩ is at most 2 (⟨ ,N⟩)− �N� 24 (again, the other cases cannot arise since f ∉ { , −1 } ). Therefore (where 3|N| is the number of choices for the restriction f N ∶ N → N of f to N , and |R/N| is the number of choices for N ∈ R∕N). ◻ Combining these results, we are able to bound |T N |.
Proof of Theorem 1.5 Since the initial statement excludes S 1 N , its proof follows by adding the bounds produced in Propositions 3.7, 3.8, 3.9 and 3.10 for |T i N | , for each 1 ≤ i ≤ 4 . If we drop the condition R = Aut(Γ(R,S)) (R) , then we must also add the bound produced in Proposition 1.14 for S 1 N (which has no effect on the bound we have given). Using Proposition 1.14 requires us to exclude groups that are either abelian of exponent greater than 2, or generalised dicyclic. ◻

Groups with a "small" normal subgroup
We begin this section of our paper with a counting result that we will need. The flavour of this result is quite distinct from most of the rest of the paper, and we have placed it in advance of the introduction of the notation and situational information that we will be using for the rest of this section.

exists a subset I ⊆ X such that
• I is f-and g-invariant (that is, I f = I and I g = I), Proof We denote by F and by G the permutation matrices of f and g, respectively. Therefore, F and G are |X| × |X|-matrices with {0, 1} entries, with rows and columns indexed by the set X and such that Let A ∶= F − G . For any S ⊆ X , let S ∈ ℤ X be the "indicator" vector of the set S, that is, Finally, let ⟨⋅, ⋅⟩ ∶ ℚ X × ℚ X → ℚ be the standard scalar product and let (e x ) x∈X be the canonical basis of ℚ X .
With the notation above, for every subset S of X, we have Therefore, For simplicity, we write Δ ∶ {0, 1} X → ℚ for the mapping defined by ↦ Δ( ) = ⟨ , A ⟩ , for every ∈ {0, 1} X . Suppose first that, there exist i, j ∈ X with i ≠ j and A i,j + A j,i ≠ 0 . Fix x ∈ {0, 1} arbitrarily for every x ∈ X ⧵ {i, j} , and let ∶= ∑ x∈X⧵{i,j} x e x . By restricting Δ , we define the function Δ � ∶ {0, 1} × {0, 1} → ℚ by setting A computation yields In particular, at least one out of the four choices ( i , j ) ∈ {(0, 0), (0, 1), (1, 0), (1, 1)} gives rise to a non-zero value for Δ( + i e i + j e j ) . Therefore, for every choice of x ∈ {0, 1} with x ∈ X ⧵ {i, j} , we have at most three more choices for i , j ∈ {0, 1} , for constructing a vector ∈ {0, 1} X with Δ( ) = 0 . Therefore, and (1) holds. Suppose that for every i, j ∈ X with i ≠ j , we have A i,j + A j,i = 0 . In this case, If A i,i ≠ 0 for some i ∈ X , then we may use the same argument as in the previous paragraph by fixing x ∈ {0, 1} arbitrarily for every x ∈ X ⧵ {i} , and by considering the restriction of Δ as a function Δ � ( i ) of i ∈ {0, 1} only. In this case, we see that one of the two choices for i gives rise to a vector ∈ {0, 1} X with Δ( ) = 0 . Therefore, and (1) holds. Suppose now that for every i, j ∈ X with i ≠ j , we have A i,j + A j,i = 0 and A i,i = 0 , that is, A is antisymmetric. Let I be the set of rows of A = F − G that are zero. From the fact that A is antisymmetric and from the definition of A, we see that I is f-and g-invariant, f |I = g |I and f |X⧵J = g −1 |X⧵J . In particular, (2) holds. ◻ Incidentally, we observe that if (2) holds in Lemma 4.1, then |S ∩ S f | = |S ∩ S g | , for every subset S of X. We find this quite interesting on its own. For instance, f ∶= (1 2 3 4 5)(6 7 8)(9 10 11 12) and g ∶= (1 5 4 3 2)(6 7 8)(9 12 11 10) have the property that |S ∩ S f | = |S ∩ S g | , for every subset S of {1, … , 12} . This condition seems very much related to the condition defining spreading groups. (For defining properly spreading groups, one needs some technical notation concerning multisets. A multiset of Ω is a function from Ω to the non-negative integers. A multiset is said to be trivial if it is the zero function. Given a multiset A ∶ Ω → {0, 1, …} , the multiplicity of i ∈ Ω in the multiset A is by definition A(i). The cardinality of A is then defined by Clearly, every subset of Ω can be regarded as a multiset, by considering its characteristic function; conversely, a multiset A of Ω is said to be a subset of Ω if A(i) ∈ {0, 1} for every i ∈ Ω . The product of two multisets A and B of Ω is the multiset A * B defined by (A * B)(i) = A(i)B(i) , for every i ∈ Ω . In particular, when A and B are subsets of Ω , A * B is the usual intersection of A with B. The image of a multiset A under a permutation g of Ω is defined by A g (i) ∶= A(i g −1 ) , for every i ∈ Ω . Now, with all of these definitions, we are ready to define spreading permutation groups. A transitive permutation group G on Ω is said to be non-spreading, if there exist two non-trivial multisets A and B of Ω and there exists a positive integer with • |A * B g | = , for every g ∈ G, • B is a set, • |A| divides |Ω|.

3
A transitive permutation group is said to be spreading if it is not non-spreading. Although the definition of spreading permutation groups might seem a bit artificial and technical, it has been introduced as a valuable tool for classifying synchronizing permutation groups, see for instance [1] for more details.) We are not sure whether Lemma 4.1 can play any role in the study of spreading permutation groups, or whether the analogy between Lemma 4.1 and the defining condition of spreading permutation groups is only superficial.

Specific notation
Henceforth, let R be a finite group of order r acting regularly on itself via the right regular representation: here, we identify the elements of R as permutation in Sym(R) . Let N denote a non-identity proper normal subgroup of R. We let b ∶= |R ∶ N| and we let 1 , … , b be coset representatives of N in R. Moreover, we choose 1 ∶= 1 to be the identity in R. Observe that R/N defines a group structure on {1, … , b} by setting ij = k for every i, j, k ∈ {1, … , b} with Write v 0 ∶= 1 where v 0 has to be understood as a point in the set R.
Observe that the O i s are the orbits of N on R, the group N acts regularly on O i and |O i | = |N|.
For an inverse-closed subset S of R, we let Γ(R, S) be the Cayley graph of R with connection set S, and we denote by F S the largest subgroup of Aut(Γ(R, S)) under which each orbit of N is invariant. In symbols we have (The subscript S in F S will make some of the later notation cumbersome to use, but it constantly emphasizes that the definition of "F" depends on S.) Similarly, we define As above, let S be an inverse-closed subset of R. For a vertex u of Γ(R, S) in O i , let (S, u, j) denote the neighbours of v 0 and u lying in O j . See Fig. 2. It is clear that In the results that follow, we use the notation that we have established here. Our aim with the next few results is to show that |Ψ({u, v}, j)| is at most 3 4 ⋅ 2 (R) . This will subsequently be used to bound the number of graphs admitting automorphisms that fix the vertex 1 and also fix each O i setwise while mapping u to v. We generally end up with some other possibilities that we gradually eliminate by introducing additional assumptions.
N is abelian and y j = y −1 for every y ∈ N, N is abelian and y ji −1 = y −1 for every y ∈ N, Proof We divide the proof in various cases.
Case Therefore, we may suppose that Lemma 4.1 (2) holds. Therefore, there exists an f-and g-invariant subset I of N j such that f |I = g |I and f |N j ⧵I = (g −1 ) |N j ⧵I . If I ≠ ∅ , then there exists x ∈ I and hence Simplifying and −1 i , we obtain xk −1 u = xk −1 v . This yields k u = k v , contradicting the fact that u ≠ v . Therefore I = � and hence f = g −1 .
This means that for every x ∈ N j , we have By writing x = y j with y ∈ N , we deduce Since y is an arbitrary element of N, we get that i k v centralizes N. From this and from (4.5) and (4.6) we see that (2) holds. ◻ For the rest of the proof, we suppose j 2 ≠ i . From (4.1), we obtain (4.7) of size having parity uniquely determined by the parity of |X|. Therefore we have 2 |N|−(a−b) 2 (a−b)−1 = 2 |N|−1 choices for S ji −1 . Altogether we have As o(ji −1 ) > 2 , from (4.8), we have |N| + ( j N ∪ −1 j N) = (R j,i ) and hence, from (4.10) (noting that if o(j) > 2 then ( j N ∪ −1 j N) = |N| , and otherwise j N ∪ −1 j N = j N ), we get and hence (1) holds in this case. Assume Since the left-hand side is at most |N|/2 and since the right-hand side is at least |N|/2, this implies o(k −1 v k u ) = 2 and Therefore N j ∩ I(R) = � , N j contains no involutions and ( j N) = |N|∕2 . Under these strong conditions, we refine the upper bound in (4.11) by first improving our upper bound in (4.10).
As o(j) = 2 , N j is inverse-closed. Recall that t 1 is the number of inverse-closed subsets S j ⊆ N j with S k −1 v k u j = S j . Consider the permutation ∶ j N → j N defined by mapping for each y ∈ N , and consider the permutation ∶ j N → j N defined by mapping for each y ∈ N . Observe that and are involutions with no fixed points: has no fixed points because j N contains no involutions and is an involution because o(k −1 v k u ) = 2 . In this new setting, .
where o is the number of orbits of ⟨ , ⟩ ≤ Sym( j N) . Each orbit of ⟨ , ⟩ has even length, because has order 2 and has no fixed points. Suppose ⟨ , ⟩ has at least one orbit of length greater then 2. Then o ≤ |N|∕2 − 1 (the upper bound is achieved when ⟨ , ⟩ has |N|∕2 − 2 orbits of length 2 and one of length 4). Thus, in this case, Using this slight improvement on x and ( j N) = |N|∕2 , we obtain As (R j,i ) = |N| + ( j N) = 3|N|∕2 (see (4.8)), we obtain In particular, from (4.9) and (4.12), we see that (1) holds. It remains to suppose that each orbit of ⟨ , ⟩ has length 2; this means = , that is, Applying this equality with y = 1 , we get −1 j = j z and hence 2 j = z because z has order 2. Thus we have y −1 −1 j = j y −2 j and hence j y −1 j = y −1 . This shows that the element j acts by conjugation on N inverting each of its elements. Therefore, N is abelian.
Case o(ji −1 ) = o(j) = 2 . This is the only remaining option. ◻ For three distinct vertices u, v, w ∈ O i and j ∈ {1, … , b} , let ( j y) = ( j y) , ∀y ∈ N. Proof If o(i) is odd, then R/N is not an elementary abelian 2-group, so we may assume this throughout the proof. We define an auxiliary graph X: the vertex-set of X is {{j, j −1 } | j ∈ R∕N} and the vertex {j, j −1 } is declared to be adjacent to In particular, X is a graph with (R∕N) vertices and where each vertex has valency at most 4. Observe that some vertex {j, j −1 } might have valency less than four, because the ele- We now use the bounds we have achieved, to show that the number of graphs admitting automorphisms that fix every orbit O k setwise, but act nontrivially on some O i is a vanishingly small fraction of the 2 (R) Cayley graphs on R, as long as either o(i) is odd, or the orbit on O i has length at least 3. Actually, these formulas only produce results that are vanishingly small if |N| is small enough relative to |R| that |R|/|N| grows with |R|, so this is the point at which it starts to become clear that we need to be assuming that |N| is relatively small, in order to apply the results in this section. The result involving an orbit of length 3 does not work in the case that R/N is an elementary abelian 2-group; this case will need to be handled separately. Our next result deals specifically with the case that R/N is an elementary abelian 2-group. (We refer to Sect. 4.1 for the definition of B S .) Lemma 4.6 (Recall the notation in Sect. 4.1.) Suppose R is not an abelian group of exponent greater than 2 that R is not a generalized dicyclic group and that R/N is an elementary abelian 2-group. Then Observe that the definition of B S immediately yields B S ⊴ Aut(Γ(R, S)) . In particular, RB S is a group of automorphisms of Γ(R, S) acting transitively on the vertex set R and normalizing N. Since R is also transitive on the vertex set, the Frattini argument gives RB S = R(B S ) v 0 . Let Since R is not an abelian group of exponent greater than 2 and since R is not a generalized dicyclic group, Proposition 1.14 yields In particular, |S � | ≤ 2 (R)− |R| 96 +(log 2 |R|) 2 . For each S ∈ S �� , choose G S a subgroup of RB S with R < G S and with R maximal in G S . Observe that RB S ∕N (R∕N) = R∕N , because RB S (R) = R.
Let K be the core of R in G S . Then Since R is maximal in G S , G S ∕K acts primitively and faithfully on the set of right cosets of R in G S . The stabilizer of a point in this action is R/K. As N ≤ K , we deduce that R/K is an elementary abelian 2-group. From [18, Lemma 2.1], we deduce |G S ∶ R| = |(G S ) v 0 | is a prime odd number and |R ∶ K| = 2.
We now partition the set S ′ further. We define In what follows, we obtain an upper bound on the cardinality of C and C ′ . For each S ∈ C , let S ∶ (G S ) v 0 → Aut(K) the natural homomorphism given by the conjugation action of (G S ) v 0 on K. For each ∈ Aut(K) ⧵ {id K } , let C ∶= {S ∈ C | ∈ S ((G S ) v 0 )} . In other words, C consists of the connection sets S such that (G S ) v 0 contains an element acting by conjugation on K as the automorphism . With this new setting, Since |(G S ) v 0 | is odd, then ∈ S ((G S ) v 0 ) has odd order. Using this and applying Theorem 1.13 to the group K, we deduce that C ∶= {S ∈ S �� | (G S ) v 0 does not act trivially by conjugation on K}, From Proposition 1.14, Lemma 4.5 and Lemma 4.6, we have explicit bounds for S 1 , S 2 , S 3 and S 4 , and hence we may consider only the set S 5 . Let S ∈ S 5 . Since S ∉ S 4 , R/N is not an elementary abelian 2-group. Since S ∉ S 3 , (F S ) v 0 has orbits of cardinality at most 2, and so does (B S ) v 0 . Therefore, (F S ) v 0 and (B S ) v 0 are elementary abelian 2-groups. Now let L S = { j ∶ (F S ) v 0 is trivial on O j } . Notice that L S is in fact a group. Since (F S ) v 0 is nontrivial, then L S is a proper subgroup of R. Since S ∉ S 2 , i ∈ L S for every i with o(i) odd. Therefore NL S contains all elements of R of odd order. Let be the core of NL S in RB S . Since all conjugates of NL S in R also contain all elements of R of odd order, we deduce that K also contains all elements of R of odd order and hence R/K is a 2-group. As (B S ) v 0 is also a 2-group, we obtain that RB S ∕K is a 2-group. Therefore RB S ∕K (R∕K) > R∕K . However, this implies that RB S (R) > R , but this contradicts the fact that S ∉ S 1 . This shows that S 5 = � . Now, adding the bounds produced for S i for each 1 ≤ i ≤ 4 , we get the result. Indeed, using the first bound in Lemma 4.5 and the fact that |R| ≥ 2|N| ≥ 4, we get Further, if |R| < 8, then |R| ≠ 7 (because N is a nontrivial proper subgroup), that is |R| ≤ 6. Consequently, If |R| ≥ 8, then Using these, and the second bound in Lemma 4.5 we get This together with Proposition 1.14, and Lemma 4.6, yields S 1 ∶={S ∈ S | R < Aut(Γ(R,S)) (R)}, S 2 ∶={S ∈ S ⧵ S 1 | ∃i ∈ {2, … , b} with o( i) odd such that (F S ) v 0 has a nontrivial orbit on O i }, R∕N not an elementary abelian 2-group, ∃i ∈ {2, … , b} such that (F S ) v 0 has an orbit of cardinality at least 3 on O i }, R∕N is an elementary abelian 2-group, (B S ) v 0 ≠ 1}, log 2 (|R||N| 2 ∕6) ≤ 2 log 2 |R| − 2 ≤ (log 2 |R|) 2 − 2.
As in the proof of Theorem 1.5, we do not need to include the bound from Proposition 1.14 if we include the condition R = Aut(Γ(R,S)) (R) . If we omit this condition, then we include this extra piece (which does not affect the overall bound as we have stated it) but must not allow groups that are either abelian of exponent greater than 2, or generalised dicyclic. ◻ Funding Open access funding provided by Università degli Studi di Milano -Bicocca within the CRUI-CARE Agreement.
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