Thank you, it's cool. Used for the first time because only then was the right drug. Very revealing detail: ordered Express shipping to get the order within 3 hours because didn't know how to get the order faster buy antibiotics online The drug is authentic exactly. Consistent with the stated prices the Staff is knowledgeable.

## 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

HUMAN SERVICES COMMITTEE December 7, 2009 Present: Doug Paddock, Dan Banach, Donna Alexander, Tim Dennis, Don House, Bob Multer, Taylor Fitch, Sarah Purdy, Connie Hayes, Nancy Gates, Earle Gleason, Amy Miller, Mark Morris, Leslie Church, Katie Smeenk, Pam Larnard, Deb Minor. Doug and Tim will do the audit. Minutes of the November meeting were approved as presented. PUBLIC HEALTH: De

Gabapentin Important drug interactions Clinical use Anti-epileptic – adjunctive treatment of partial• Antidepressants: antagonism of anticonvulsiveseizures with or without secondary generalisation Administration Dose in normal renal function 300 mg on day 1, 300 mg twice daily on day 2,300 mg three times daily on day 3, then increasedaccording to response to 1.2 g daily (in t