Digraph representations of 2-closed permutation groups with a normal regular cyclic subgroup

In this paper, we classify 2-closed (in Wielandt’s sense) permutation groups which contain a normal regular cyclic subgroup and prove that for each such group G, there exists a circulant Γ such that Aut(Γ) = G.


Introduction
In 1969, Wielandt [15] introduced the concept of the 2-closure of a permutation group.Let G be a finite permutation group on a set Ω, the 2-closure G (2) of G on Ω is the largest subgroup of Sym(Ω) containing G that has the same orbits as G in the induced action on Ω × Ω, and we say G is 2-closed if G = G (2) .It seems impossible to classify all 2-closed transitive permutation groups.However, certain classes of 2-closed transitive groups have been determined.For example, in [16,17] the author determined all 2-closed odd-order transitive permutation groups of degree pq where p, q are distinct odd primes.In this paper, one of our main purposes is to classify all 2-closed permutation groups with a normal regular cyclic subgroup, see Theorem 1.2.Recall that a permutation group is regular if it is transitive and the only element that fixes a point is the identity.And for more information about the 2-closures of permutation groups containing a cyclic regular subgroup, see also [7].
Another research topic of this paper is the study of the automorphism groups of (di)graphs.The full automorphism group of a (di)graph Γ must be 2-closed since any permutation of the vertex set that preserves the orbits of Aut(Γ) on ordered pairs preserves adjacency.However, not every 2-closed permutation group is the full automorphism group of some (di)graph.Therefore, the concept of 2-closed groups is more general than the concept of the full automorphism groups of (di)graphs, and the classification of 2-closed groups is closely related to the study of the full automorphism groups of the corresponding digraphs.In this paper, in order to determine 2-closed groups that contain a normal regular cyclic subgroup, we also study circulant digraphs, that is Cayley digraphs of cyclic groups.See Section 2 for a more detailed explanation.Furthermore, we discuss the following representation problem.A digraph Γ with vertex set Ω is said to represent a permutation group G Sym(Ω) if Aut(Γ) = G.In this case, we also say that the permutation group G has a digraph representation Γ.
Digraph representation problem: given a 2-closed group G, is there a digraph Γ that represents G?
Suppose the digraph Γ represents a 2-closed group G Sym(Ω).Then for any g ∈ Sym(Ω), to determine whether g lies in G we only need to test if g preserves the single 2-relation given by the arc set of Γ, instead of checking all G-invariant 2-relations.We say a digraph Γ is arc-transitive if Aut(Γ) is transitive on the arc set of Γ.This means, the arc set of Γ is actually a minimal Aut(Γ)-invariant 2-relation.Suppose further that the 2-closed group G can be represented by an arc-transitive digraph Γ.Then a permutation g lies in G if and only if g leaves invariant the minimal G-invariant 2-relation given by the arc set of Γ.We will show that there are arc-transitive digraph representations for most 2-closed groups that contain a normal regular cyclic subgroup, see the remark after Lemma 3.12.
Replacing digraph with graph, we obtain the graph representation problem which asks for an undirected graph to represent a 2-closed group.These two questions have previously appeared in the literature, see for example [1,4].Clearly, the graph version problem is much more complicated than the digraph one.Since we are interested in understanding the concept of 2-closed groups, we concentrate on the digraph representation problem in this paper.
A regular permutation group is 2-closed, and in 1980, Babai [2] proved that with five exceptions, every finite regular permutation group occurs as the automorphism group of a digraph.This is the famous DRR (digraphical regular representations) problem [2].It is proved in [14] that for any prime power q, the semilinear group ΓL(1, q) can be represented by an arc-transitive circulant digraph.Moreover, it is shown in [16,17] that every 2closed odd-order transitive permutation group of degree pq has a tournament digraph representation.As for graphical representation problem, see for example [3,6,8,9,10,13].
In this paper, we will prove that every 2-closed permutation group G with a normal regular cyclic subgroup is the full automorphism group of a circulant digraph.We may suppose that G = Z n G 0 acting on Z n naturally where G 0 Aut(Z n ).We first describe the necessary and sufficient condition for G 0 such that G is 2-closed.For the detailed explanation of notation, see Section 2 and Section 3.3.1.  .Note that the induced action D 1 on the subgroup Z 2 d 1 is permutation isomorphic to (−1) * × 5 * (d 1 3), the multiplicative group of units of the ring Z 2 d 1 acting on the additive group Z 2 d 1 , let φ : (−1) * × (5) * → D 1 be the corresponding group isomorphism. Let The main result of this paper is the following theorem.
Therefore G is a semidirect product Ẑn G 0 for some subgroup G 0 Aut(Z n ) acting naturally on Z n .Since Ẑn ∼ = Z n , we may also write G = Z n G 0 directly.Our goal is to determine all such 2-closed groups.
The mail tool used in this paper is the Kovács-Li classification of arc-transitive circulants [11,12].Praeger and the author [14] refined the Kovács-Li classification and obtained the following theorem.
Then, up to isomorphism, there is a unique connected Z n -circulant Γ on which G acts arc-transitively.Moreover either Aut(Γ) = G or one of the following holds.
We point out that up to isomorphism, in the above theorem Γ can be defined as Cay(Z n , z G 0 ) where z is a generator of Z n and z G 0 is the orbit of z under G 0 .Moreover, if case (b) happens, then the group Z n has a subgroup Y of order b, and Γ = Cay(Z n , S) where S is a union of Y -cosets each consisting of generators for Z.
As a simple application of Theorem 2.1, we determine the 2-closed transitive permutation groups of degree p where p is a prime.Corollary 2.2.Let p be a prime.Let G Sym(Ω) be a 2-closed transitive permutation group of degree p. Then there exists a digraph representing G.Moreover, G is one of the following.
1.The symmetric group S p (p 2) which is 2-transitive on Ω.
Conversely, each group of the above two types is 2-closed.
Proof.Suppose G is a 2-closed transitive permutation group of degree p.By a classical result of Burnside, G is either 2-transitive or is affine.
For the converse, note that S p is the full automorphism group of the complete graph K p and so S p is indeed 2-closed.Next, let G = Z p Z k where p 3, 1 k < (p − 1) and k|(p − 1).By Theorem 2.1, there is a connected arc-transitive circulant Γ of order p such that Aut(Γ) = G, and so G is 2-closed.
We also need the following theorem.
the electronic journal of combinatorics 22(4) (2015), #P4.31 Finally, we fix the following notation.Let A Sym(Ω).Suppose that A B is the setwise stabilizer of B ⊆ Ω and g ∈ A B , we denote A B B to be the induced permutation group on B by A B and denote g B to be the induced permutation on B by g.

2-closed groups containing a normal regular cyclic group
In this section we classify 2-closed groups G that contain a normal regular cyclic group Z n .With notation in Section 2, we may suppose that G = Z n G 0 Z n Aut(Z n ) acting naturally on Z n .We first handle the special case that n is a prime power in Subsection 3.1 and Subsection 3.2.The notation needed for the statement of Theorem 1.2 is given in Subsection 3.3.1 and the proof is given in Subsection 3.3.2.

The case n = p d with p an odd prime
Let n = p d where p is an odd prime and d 2 is an integer.Then Aut( Since the elements in B 2 , . . ., B p are of order p d , γ fixes each block setwise and γ B i is a p d−1 −cycle for each i 2. However, γ fixes the point 1 ∈ H = B 1 , so the order of γ B 1 is strictly less than p d−1 .It then follows that α fixes B 1 pointwise and is fixed point free on each B i for i 2.
On the other hand, let N B i B i be the induced permutation group of the setwise stabilizer ), ( Ẑp d−1 is defined in equation (1)).For each i 2, since γ B i is fixed point free, we have that γ where a i2 is some element in Ẑp d−1 .By induction, we have that for each k 1, (γ i for some x i ∈ Z n with order p.Note that x i may not equal x j for 2 i < j p, but they are all of order p.We have proved the following lemma.

Proof. Let Aut(Z
Let B = {B 1 = H, B 2 , . . ., B p } be the cosets of the subgroup H where H < Z p d is of order p d−1 .Then it is easy to show that µ fixes B 1 setwise, and permutes B 2 , . . ., B p as a (p − 1)-cycle.
By Lemma 3.1, if α ∈ G 0 then zZ p ⊆ z G 0 .Conversely, suppose that zZ p ⊆ z G 0 .Note that the generator z ∈ B k for some k 2 and zZ p ⊆ B k .By the action of µ and γ, we conclude that α ∈ G 0 .Proposition 3.3.Let n = p d where p is an odd prime and d 2. Suppose first that G 0 Z p−1 , that is p||G 0 |, then α ∈ G 0 .By Lemma 3.1, α fixes B 1 = H pointwise and for each i 2, α B i is xB i i for some x i ∈ Z n with order p.Let 1 = β ∈ Sym(Z n ) such that β fixes every element of B 1 , . . ., B p−1 and β Bp = α Bp .That means β Bp = xBp p , (recall that x : z → zx for any z ∈ Z n ).We claim that β ∈ (Z p d α ) (2) and so β ∈ G (2) .Take any pair (y 1 , y 2 ) ∈ Z n × Z n .If both y 1 and y 2 belong to B p , then (y 1 , y 2 ) β = (y 1 x p , y 2 x p ) is in the orbital (y 1 , y 2 ) G .Suppose next that exactly one of {y 1 , y 2 } lies in B p , say y 2 ∈ B p .Since the stabilizer G y 1 is the conjugate of G 0 in G by an element in Ẑn , a conjugate of α, say ρ, is in G y 1 .Therefore β Bp equals (ρ j ) Bp for some j ∈ {1, . . ., p − 1}, and so (y 1 , y 2 ) β ∈ (y 1 , y 2 ) G .It then follows that β ∈ (Z p d α ) (2) G (2) .However, since β fixes B 1 and B 2 pointwise, β / ∈ Z p d Aut(Z p d ), and so β / ∈ G and G is not 2-closed.Suppose next that G 0 Z p−1 .Let S = z G 0 where z ∈ Z p d is an element of order p d and let Γ = Cay(Z n , S).Since (p, |G 0 |) = 1, p |S| and so S is not a union of cosets of any subgroup of Z n .By Theorem 2.1, Aut(Γ) = G and so G is 2-closed.This completes the proof.

Remark:
In above proof, note that β is in (Z p d α ) (2) .Hence we actually proved that (Z p d α ) (2) Z p d Aut(Z p d ) where α ∈ Aut(Z p d ) is of order p.

3.2
The case n = 2 d for d 2 Notation: For convenience, in this subsection we write Z n additively as the group Z n of integers modulo n, so in this case Moreover Aut(Z n ) is the multiplicative group Z * n so that i * ∈ Aut(Z n ) denotes the map j → ij.

d 3:
We first study the action of 5 * .By computation 5 * preserves each block B i , we determine the induced permutation (5 * ) B i next.Since B 1 ∪ B 3 consists of all elements of order 2 d , (5 * ) B 1 and (5 is trivial).Therefore the orders of (5 * ) B 1 and (5 * ) B 3 are 2 d−2 , the order of (5 * ) B 2 is 2 d−3 , and the order of (5 * ) B 0 is 2 d−4 (if d = 3, then the order is 1).Proof.Note that both z and z + Z 2 are contained in B 1 or B 3 and (−1) * interchanges two blocks B 1 and B 3 .The result follows from the analysis of the actions of (−1) * and 5 * easily.
) and G 0 = 5 * , then G is 2-closed and can be represented by an arc-transitive circulant.
3. if G 0 = 5 * , then G is 2-closed and can be represented by a circulant.
Case 2: d 4 Let α = (5 * ) 2 d−4 be an element of order 4 in 5 * .By the analysis of action of 5 * , we deduce that α fixes B 0 pointwise and o(α Suppose first that d = 4, then α = 5 * .By direct computation, α i where a i2 is some element in Ẑ2 d−2 .By induction, we have that for each k 1, ((5 . By the order of α B i , we have that α B i = xB i i , where x 1 , x 3 ∈ Z n are of order 4 and x 2 = 2 d−1 is the unique involution in Z n .In addition, 2x 1 = 2x 3 = 2 d−1 .Therefore we have proved the following lemma.
Lemma 3.10.With above notation, let n = 2 d for d Then G is the full automorphism group of an arc-transitive circulant and so G is 2-closed.
Proof.Let S = 1 G 0 be the orbit of 1 under G 0 , and let Γ = Cay(Z n , S).Since α 2 / ∈ G 0 , it follows from corollary 3.8 that S is not a union of cosets of any subgroup of Z n .By Theorem 2.1, Aut(Γ) = G as required.
It remains to show that if G = Z n G 0 where α 2 ∈ G 0 but α / ∈ G 0 then G is the full automorphism group of some circulant.We will prove this in Proposition 3.15 when we handle the more general case.
3.3 The general case.

3.3.1
The notation for the main theorem.
We explain Conditions 1.1 in more detail first.
Let Aut(Z n ).In order to reduce the proof in the general case to the prime power case, we choose the product action form to describe G. Let Z m be the unique subgroup of Z n of order m for m|n.Then we may write , where p 1 = 2}.
The normalizer of Ẑn in Sym(Z n ) is acting on Z n by the natural product action.Therefore G = Ẑn G 0 N has the natural product action.We need the following two easy observations in the proof below.(1) Note that when i 2, Aut(Z ( Proof.If condition (i) does not hold, then there exists an odd prime p i 5 where i 2 such that p i ||n and D i G 0 .In this case we take K = Ẑp i D i .By hypothesis, K is the subgroup of G which fixes each component of elements of Z n except for the i-th component.
Hence the action of K on Z n is the product action of K × {1} on Z n = Z p i × Z n p i where K ∼ = K acts on Z p i naturally.It follows from Theorem 2.3 that K (2) = (K) (2) × {1}.By the remark after Corollary 2.2, (K) (2) Z p i Aut(Z p i ).Since G (2) K (2) , we have that G is not 2-closed in this case.
If condition (ii) does not hold, then there exists an odd prime p i where i 2 such that ).The same argument as above proves that G is not 2-closed in this case either.Suppose 2 d 1 ||n and d 1 3, suppose also that either condition (iii) or (iv) fails.Take . By the same argument as above, it follows from Proposition 3.6(1) and Proposition 3.9 that G is not 2-closed.Moreover, in the latter three cases, Γ = Σ[K 2 ] is a lexicographic product and the pointwise stabilizer of {1, z} in Aut(Γ) preserves each coset of Z 2 .
Proof.Suppose that G is not the full automorphism group of Γ.By the condition (i), for any odd prime p i 5 such that p i ||n, we have G 0 = Aut(Z p i ) × H for some H Aut(Z n/p i ).It then follows from Recall that n = 2 d 1 p d 2 2 • • • p dt t .Suppose that p j |b for some j ∈ {1, . . ., t}.Then zZ p j ⊆ S where Z p j is the subgroup of order p j and d j 2 by Theorem 2.1 (b).Let z = (z 1 , . . ., z t ) where z i is a generator of Z p d i i for each i.Thus zZ p j ⊆ S = z G 0 implies that z j Z p j ⊆ z D j ∩G 0 j in the j-th component.By Corollary 3.2, the condition (ii) implies that b = 2 l is a power of 2. Similarly, by Corollary 3.8, Lemma 3.5 and the action of Aut(Z 4 ), the condition (iii) and (iv) imply that b must be 2 and one of cases 2-4 happens.
Remark: Suppose G satisfies Conditions 1.1.By the above lemma, G can be represented by an arc-transitive circulant if and only if G does not arise in any of the cases 2-4 of Lemma 3.12.
Next we will show that if one of cases 2-4 occurs then there exists a circulant Γ which is not arc-transitive such that Aut(Γ) = G.We discuss case 4 first.Lemma 3.13.Suppose n = 4m where m > 1 is odd and G = Z 4m G 0 where G 0 = Aut(Z 4 ) × K and K Aut(Z m ).Suppose further that G 0 satisfies Conditions 1.1.Then G is 2-closed and can be represented by a circulant.
where D i is the direct factor subgroup of Aut(Z n ) that fixes each component the electronic journal of combinatorics 22(4) (2015), #P4.31 of the elements of Z n except for the i-th component.So D i ∼ = Aut(Z p d i i ) for each i.In fact D i induces a faithful action on the subgroup Z p d i i

Lemma 3 . 1 .Corollary 3 . 2 .
Let α ∈ Aut(Z p d ) with order p.Let B = {B 1 = H, B 2 , . . ., B p } be the cosets of the subgroup H where H < Z p d is of order p d−1 .Then α fixes B 1 = H pointwise and for each i 2, α B i is xB i i for some x i ∈ Z n with order p.Let n = p d and Z n = z .Let Z p Z n be the subgroup of order p. Suppose that G = Z n G 0 where G 0 Aut(Z n ).Then the coset zZ p ⊆ z G 0 if and only if p||G 0 |.Remark: Let S = z G 0 and Γ = Cay(Z n , S).If case (b) of Theorem 2.1 occurs for Γ, then zZ p ⊆ z G 0 .That is why we consider this corollary.
be represented by an arc-transitive circulant.Proof.As defined at the beginning of Subsection 3.1, let α ∈ Aut(Z p d ) be an element of order p.Let B = {B 1 = H, B 2 , . . ., B p } be the cosets of the subgroup H where H < Z p d is of order p d−1 .
Theorem 2.1 that case (b) of Theorem 2.1 occurs for Γ.That is n = bk > 4 where b 2 and Γ = Σ[K b ].Moreover, the group Z n has a subgroup Y of order b and S is a union of Y -cosets each consisting of generators for Z n .
d t 1, t 1 where p 2 , . . ., p t are distinct odd primes (also write p 1 = 2).And let Aut( Theorem 1.2.Suppose G = Z n G 0 acting on Z n naturally whereG 0 Aut(Z n ).Then G is 2-closed if and only if G 0 satisfies Conditions 1.1.Moreover, if G is 2-closed then G can be represented by a circulant digraph.First we introduce some concepts and notation concerning Cayley digraphs.Given a finite group H, and a subset S ⊂ H \{1}, the Cayley digraph Γ = Cay(H, S) with respect to S is defined as the directed graph with vertex set H and arc set AΓ = {(g, sg) | g ∈ H, s ∈ S}.