## Math.pku.edu.cn

A note on permutation groups with a regular subgroup
Dedicated to Cheryl Praeger on the occasion of her 60th birthday
In this note we first prove that for a positive integer

*n > *1 with

*n *=

*p *or

*p*2 where

*p *is a prime, there exists a transitive group of degree

*n *without regular subgroups.

Then we look at 2-closed transitive groups without regular subgroups, and pose twoquestions and a problem for further study.

We first define four subsets of positive integers:

*N R *=

*{n ∈ *N there is a transitive group of degree

*n *without a regular subgroup

*}*
*N*2

*R *=

*{n ∈ *N there is a 2-closed transitive group of degree

*n *without a regular subgroup

*}*
*N D *=

*{n ∈ *N there is a vertex-transitive digraph of order

*n *which is non-Cayley

*}*
=

*{n ∈ *N there is a vertex-transitive graph of order

*n *which is non-Cayley

*}*
In the literature there is much work studying the set

*N C*; see [5, 6, 7, 8, 9] for example.

*∈ N C *by [7, Theorem 3], but 12

*∈ N R *since

*M*11, acting on 12 points, has no regular
subgroup by [3]. And it is easy to see that 6 is the smallest number in

*N R \ N C *since

*A*6has no regular subgroups. In the first part of this note we shall determine the set

*N R*.

It is well-known that any prime number

*p *does not belong to any one of the four sets
above. Moreover, Maruˇsiˇc [5] proved that

*p*2

*/*
*∈ N C*. In fact, we have

*p*2

*/*
Proposition 1

*Any transitive group G of degree p*2

*on *Ω

*has a regular subgroup. Hencep*2

*/*
Proof Take a minimal transitive subgroup

*P *of

*G*. Then

*P *is a

*p*-group and every maximalsubgroup

*M *of

*P *is intransitive. For any

*α ∈ *Ω, we have

*|Pα| *=

*|P |/p*2 and

*|Mα| > |M|/p*2,so

*Mα *=

*Pα*. It follows that

*Pα ≤ M *and hence

*Pα ≤ *Φ(

*P *). If

*|P *: Φ(

*P *)

*| *=

*p*, then

*P *iscyclic and is regular. If

*|P *: Φ(

*P *)

*| *=

*p*2, then

*Pα *= Φ(

*P *). Since Φ(

*P *) is normal in

*P *and

*Pα *is core-free, we have

*Pα *= 1 and hence

*P ∼*
The following example shows that

*p*3

*∈ N R*. However, it was proved that

*p*3

*/*
[5, 6]. Therefore

*p*3

*∈ N R \ N C*.

Example 2 (1)

*Let p be an odd prime and let G be the group of order p*4

*presented by*
*G *=

*a, b ap*2 =

*bp *=

*cp *= 1

*, *[

*a, b*] =

*c, *[

*c, a*] =

*ap, *[

*c, b*] = 1

*.*
*Let H *=

*c . Consider the transitive permutation representation ϕ of G acting on the cosetspace *[

*G *:

*H*]

*. Then ϕ*(

*G*)

*is a transitive group of degree p*3

*, and ϕ*(

*G*)

*has no regularsubgroups.*
*G *=

*a, b, c, d a*2 =

*b*2 =

*c*2 =

*d*4 = 1

*, *[

*a, b*] = [

*b, c*] = [

*c, a*] = 1

*, ad *=

*ab, bd *=

*bc, cd *=

*c .*
Z4

*has order *25

*. Let H *=

*b, d*2

*and ϕ be the transitive permutation*
*representation of G acting on the coset space *[

*G *:

*H*]

*. Then ϕ*(

*G*)

*is a transitive group ofdegree *23

*and has no regular subgroup.*
Proof (1) Since [

*c, a*] =

*ap*,

*c*
*G*. Since Ker

*ϕ *= core

*G*(

*H*) = 1, the action is faithful.

=

*G*. Suppose that

*ϕ*(

*G*) has a regular subgroup, say

*ϕ*(

*R*). Then

*R *is maximal in

*G*, and

*RH *=

*G *by the Frattini argument. However,

*H ≤ G ≤ *Φ(

*G*)

*≤ R*, a contradiction.

(2) Similar to (1), prove that

*H *is core-free and contained in Φ(

*G*). Details omitted.

*✷*
Now we are ready to determine the set

*N R*. We first need the following proposition.

Proposition 3

*Let p < q be two primes. Then pq ∈ N R.*
Proof Let

*W *= Z

*p *Z

*q *=

*a*
*b *, viewed as an imprimitive group of degree

*pq*. Since the
action of

*b *on the base group Z

*qp *is nontrivial, we may take a

*b *-invariant subgroup

*H *of
the base group such that the action of

*b *on

*H *is also nontrivial and

*H *is smallest subjectto this property. Then

*b *is irreducible on

*H*. Let

*G *=

*H*
*b *. Since

*p < q*,

*|H| *=

*pk > p*.

*H*. Consider the transitive permutation representation

*ϕ *of

*G *acting on the coset
space [

*G *:

*M *]. Since

*H *is a minimal normal subgroup of

*G*, core

*G*(

*M*) = 1 and

*ϕ *is faithful.

Since

*b *is a Sylow

*q*-subgroup and maximal in

*G *by the irreducibility of

*b *on

*H*,

*G *has nosubgroup of order

*pq*. Hence

*ϕ*(

*G*) has no regular subgroups. It follows that

*pq ∈ N R*.

*✷*
Theorem 4

*Let n be a positive integer greater than *1

*. Then n ∈ N R unless n *=

*p or p*2

*for a prime p.*
Proof This theorem follows from Proposition 1, Example 2, Proposition 3 and the factthat if

*m ∈ N R *then

*km ∈ N R *for any positive integer

*k*.

In the second part of this note we look at the set

*N*2

*R*. The next proposition shows that

*∈ N*2

*R*, while Maruˇsiˇc [5] proved that

*p*3

*/*
Proposition 5

*Any *2

*-closed transitive group G of degree p*3

*on *Ω

*has a regular subgroup.*
To prove the above proposition, we need the concept of 2-closures of permutation groups
Let

*G *be a permutation group acting on Ω. Suppose that ∆0

*, *∆1

*, . . . , *∆

*r−*1, are orbits
of

*G *acting on Ω

*× *Ω. The 2-closure

*G*(2) of

*G *is defined by

*G*(2) =

*{x ∈ *Sym(Ω)

*|*∆

*xi *= ∆

*i, i *= 0

*, *1

*, . . . , r − *1

*}.*
Obviously,

*G*(2)

*≥ G*; if

*G*(2) =

*G*, we say that

*G *is 2-closed. The following lemma is quotedfrom [10, Exercise 5.28].

Lemma A:

*Suppose that G is a *2

*-closed group and p a prime. Then the Sylow p-subgroupP of G is also *2

*-closed.*
Theorem B: (Wielandt’s dissection theorem)

*Let G be a permutation group acting on *Ω

*,and H a subgroup of G. Suppose that *Ω = ∆

*∪ *Γ

*, *∆

*∩ *Γ =

*∅, *∆ =

*∅, *Γ =

*∅ and *∆

*H *=∆

*, *Γ

*H *= Γ

*. If for any δ ∈ *∆

*, H and Hδ have the same orbits on *Γ

*, then H*∆

*× H*Γ

*≤ G*(2)

*.*
This theorem follows from [10, Theorem 6.5] and the following obvious fact: if

*H ≤ G*,
Let

*P ∈ *Syl(

*G*). Then

*P *is also transitive on Ω. Take an
element

*z ∈ Z*(

*P *) with

*o*(

*z*) =

*p*. Let

*B *=

*{B*1

*, . . . , Bp*2

*} *be the set of orbits of

*z *. Then

*B*is a complete block system of

*P *. Assume that

*K *=

*PB *is the kernel of

*P *acting on

*B*. Since

*KBi *= Z

*p*,

*K *is elementary abelian. Set

*P *=

*P/K*.

*P *is a transitive group on

*B*.

Take 1 =

*x ∈ K *such that the support supp(

*x*) of

*x *has the minimum size. We claim
that supp(

*x*) is a block of

*P *. Since

*K *is elementary abelian,

*x *is of order

*p*. If supp(

*x*)were not a block of

*P *, then we could find an

*h ∈ P *such that supp(

*x*)

*h *= supp(

*x*) and

*D *= supp(

*x*)

*∩ *supp(

*x*)

*h *=

*∅*. Since every

*Bi *is a block of size

*p *and

*p *a prime, supp(

*x*),supp(

*x*)

*h *= supp(

*xh*) and

*D *are unions of several entire blocks of

*P *in

*B*. Set

*J *=

*x, xh *.

Then the nontrivial orbits of

*J *are precisely the blocks contained in supp(

*x*)

*∪ *supp(

*xh*).

It is not difficult to check that for any

*g ∈ *Ω

*− D*,

*J *and

*Jg *have the same orbits in

*D*.

So, by Theorem B,

*JD × *1Ω

*−D ≤ P *(2). (In what follows, for brevity, we use

*JD *to denote

*JD × *1Ω

*−D*.) Since

*P *is 2-closed,

*P *=

*P *(2). Then

*JD ≤ P *and hence

*JD ≤ K*. Thus thereexists an element

*y ∈ JD ≤ K *of order

*p *such that

*|*supp(

*y*)

*| < |*supp(

*x*)

*|*, which contradictsthe choice of

*x*.

Now we distinguish three cases: (1)

*|*supp(

*x*)

*| *=

*p*; (2)

*|*supp(

*x*)

*| *=

*p*2 and (3)

*|*supp(

*x*)

*| *=
If

*|*supp(

*x*)

*| *=

*p*, then supp(

*x*) =

*Bi *for some

*i*. Hence

*P *= Z

*p P *. Since

*P *is a transitive
group of degree

*p*2, it has a regular subgroup

*H *isomorphic to Z2

*p *or Z

*p*2. Thus

*P *has a
subgroup isomorphic to Z

*p H*, which has an abelian regular subgroup.

If

*|*supp(

*x*)

*| *=

*p*3, then

*K *=

*z *is semiregular. Assume that

*H/K *is a regular subgroup
of

*P *=

*P/K*. Then

*H *is a regular subgroup of

*P *.

Finally we assume that

*|*supp(

*x*)

*| *=

*p*2. Then supp(

*x*) is a block of

*P *of length

*p*2 which
is a union of

*p Bi*s. Assume that

*C *=

*{C*1

*, . . . , Cp} *is a block system of

*P *with supp(

*x*) =

*C*1.

Then

*K *=

*KC*1

*× · · · × KCp ∼*
If

*P *=

*P/K *has an element

*aK *of order

*p*2, then

*a, z *is a regular subgroup of

*P *.

So we may assume that exp

*P *=

*p*. Take a regular subgroup

*H/K *=

*uK, vK *of

*P *suchthat

*v ∈ PC − K*. (Since

*PC *has index

*p *in

*P *, this is possible.) We have

*HC *=

*v, K *and

*u /*
*∈ HC*. Without loss of generality we assume that

*Cui *=

*Ci*+1, for all

*i *(mod

*p*). Define a

*w *=

*vC*1(

*vu*)

*C*2(

*vu*2)

*C*3

*· · · *(

*vup−*1)

*Cp.*
Since [

*v, u*] =

*k ∈ K*,

*vu *=

*vk*. So

*w *=

*vC*1(

*vk*)

*C*2(

*vkku*)

*C*3

*· · · *(

*vkku · · · kup−*2)

*Cp*
= (

*vC*1

*vC*2

*vC*3

*· · · vCp*)(1

*C*1

*kC*2(

*kku*)

*C*3

*· · · *(

*kku · · · kup−*2)

*Cp*)=

*v*¯

*k *= 1

*C*1

*kC*2(

*kku*)

*C*3

*· · · *(

*kku · · · kup−*2)

*Cp ∈ K*, as

*K *=

*KC*1

*× · · · × KCp*. So

*w ∈ HC*.

Since

*up ∈ K*,

*vup *=

*v*. So

*u *centralizes

*w*, and then

*R *=

*u, w *is abelian and

*RK/K *istransitive on

*B*. If

*|R| > p*2, then

*R *is regular; if

*|R| *=

*p*2, then

*R × z *is regular. Thiscompletes the proof of this proposition.

It is known that not all 2-closed transitive groups are the full automorphism groups of
a (di)graph. For example, the regular representation of a finite group that has no GRR orDRR is such an example since regular groups are obviously 2-closed. (For GRRs and DRRsof finite groups, see [1, 2, 4].) Now we would like to pose the following questions.

Question 1

*Is N*2

*R *=

*N C?*
To study Question 1, we should first find nonregular 2-closed groups which are not the
full automorphism groups of (di)graphs. We do not know such examples.

To end this note, we propose a problem, we first define one more subset of positive

*PN R *=

*{n ∈ *N there is a primitive group of degree

*n *without a regular subgroup

*}*
Different from the set

*N R*, we know that

*pn /*
*∈ PN R *for any prime

*p *and any positive
integer

*n*; see [11, Theorem]. Hence, determining the set

*PN R *should be much harder than

*N R*.

The author is grateful to Dr. Haipeng Qu and Dr. Jing Xu for their valuable discussions
This work was supported by NSFC (proj. no. 10671114).

[1] L´aszl´o Babai, On the abstract group of automorphisms,

*Combinatorics *(Ed. H. N. V.

Temperley), London Math. Soc. Lecture Note Series 52, 1981; 1–40.

[2] L´aszl´o Babai, Finite digraphs with regular automorphism groups,

*Periodica Math.*
[3] Michael Giudici, Factorisations of sporadic simple groups,

*J. Algebra*, 304 (2006), 311–
[4] Chris D. Godsil, GRR’s for non-solvable groups,

*Colloq. Math. Soc. Jannos Bolyai*,
[5] Dragan Maruˇsiˇc, Vertex-transitive graphs and digraphs of order

*pk*,

*Ann. of Discrete*
[6] Brendan D. McKay, Cheryl E. Praeger, Vertex-transitive graphs which are not Cayley
graphs, I,

*J. Austral. Math. Soc.(A) *56(1994), 53–63.

[7] Brendan D. McKay, Cheryl E. Praeger, Vertex-transitive graphs which are not Cayley
graphs, II,

*J. Graph Theory *22(1996), 321–334.

[8] A. A. Miller, Cheryl E. Praeger, Non-Cayley, vertex transitive graphs of order twice
the product of two distinct odd primes,

*J. Alg. Combin. *3(1994), 77–111.

[9] M. A. Iranmanesh, Cheryl E. Praeger, On non-Cayley vertex-transitive graphs of order
a product of three primes,

*J. Combin. Theory Ser. B *81(2001), 1–19.

[10] H. Wielandt,

*Permutation Groups Through Invariant Relations And Invariant Func-*
*tions*, Ohio State University, Columbus, 1969.

[11] Ming-Yao Xu, Vertex-primitive digraphs of prime-power order are hamiltonian,

*Discrete*
Source: http://www.math.pku.edu.cn:8000/var/preprint/7209.pdf

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