Asymptotic enumeration of graphical regular representations

We estimate the number of graphical regular representations (GRRs) of a given group with large enough order. As a consequence, we show that almost all finite Cayley graphs have full automorphism groups ‘as small as possible’. This confirms a conjecture of Babai–Godsil–Imrich–Lovász on the proportion of GRRs, as well as a conjecture of Xu on the proportion of normal Cayley graphs, among Cayley graphs of a given finite group.


Introduction
Let R be a finite group and S a subset of R. The Cayley digraph of R with connection set S, denoted by Cay(R, S), is defined as the digraph with vertex set R such that for u, v ∈ R, (u, v) is an arc if and only if vu −1 ∈ S. Note that 1 ∈ S is allowed in our definition of Cayley digraphs and that a Cayley digraph Cay(R, S) is a graph if and only if S = S −1 .For convenience, we identify a group R with its right regular permutation representation, which is obviously a subgroup of Aut(Cay(R, S)).A digraph Γ is said to be a digraphical regular representation (DRR for short) of a group R if its full automorphism group Aut(Γ) is isomorphic to R and acts on the vertex set of Γ as a regular permutation group.Thus DRRs of R are precisely Cayley digraphs of R with full automorphism group as small as R. If a DRR of R is a graph, then it is called a graphical regular representation (GRR for short) of R.
It is natural to ask which finite groups admit DRRs or GRRs.The question for GRRs was studied in a series of papers [4,13,14,20] before a complete answer was given by Godsil [7] in 1978: Apart from two infinite families and thirteen small solvable groups (of order at most 32), every finite group admits a GRR.Here one of the two infinite families consists of abelian groups of exponent greater than 2 (see [4,14]).The other family is the so called generalized dicyclic groups, namely, non-abelian groups R that have an abelian normal subgroup A of index 2 and an element x ∈ R \ A of order 4 such that x −1 ax = a −1 for all a ∈ A (see [13,20]).The situation for DRRs is simpler, as shown by Babai [1]: C 2 2 , C 3 2 , C 4 2 , C 2 3 and Q 8 are the only five groups without DRRs.
It was conjectured by Babai and Godsil [2,8] that almost all finite Cayley digraphs are DRRs.This conjecture has been confirmed recently by Morris and Spiga [11].In fact, they proved the following quantified version: Theorem 1.1 (Morris-Spiga).Let R be a group of order r, where r is sufficiently large.The proportion of subsets S of R such that Cay(R, S) is a DRR of R is at least 1 − 2

+2
, where b is an absolute constant.
A Cayley digraph Cay(R, S) is said to be normal if R is normal in Aut(Cay(R, S)).Studying the normality of Cayley digraphs is a helpful approach to understand the symmetry of Cayley digraphs.In [22], Xu conjectured that almost all Cayley (di)graphs are normal.The digraph part of this conjecture has been confirmed by Morris and Spiga in [11] as an immediate corollary of Theorem 1.1, while the graph part of this conjecture, stated below, has been open.Conjecture 1.5 (Xu).Let R be a group of order r such that R ≇ Q 8 × C m 2 for any nonnegative integer m.The proportion of inverse-closed subsets S of R such that Cay(R, S) is a normal Cayley graph of R approaches 1 as r approaches infinity.
We remark that the excluded groups Q 8 × C m 2 are the Hamiltonian 2-groups, which are proved in [19, Lemma 5] to be the groups other than C 4 × C 2 that admit no normal Cayley graphs.It is also worth noting that almost all Cayley graphs on finite groups R without GRRs have full automorphism groups 'as small as possible', see Dobson-Spiga-Verret [5,Theorem 1.5] for abelian groups and Morris-Spiga-Verret [12,Theorem 1.4] for generalized dicyclic groups.Combining these results with our Theorem 1.3, we confirm Conjecture 1.5 by the quantified version in the next corollary.In fact, the equivalence between Conjectures 1.2 and 1.5 was already known to Spiga [17].
Corollary 1.6.Let R be a group of order r such that R ≇ Q 8 × C m 2 for any nonnegative integer m.For large enough r, the proportion of inverse-closed subsets S of R such that Cay(R, S) is a normal Cayley graph of R is at least 1 − 2 − r 0.499 8 log 3 2 r +log 2 2 r+3 .
The rest of this paper is devoted to the proof of Theorem 1.3 and Corollaries 1.4 and 1.6.The strategy we apply to prove Theorem 1.3 is initiated in [2], followed by [5,12,15,16], and further developed in [11].We also make use of the substantial progress on enumeration of GRRs achieved very recently by Spiga [18] and Morris-Moscatiello-Spiga [10].Our main contribution to the solution of Conjecture 1.2 lies in the results in Section 3, which eventually leads us to Theorem 1.3 from several important pieces of work by various authors.

Preliminaries
For a subset X of a group, denote by I(X) the set of elements of X of order at most 2 and let we see that, for a given nonempty set, the number of subsets of odd size equals the number of subsets of even order.In particular, m k ≤ 2 m−1 for all k ∈ {0, 1, . . ., m}.
(2.3) Lemma 2.2.Let R be a group, and let X be a nonempty inverse-closed subset of R. The number of inverse-closed subsets of X with a given size is at most 2 c(X)−1 .
Proof.Let n = |X|.For a nonnegative integer k, denote the number of inverse-closed k-subsets of X by n k .Case 1: Case 2: n is even and I(X) = ∅.Take an element x in I(X).Then X \ {x} is an inverse-closed subset of size n − 1, which is odd.By the conclusion of Case 1, we know that the number of inverse-closed k-subsets of For a nonnegative integer ℓ, let C ℓ = {S ⊆ X : S = S −1 , x / ∈ S and |S| = ℓ}, D ℓ = {S ⊆ X : S = S −1 , x ∈ S and |S| = ℓ}, and let g ℓ : C ℓ → D ℓ+1 be the map such that g ℓ (S) = S ∪ {x}.
It is clear that g ℓ is a bijection between C ℓ and D ℓ+1 .Hence for each positive integer k, the number of inverse-closed k-subsets of X that contains x is exactly the number of inverse-closed (k − 1)-subsets of X \ {x}, which is at most 2 c(X)−2 by (2.4).Combining this with (2.4), we obtain n k ≤ 2 c(X)−2 + 2 c(X)−2 = 2 c(X)−1 for every nonnegative integer k.
Case 3: n is even and I(X) = ∅.
Suppose n = 2m and write X = {x 1 , x 2 , . . ., x 2m } with x −1 i = x i+m for i ∈ {1, . . ., m}.Note that an inverse-closed subset S of X has even order and is determined by S ∩ {x 1 , x 2 , . . ., x m }, whose size is exactly |S|/2.Hence for a nonnegative even integer k, we have where the second inequality is from (2.3).Since n k = 0 when k is odd, it follows that n k ≤ 2 c(X)−1 for every nonnegative integer k.
The following result, proved in [18,Section 4], is useful in estimating the number of Cayley graphs on R whose full automorphism group contains a given overgroup of R.
Lemma 2.3 (Spiga).Let G be a transitive group properly containing a regular subgroup R on R, where the group R, identified with its right regular permutation representation, is neither abelian of exponent greater than 2 nor generalized dicyclic.Let G 1 be the stabilizer of 1 ∈ R in G, and let ι be the permutation on R sending every element to its inverse.The number of G 1 , ι -orbits on R is at most c(R) − |R|/96.In particular, the number of (labeled) Cayley graphs Γ on R with The next lemma is from [10, Theorem 1.6], whose proof involves applying results from [17].
Lemma 2.4 (Morris-Moscatiello-Spiga).Let R be a finite group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic, and let N be a nontrivial proper normal subgroup of R. The number of inverse-closed subsets S of R such that there exists a non-identity g ∈ N Aut(Cay(R,S)) (N ) that fixes the vertex 1 and stabilizes each N -orbit is at most Let G be a group that acts transitively on a set Ω with |Ω| > 1.A nonempty subset B of Ω is called a block of G if, for each g ∈ G, either B g = B or B g ∩ B = ∅.In this case, we call {B g : g ∈ G} a block system of G.It is clear that the group G has Ω and the singletons as blocks, which are viewed as the trivial ones.The group G is said to be primitive if it only has trivial blocks.It is easy to see that G is primitive if and only if a point stabilizer in G is a maximal subgroup of G.Note that there are eight O'Nan-Scott types of the finite primitive permutation groups: HA, HS, HC, SD, CD, TW, AS, PA, as divided in [3].Lemma 2.5.([12, Lemmas 2.4 and 6.1])A primitive permutation group with an abelian point stabilizer or a generalized dicyclic point stabilizer is of type HA.
Let Γ be a digraph on vertex set V .A partition of V into sets C 1 , C 2 , . . ., C t is said to be equitable if for a vertex v ∈ C i , the number of outneighbours of v in C j depends only on the choice of C i and C j , that is, the number of outneighbours of any v ∈ C i in C j is a constant b i,j .The following observation shows that if G is a subgroup of Aut(Γ), then the partition formed by the G-orbits on V is equitable.Lemma 2.6.Let Γ be a digraph, and let G ≤ Aut(Γ).Given G-orbits B and C on the vertex set of Γ, vertices in B are adjacent to the same number of vertices in C.
Proof.For a vertex v of Γ, denote the set of outneighbours of v by N + (v).Let u, w ∈ B. There is an element ϕ ∈ G ≤ Aut(Γ) such that u ϕ = w and (N + (u) which means u and w have the same number of outneighbours in C.
We close this section with a somewhat technical lemma.
Lemma 2.7.Let R < G ≤ Sym(R) such that R, identified with its right regular permutation representation, is a maximal subgroup of G, and let L be a normal subgroup of G.The L-orbit on R containing 1 is a subgroup of R. If it is further a normal subgroup of R, then either L ≤ R or every L-orbit is stabilized by G 1 .
Proof.Since L is normal in G, the L-orbits on R form a block system of G. Let Q = 1 L be the L-orbit containing 1. Since R acts regularly by right multiplication, we have Q ≤ R. Hence the block system formed by the L-orbits coincides with the set of right

Subsets evenly intersecting cosets
Let R be a group with a subgroup Q, and let ∆ be a union of some right Q-cosets in R, say, Λ 1 , Λ 2 , . . ., Λ b (so that |∆| = b|Q|).We say that a subset S of R intersects ∆ evenly if We are interested in the case when ∆ comes from an orbit of some group acting on the set R/Q of right Q-cosets in R. For the group actions in the two subsections below, we estimate the number of such S.

3.1.
Cosets from a double coset.For a subgroup Q of R, under the action of Q on R/Q by right multiplication, an orbit of Q is a double coset of Q in R.This subsection is devoted to the proof of the following proposition, which estimates the number of inverse-closed subsets in a group that intersect evenly with all the double cosets of a given subgroup.Recall the notation In what follows we are going to provide a series of lemmas, which will lead to a proof of Proposition 3.1 at the end of this subsection.
Proof.Since QxQ is inverse-closed and {Φ 1 , Φ 2 , . . ., Φ b } forms a partition of QxQ, we have Suppose for a contradiction that Write Φ i = Qy i for each i ∈ {1, . . ., k}.Note that Then we derive that As a consequence, This implies that b|Q| = |QxQ| ≤ k|Q|, contradicting to the condition that 2k ≤ b.
Suppose that, for some ℓ ∈ {2, . . ., ⌊b/2⌋}, there is an ordering (Note that this supposition holds for ℓ = 2 by letting Φ 2,j = Φ j for j ∈ {1, . . ., b}.) Applying Lemma 3.2 to the ordering (Φ ℓ,1 , Φ ℓ,2 , . . ., Φ ℓ,b ), we have We prove in the next paragraph that b j=i As for i = ℓ + 1, according to (3.3) and the relation Φ ℓ+1,ℓ+1 = Φ ℓ,s , we have b j=ℓ+1 satisfies the requirement of the lemma.Lemma 3.4.Let R be a group, let QxQ be an inverse-closed double coset of a subgroup Q in R with |QxQ| = b|Q|, and let Proof.If b = 1, then the set M is exactly the set of inverse-closed subsets of QxQ.In this case, we have |M| = 2 c(QxQ) , which meets the upper bound.For the rest of the proof we assume The reader may find Figure 1 helpful.
| and every element of order at most 2 in QxQ lies in some Λ i,i , one has To determine each S ∈ M (and count the number of possibilities for S), we determine S ∩ Λ 1 , S ∩ Λ 2 , . . ., S ∩ Λ b in turn.First we determine S ∩ Λ 1 .Note that this is further determined by S ∩ Λ 1,1 and S ∩ ∪ b j=2 Λ 1,j .We first choose an inverse-closed subset of Λ 1,1 and then choose an arbitrary subset of ∪ b j=2 Λ 1,j .Thus the number of choices for S ∩ Λ 1 is In the following, we estimate the number of choices for S ∩ Λ i with i ≥ 2. Note that once S ∩ Λ 1 has been determined, the size of S ∩ Λ i for every i ∈ {2, . . ., b} has to be |S ∩ Λ 1 | since we require that S intersects QxQ evenly.Now assume that {S ∩ Λ 1 , . . ., S ∩ Λ i−1 } has been determined for some i ∈ {2, . . ., b}.For an inverse-closed subset S of R, we have S ∩ Λ i,j = S −1 ∩ Λ −1 j,i = (S ∩ Λ j,i ) −1 for all j ∈ {1, 2, . . ., b}.In particular, for each j ∈ {1, . . ., i − 1}, the set S ∩ Λ i,j is determined as S ∩ Λ j,i is already determined, see Figure 2. Therefore, to determine S∩Λ i , we only need to determine This means that we must choose a subset ∪ b j=i (S ∩ Λ i,j ) of ∪ b j=i Λ i,j of determined size.We estimate the number of such choices in the following two paragraphs, according to 2 ≤ i ≤ ⌊b/2⌋ + 1 or i > ⌊b/2⌋ + 1, respectively.j=i+1 Λ i,j = ∅, then Λ i,i = ∅ and we only need to choose an inverse-closed subset of Λ i,i of determined size.In this case, by Lemma 2.2, the number of choices for S ∩ Λ i is at most If ∪ b j=i+1 Λ i,j = ∅, then we need to first choose an inverse-closed subset of Λ i,i (no matter whether Λ i,i is empty or not) and then choose a subset of the nonempty set In this case, by (2.3), the number of choices for S ∩ Λ i is at most In either case, we have at most Next assume that i > ⌊b/2⌋+1.To determine ∪ b j=i (S ∩Λ i,j ), we need to first choose an inverseclosed subset of Λ i,i and then choose a subset of ∪ b j=i+1 Λ i,j .Hence the number of choices for S ∩ Λ i is at most Now combining the bounds in (3.6), (3.7) and (3.8) for i ∈ {1, 2, . . ., b}, we obtain This together with (3.5) yields that |M| ≤ 2 c(QxQ)− as desired.
Lemma 3.6.Let R be a group, let Q be a subgroup of R with |Q\R/Q| = ℓ, and let Proof.We divide Q\R/Q into the following two subsets: For an inverse-closed subset S of R, we have (S ∩ ∆) According to Lemma 3.4, for each i ∈ {1, . . ., s}, there are at most 2 c(A i )− 1 2 a i + 1 2 choices for S ∩ A i .Hence the number of choices for S ∩ A 1 , . . ., S ∩ A s is at most . By Lemma 3.5, for each i ∈ {1, . . ., t}, we have at most 2 |B i |−b i +1 choices for S ∩B i .This implies that the number of choices for S ∩ B 1 , . . ., S ∩ B t is at most .
We are now ready to prove the main result of this subsection.
Proof of Proposition 3.1.Let Ω = R/Q = {Qx : x ∈ R} be the set of right cosets of Q in R.
Then R acts transitively by right multiplication on Ω with stabilizer Q.Note that the stabilizer Q does not act trivially on Ω, for otherwise we would have (x −1 Qx) ∩ Q = Q for every x ∈ R, contradicting the assumption that Q is non-normal in R.This implies that R is not regular on Ω.Let ℓ be the number of orbits of Q on Ω, that is, the number of double cosets of Q in R.

3.2.
Cosets from an orbit of a stabilizer.Let G be a finite transitive permutation group on R that properly contains the right regular representation of R, let G 1 be the stabilizer of be the core of R in G, and let : R → R/K be the quotient modulo K. Then G/K acts naturally on R/K and induces a transitive permutation group containing the right regular representation of R/K, and is the stabilizer of 1 ∈ R/K in G/K.For a subset ∆ of R/K, the full preimage of ∆ in R under , denoted by ∆, is the union of the right K-cosets in ∆.
Proposition 3.7.With the above notation, suppose that R/K is neither abelian of exponent greater than 2 nor generalized dicyclic, and let L = {S ⊆ R : S = S −1 , S intersects ∆ evenly for each H-orbit ∆ on R/K}.
In the rest of this subsection we embark on the proof of Proposition 3.7 and fix the notation R, K, H, and .Moreover, let ι be the permutation on R/K sending every element to its inverse.
Lemma 3.8.Let ∆ ⊆ R/K be an H-orbit on R/K such that ∆ ι = ∆, and let Proof.Write I(∆) = {x 1 , . . ., x n } and ∆ \ I(∆) = {y 1 , . . ., y 2m } with (y j ) −1 = y m+j for j ∈ {1, . . ., m}.It is clear that and that For i ∈ {1, . . ., n}, since x i has order at most 2, we have First assume n ≥ 1.Consider S in M. To determine S ∩ Kx 1 , we may choose an inverseclosed subset of Kx 1 , which means that there are 2 c(Kx 1 ) choices for S ∩ Kx 1 .Since S intersects ∆ evenly, it follows that, once S ∩ Kx 1 is determined, the size of S ∩ Kz for every z ∈ ∆ \ {x 1 } is also determined.Thus, Lemma 2.2 implies that there are at most 2 c(Kx i )−1 choices for each S ∩ Kx i with i ∈ {2, . . ., n}, and (2.3) implies that there are at most 2 |Ky j |−1 choices for each S ∩ Ky j with j ∈ {1, . . ., m}.Consequently, Now assume n = 0 and consider S in M. To determine S ∩ Ky 1 , one may choose a subset of Ky 1 , which means that there are 2 |Ky 1 | choices for S ∩ Ky 1 .Note that once S ∩ Ky 1 is determined, the size of S ∩ Kz for every z ∈ ∆ \ {y 1 } is also determined.Thus, by (2.3), there are at most 2 |Ky j |−1 choices for each S ∩ Ky j with j ∈ {2, . . ., m}.This leads to Lemma 3.9.Let ∆ ⊆ R/K be an H-orbit on R/K, and let Proof.Write ∆ = {x 1 , . . ., x n }, where n = |∆|.Consider S in M. As S intersects ∆ evenly, we derive that, once S ∩ Kx 1 is determined, the size of S ∩ Kx for every x ∈ ∆ \ {x 1 } is also determined.Hence we deduce from (2.3) that Lemma 3.10.Let κ be the number of H, ι -orbits on R/K, and let Proof.Let D be the set of H-orbits on R/K.It is well known and not difficult to prove (see, for example, [21, Theorem 24.1]) that the set of H-orbits on R/K is ι-invariant.We divide D into the following two subsets: Note that the elements in D 2 are paired by ι.Write D 1 = {A 1 , . . ., A s } and D 2 = {B 1 , . . ., B 2t } with (B j ) ι = B t+j for j ∈ {1, . . ., t}.Clearly, For an inverse-closed subset S of R, we have (S ∩ ∆) −1 = S ∩ ∆ −1 for every ∆ ∈ D, which means that S ∩ ∆ −1 is determined by S ∩ ∆ when ∆ ∈ D 2 .Thus a subset S ∈ L is determined by Since S intersects ∆ evenly for all ∆ ∈ D, we conclude by Lemmas 3.8 and 3.9 that which completes the proof.
Proof of Proposition 3.7.It follows from Lemma 2.3 that the number κ of H, ι -orbits on R/K is at most c(R/K) − 1 96 |R/K|.Applying Lemma 3.10, we obtain completing the proof.

Reduction
Let R be a group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic.We also use R to represent the right regular representation of the group R. Let Clearly, Z is the set of inverse-closed subsets S of R such that Cay(R, S) is not a GRR of R, and the key to prove Theorem 1.3 is to establish a suitable upper bound for |Z|.
Let S be a subset of R in Z.There exists some group G ≤ Aut(Cay(R, S)) such that R is a maximal subgroup of G. Denote the stabilizer of the vertex 1 in G by G 1 , and denote the core of It is clear that G/Core G (R) acts primitively and faithfully by right multiplication on the set To sum up, for each S ∈ Z, there exists a group G = G(S) such that G/Core G (R) is a primitive permutation group on G/R with point stabilizer R/Core G (R) and a regular subgroup Now we have chosen a group G(S) for each S ∈ Z.For ε ∈ (0, 0.5) let We omit the subscript ε when we take ε = 0.001.The following result is obtained by combining Lemma 2.3 with [11,Lemma 3.2].It is worth remarking that [11,Lemma 3.2] depends upon the following asymptotic enumeration due to Lubotzky [9] (see the third line on page 416 of [11]): The number of isomorphism classes of groups of order n that are d-generated is at most n 2(d+1) log 2 n .Lemma 4.1.Let R be a group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic, and let ε ∈ (0, 0.5).If r is large enough, then The following lemma is obtained by adapting the proof of [11,Theorem 3.4].
Lemma 4.2.Let R be a group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic, and let ε ∈ (0, 0.5).If r is large enough, then Proof.Choose r ε ∈ N such that r 0.5−ε > log 2 2 r whenever r ≥ r ε , and assume r ≥ r ε for the rest of the proof.Let one can read off from the proof of [11,Theorem 3.4] that From the inequality n! ≤ n(n/e) n we derive that, for k > 96 log 2 r, It follows that Then by Lemma 2.3, we obtain Recall the notation introduced before Lemma 4.1.In particular, Let R be a finite group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic.For large enough r we have Proof.As r is large enough, we have r > 96 log 2 r.Let N be the set of nontrivial proper normal subgroups of R of order at most 96 log 2 r.For N ∈ N , we denote by M N the set of inverse-closed subsets S of R such that there exists a non-identity g ∈ N Aut(Cay(R,S)) (N ) that fixes the vertex 1 and stabilizes each N -orbit.According to Lemma 2.4, Take an arbitrary Consequently, any non-identity element of G 1 lies in N Aut(Cay(R,S)) (K) and stabilizes every orbit of K.In other words, S ∈ M K .This leads to Since every subgroup of R has a generating set of size at most log 2 r, we have at most r log 2 r choices for a subgroup N of R. Combining this with (4.1) and (4.2), we conclude that Combining Lemmas 4.1, 4.2 and 4.3, we obtain the following result.
Proposition 4.4.Let R be a finite group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic.For large enough r we have Proof.As r is large enough, we see from Lemmas 4.1, 4.2 and 4.3 that Observe that a 3 = max{a 1 , a 2 , a 3 } for large enough r.It follows that as the proposition asserts.For each S ∈ T , let K(S) = Core G(S) (R).It is known from the discussion in the previous section that G(S)/K(S) is a primitive permutation group on G(S)/R with point stabilizer R/K(S) and a regular subgroup G(S) 1 K(S)/K(S).For each C ∈ {HA, HS, HC, SD, CD, TW, AS, PA}, let T C be the set of elements S ∈ T such that G(S)/K(S) is of type C in its primitive action on the set of right cosets of R in G(S).Then Note that there are two actions of G(S), namely the primitive action on the set G(S)/R of right cosets of R in G(S) and the transitive action on the vertex set R of the Cayley graph Cay(R, S).
In the following, we will always emphasize the primitivity when referring to the first action, and the reader should keep in mind which action we are considering.
Lemma 5.1.Suppose that r is large enough.For each S ∈ T HA ∪ T SD ∪ T TW , there exists a non-normal subgroup Q of R with |Q| < r 0.501 log 3 2 r such that the right Q-cosets in R are precisely the orbits of some subgroup of Aut(Cay(R, S)).
Proof.Take an arbitrary S ∈ T HA ∪ T SD ∪ T TW .Let G = G(S), let K = K(S) = Core G (R), and let : G → G/K be the quotient modulo K. Since K is normal in G, the set R/K of right K-cosets in R forms a block system of G. Let L be the kernel of the induced action of G on R/K.Then the L-orbits are precisely the K-orbits, and hence some L-orbit is not stabilized by G 1 as S ∈ T .Thereby we obtain from Lemma 2.7 that L ≤ R. It follows that L = K, and so G = G/K acts faithfully on R/K.
Let N be the full preimage of Soc(G) under .Clearly, we have K ≤ N .Since G is transitive on the vertex set of Cay(R, S) and N is normal in G, the N -orbits on the vertices of Cay(R, S) form a block system of G. Let Q = 1 N be the N -orbit containing the vertex 1.Then K = 1 K ⊆ 1 N = Q.Since R acts regularly on the vertex set of Cay(R, S) by right multiplication, we conclude that Q is a subgroup R. Hence the block system formed by the N -orbits coincides with the set of right Q-cosets of R.
If K = Q, then every N -orbit has the form x N = Qx = Kx = xK = x K and thus is a K-orbit.This would imply that N stabilizes every K-orbit, that is, N is trivial, which is impossible as For g ∈ N and x ∈ R, we have Recall that the N -orbits on R, which are the right Q-cosets in R, form a block system of G. Since (Qx) g is a block in this block system that contains x g , it follows that (Qx) g = Qx g and hence x gQ = x Qg .This means that gQ = Qg for all g ∈ N , that is, Thus Q is non-normal in R. Denote the identity element of R/K by 1.We have This together with [11,Proposition 5.7] gives |Q| ≤ |R/K| .x ∈ R, the neighbourhood of x in Cay(R, S) is Sx.For each q ∈ Q, Lemma 2.6 asserts that 1 and q have the same number of neighbours in Λq, which means This implies that for every right Q-coset contained in ∆, say Λ ′ , we have |S ∩ Λ ′ | = |S ∩ Λ|.Since this ∆ is chosen arbitrarily, the set S must intersect ∆ evenly for every ∆ ∈ D. Hence by Proposition 3.1, we obtain Since |Q| < r 0.501 log 3 2 r, it follows that Note that R has at most 2 log 2 2 r subgroups, which implies that |Q| ≤ 2 log 2 2 r .Therefore, Proof.Let K = {K(S) : S ∈ T HS ∪ T HC ∪ T AS ∪ T PA }.For N ∈ K, denote by T N the set of inverse-closed subsets X of R/N such that there exists a subgroup A of Aut(Cay(R/N, X)) with the following conditions: • R/N is maximal and core-free in A; • the vertex stabilizer in A has size greater than 2 |R/N | 0.499 ; • the primitive action of A by right multiplication on the set of right (R/N )-cosets in A is of type HS, HC, AS or PA.
Let G N be the set of (labeled) Cayley graphs on R/N with connection set in T N .For each Γ ∈ G N , let N Γ be the set of S such that the odd quotient graph of Cay(R, S) with respect to N is Γ.From the argument in [11,Pages 446 and 447] we see that for each S ∈ T HS ∪ T HC ∪ T AS ∪ T PA , there exists (not necessarily nontrivial) N in K such that the odd quotient graph of Cay(R, S) with respect to N lies in G N .This implies that Take an arbitrary N ∈ K and write R = R/N .It is known from [11,Theorems 5.2,5.3 and 5.4 (5.2) Now take Γ ∈ G N and consider |N Γ |, namely, the number of choices for S such that the odd quotient graph of Cay(R, S) with respect to N is Γ.Let S be such a choice.For each x ∈ R, According to the definition of odd quotient graph, this subset S of R must satisfy the following conditions:   for some absolute constant c 3 .
Take arbitrary Y ∈ F N and S ∈ N Y , so that Y = G(S)/N .Let H = G(S) 1 N/N .For an H-orbit ∆ on R/N , let ∆ be the union of the right N -cosets in ∆.Let N x ∈ ∆ and g ∈ G(S) 1 .Note that S is exactly the neighborhood of the vertex 1, and g is an automorphism of Cay(R, S) that fixes 1.It follows that S g = S and so This means that S intersects ∆ evenly for every H-orbit ∆ on R/N .Since Lemma 2.5 implies that R/N is neither abelian nor generalized dicyclic, it follows from Proposition 3.7 that for all Y ∈ F N .Thus, by (5.3) and (5.4), we obtain Since R has at most 2 log 2 2 r subgroups, there are at most 2 log 2 2 r choices for N .Consequently, .

Proof of main results
We are now ready to prove Theorem 1.3 and its corollaries. .
Recall that the number of inverse-closed subsets of R is 2 c(R) .Therefore, Next assume that R is a generalized dicyclic group.Since R ≇ Q 8 × C m 2 for any nonnegative integer m, we see from [12,Theorem 1.5] that the number of inverse-closed subsets S of R

Figure 1 .
Figure 1.The b right Q-cosets in QxQ
are precisely the orbits of some subgroup of Aut(Cay(R, S)).Then by Lemma 5.1 we have|T HA ∪ T SD ∪ T TW | ≤ Q∈Q |N Q |.Take arbitrary Q ∈ Q and S ∈ N Q .Denote D = Q\R/Q = {QxQ : x ∈ R}.Let ∆ ∈ D,and let Λ be a right coset of Q contained in ∆.Recall by the definition of Cayley graphs that, for

. 2 cr +log 2 2
Let Γ be a graph on a vertex set V , and let B be a partition of V .If B is an equitable partition, that is, given any B, C ∈ B, all vertices v ∈ B have the same number of neighbours in C, then we denote this number by e(B, C) for (B, C) ∈ B × B. Let B be an equitable partition of V into sets of equal size.Then e(B, C) = e(C, B) for B, C ∈ B. The odd quotient graph Γ odd B of Γ with respect to B, defined in [11, Definition 6.1], is the graph whose vertices are the sets in B, with an edge between B and C if and only if e(B, C) is odd.If B is the orbit partition of some K ≤ Aut(Γ), then we call Γ odd B the odd quotient graph with respect to K. Lemma 5.3.|T HS ∪ T HC ∪ T AS ∪ T PA | ≤ r+c 1 for some absolute constant c 1 .

. 3 )
Note that each G(S)/N ∈ F N acts faithfully on G(S)/R as a primitive group of type CD with a regular subgroup G(S) 1 N/N ∼ = G(S) 1 of order 2 r 0.499 ≥ 2 |R/N | 0.499 .By the proof of[11, Theorem  5.10], we have|F N | ≤ 2 log

2 2 r • 2 cr+c 2 log 2 2 r+log 2 r , where c 2 = 2 c 2 c
c 3 + 1 is an absolute constant.Comparing the three upper bounds in Lemmas 5.2, 5.3 and 5.4, it is clear that for large enough r, the bound the largest.Thereby, in view of (5.1), we obtain the following result.Proposition 5.5.Let R be a finite group of order r that is neither abelian of exponent greater than 2 nor generalized dicyclic.For large enough r we have |T | ≤

Proof of Theorem 1 . 3 . 2 c
For the group R that is neither abelian of exponent greater than 2 nor generalized dicyclic, recall from the beginning of Section 4 thatZ = {S ⊆ R : S −1 = S, Aut(Cay(R, S)) > R} = (Z \ T ) ∪ T .Let c 4 be a constant number such that r ≥ c 4 is large enough for Propositions 4.4 and 5.5 to hold.Comparing the two upper bounds in Propositions 4.4 and 5.5, it is clear that the bound the larger one when r is sufficiently large, that is, whenever r ≥ c 5 for some constant number c 5 .Let c 0 = max{c 4 , c 5 }.It follows that, whenever r ≥ c 0 , we derive an upper bound for |Z| by doubling (6.1), so that|Z| ≤ 2 c(R)− r 0