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## Imath.kiev.ua

**H. Wei **(Guangxi Teacher’s College, Zhongshan, China),

**Y. Wang **(Zhongshan Univ., China)

**c∗****-SUPPLEMENTED SUBGROUPS**

AND *p***-NILPOTENCY OF FINITE GROUPS***
**∗**
**c ****-DOPOVNENI PIDHRUPY**
**TA ***p***-NIL**\

**POTENTNIST**\

**SKINÇENNYX HRUP**
A subgroup

*H *of a finite group

*G *is said to be

*c *-supplemented in

*G *if there exists a subgroup

*K *suchthat

*G *=

*HK *and

*H ∩ K *is permutable in

*G*. It is proved that a finite group

*G *which is

*S*4-free is

*p*-nilpotent if

*NG*(

*P *) is

*p*-nilpotent and, for all

*x ∈ G\NG*(

*P *), every minimal subgroup of

*P ∩ P x ∩ GNp*is

*c *-supplemented in

*P *and, if

*p *= 2, one of the following conditions holds: (a) Every cyclic subgroup of

*P ∩ P x ∩ GNp *of order 4 is

*c *-supplemented in

*P *; (b) [Ω2(

*P ∩ P x ∩ GNp *)

*, P *]

*≤ Z*(

*P ∩ GNp *); (c)

*P*is quaternion-free, where

*P *a Sylow

*p*-subgroup of

*G *and

*GNp *the

*p*-nilpotent residual of

*G*. That will extendand improve some known results.

Pidhrupa

*H *skinçenno] hrupy

*G *nazyva[t\sq

*c *-dopovnenog v

*G, *qkwo isnu[ pidhrupa

*K *taka, wo

*G *=

*HK *ta

*H ∩ K *[ perestanovoçnog v

*G*. Dovedeno, wo skinçenna hrupa

*G, *qka [

*S*4-vil\nog, [

*p*-nil\potentnog, qkwo

*NG*(

*P *)

*p*-nil\potentna i dlq vsix

*x ∈ G\NG*(

*P *) koΩna minimal\na pidhrupaiz

*P ∩ P x ∩ GNp *[

*c *-dopovnenog v

*P *ta, qkwo

*p *= 2, vykonu[t\sq odna z nastupnyx umov: a) koΩna
cykliçna pidhrupa porqdku 4 iz

*P ∩ P x ∩ GNp *[

*c *-dopovnenog v

*P *; b) [Ω2(

*P ∩ P x ∩ GNp *)

*, P *]

*≤ Z*(

*P ∩*
*∩ GNp*); c)

*P *[ bezkvaternionnog, de

*P *— sylovs\ka

*p*-pidhrupa hrupy

*G *ta

*GNp *—

*p*-nil\potentnyjzalyßok hrupy

*G*. Tym samym poßyreno ta pokraweno deqki vidomi rezul\taty.

**1. Introduction. **All groups considered will be finite. For a formation

*F *and a group

*G, *there exists a smallest normal subgroup of

*G, *called the

*F*-residual of

*G *and denoted

by

*GF , *such that

*G/GF ∈ F *(refer [1]). Throughout this paper,

*N *and

*Np *will denote

the classes of nilpotent groups and

*p*-nilpotent groups, respectively. A 2-group is called

quaternion-free if it has no section isomorphic to the quaternion group of order 8

*.*
General speaking, a group with a

*p*-nilpotent normalizer of the Sylow

*p*-subgroup
need not be a

*p*-nilpotent group. However, if one adds some embedded properties onthe Sylow

*p*-subgroup, he may obtain his desired result. For example, Wielandt provedthat a group

*G *is

*p*-nilpotent if it has a regular Sylow

*p*-subgroup whose

*G*-normalizeris

*p*-nilpotent [2]. Ballester-Bolinches and Esteban-Romero showed that a group

*G *is

*p*-nilpotent if it has a modular Sylow

*p*-subgroup whose

*G*-normalizer is

*p*-nilpotent [3].

Moreover, Guo and Shum obtained a similar result by use of the permutability of someminimal subgroups of Sylow

*p*-subgroups [4].

In the present paper, we will push further the studies. First, we introduce the

*c *-
supplementation of subgroups which is a unify and generalization of the permutability andthe

*c*-supplementation [5, 6] of subgroups. Then, we give several sufficient conditions fora group to be

*p*-nilpotent by using the

*c *-supplementation of some minimal

*p*-subgroups.

In detail, we obtain the following main theorem:

**Theorem 1.1.**
*Let G be a group such that G is S*4

*-free and let P be a Sylow p-*
*subgroup of G. Then G is p-nilpotent if NG*(

*P *)

*is p-nilpotent and, for all x ∈ G\NG*(

*P *)

*,one of the following conditions holds*:
(a)

*Every cyclic subgroup of P ∩ P x ∩ GNp of order p or *4 (

*if p *= 2)

*is c -*
* Project supported by NSFC (10571181), NSF of Guangxi (0447038) and Guangxi Education Department.

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
(b)

*Every minimal subgroup of P ∩ P x ∩ GNp is c -supplemented in P and, if p *= 2

*,*
[Ω2(

*P ∩ P x ∩ GNp*)

*, P *]

*≤ Z*(

*P ∩ GNp*);
(c)

*Every minimal subgroup of P ∩ P x ∩ GNp is c -supplemented in P and P is*
Following the proof of Theorem 1.1, we can prove the Theorem 1.2. It can be consid-
ered as an extension of the above-mentioned result of Ballester-Bolinches and Esteban-Romero.

**Theorem 1.2.**
*Let P be a Sylow p-subgroup of a group G. Then G is p-nilpotent if*
*NG*(

*P *)

*is p-nilpotent and, for all x ∈ G\NG*(

*P *)

*, one of the followings holds*:
(a)

*Every cyclic subgroup of P ∩ P x ∩ GNp of order p or *4 (

*if p *= 2)

*is permutable*
(b)

*Every minimal subgroup of P ∩ P x ∩ GNp is permutable in P and, if p *= 2

*,*
[Ω2(

*P ∩ P x ∩ GNp*)

*, P *]

*≤ Z*(

*P ∩ GNp*);
(c)

*Every minimal subgroup of P ∩ P x ∩ GNp is permutable in P and, if p *= 2

*, P is*
As an application of Theorem 1.1, we get the following theorem:

**Theorem 1.3.**
*Let G be a group such that G is S*4

*-free and let P be a Sylow p-*
*subgroup of G, where p is a prime divisor of |G| with *(

*|G|, p − *1) = 1

*. Then G isp-nilpotent if one of the following conditions holds*:
(a)

*Every cyclic subgroup of P ∩ GNp of order p or *4 (

*if p *= 2)

*is c -supplemented*
(b)

*Every minimal subgroup of P ∩ GNp is c -supplemented in NG*(

*P *)

*and, if p *= 2

*,*
Our results improve and extend the following theorems of Guo and Shum [7, 8].

**Theorem 1.4 **([7], Main theorem)

**. ***Let G be a group such that G is S*4

*-free and let P*
*be a Sylow p-subgroup of G, where p is the smallest prime divisor of |G|. If every minimalsubgroup of P ∩ GN is c-supplemented in NG*(

*P *)

*and, when p *= 2

*, P is quaternion-free,then G is p-nilpotent.*
**Theorem 1.5 **([8], Main theorem)

**. ***Let P be a Sylow p-subgroup of a group G, where*
*p is a prime divisor of |G| with *(

*|G|, p − *1) = 1

*. If every minimal subgroup of P ∩ GN ispermutable in NG*(

*P *)

*and, when p *= 2

*, either every cyclic subgroup of P ∩ GN of order*4

*is permutable in NG*(

*P *)

*or P is quaternion-free, then G is p-nilpotent.*
**2. Preliminaries. **Recall that a subgroup

*H *of a group

*G *is

*permutable *(or

*quasi-*
*normal*) in

*G *if

*H *permutes with every subgroup of

*G. H *is

*c-supplemented *in

*G *if thereexists a subgroup

*K*1 of

*G *such that

*G *=

*HK*1 and

*H ∩ K*1

*≤ HG *= Core

*G*(

*H*) [5,6]; in this case, if we denote

*K *=

*HGK*1

*, *then

*G *=

*HK *and

*H ∩ K *=

*HG*; of course,

*H ∩ K *is permutable in

*G. *Based on this observation, we introduce:

**Definition 2.1.**
*A subgroup H of a group G is said to be c -supplemented in G if*
*there exists a subgroup K of G such that G *=

*HK and H ∩ K is a permutable subgroupof G. We say that K is a c -supplement of H in G.*
It is clear from Definition 2.1 that a permutable or

*c*-supplemented subgroup must be
a

*c *-supplemented subgroup. But the converses are not true. For example, let

*G *=

*A*4

*,*the alternating group of degree 4. Then any Sylow 3-subgroup of

*G *is

*c*-supplementedbut not permutable in

*G. *If we take

*G *=

*a, b|a*16 =

*b*4 = 1

*, ba *=

*a*3

*b , *then

*b*2(

*aibj*) == (

*aibj*)9+2((

*−*1)

*j−*1)

*b*2

*. *Hence

*b*2 is permutable in

*G. *However,

*b*2 is not

*c*-supple-mented in

*G *as

*b*2 is in Φ(

*G*) and not normal in

*G.*
*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
*c *-SUPPLEMENTED SUBGROUPS AND

*p*-NILPOTENCY OF FINITE GROUPS
The following lemma on

*c *-supplemented subgroups is crucial in the sequel. The
proof is a routine check, we omit its detail.

**Lemma 2.1.**
*Let H be a subgroup of a group G. Then*:
(1)

*If H is c -supplemented in G, H ≤ M ≤ G, then H is c -supplemented in M *;(2)

*Let N ✁ G and N ≤ H. Then H is c -supplemented in G if and only if H/N is*
(3)

*Let π be a set of primes, H a π-subgroup and N a normal π -subgroup of G. If*
*H is c -supplemented in G, then HN/N is c -supplemented in G/N *;
(4)

*Let L ≤ G and H ≤ *Φ(

*L*)

*. If H is c -supplemented in G, then H is permutable*
**Lemma 2.2.**
*Let c be an element of a group G of order p, where p is a prime divisor*
*of |G|. If c is permutable in G, then c is centralized by every element of G of order p or*4 (

*if p *= 2)

*.*
**Proof. **Let

*x *be an element of

*G *with order

*p *or 4 (if

*p *= 2). By the hypotheses,

*x c *=

*c x . *Clearly, if

*x *is of order

*p, *then

*c *is centralized by

*c. *Now assume that

*p *= 2 and

*x *is of order 4

*. *If [

*c, x*] = 1

*, *then

*c−*1

*xc *=

*x−*1 and (

*xc*)2 = 1

*. *Furthermore,

*| x c | ≤ *4

*, *of course, [

*c, x*] = 1

*, *a contradiction. We are done.

**Lemma 2.3 **([9], Lemma 2)

**. ***Let F be a saturated formation. Assume that G is a*
*non-F-group and there exists a maximal subgroup M of G such that M ∈ F and G *==

*F *(

*G*)

*M, where F *(

*G*)

*is the Fitting subgroup of G. Then*:
(1)

*GF /*(

*GF *)

*is a chief factor of G*;(2)

*GF is a p-group for some prime p*;(3)

*GF has exponent p if p > *2

*and exponent at most *4

*if p *= 2;(4)

*GF is either an elementary abelian group or *(

*GF *) =

*Z*(

*GF *) = Φ(

*GF *)

*is an*
**Lemma 2.4 **([10], Lemma 2.8(1))

**. ***Let M be a maximal subgroup of a group G and*
*let P be a normal p-subgroup of G such that G *=

*P M, where p a prime. Then P ∩ M isa normal subgroup of G.*
**Lemma 2.5 **([11], Theorem 2.8)

**. ***If a solvable group G has a Sylow *2

*-subgroup P*
*which is quaternion-free, then P ∩ Z*(

*G*)

*∩ GN *= 1

*.*
**Lemma 2.6.**
*Let G be a group and let p be a prime number dividing |G| with*
(1)

*If N is normal in G of order p, then N lies in Z*(

*G*);

(2)

*If G has cyclic Sylow p-subgroups, then G is p-nilpotent*;

(3)

*If M is a subgroup of G of index p, then M is normal in G.*

**Proof. **(1) Since

*|*Aut(

*N *)

*| *=

*p − *1 and

*G/CG*(

*N *) is isomorphic to a subgroup of

Aut(

*N *)

*, |G/CG*(

*N *)

*| *must divide (

*|G|, p − *1) = 1

*. *It follows that

*G *=

*CG*(

*N *) and

*N ≤ Z*(

*G*)

*.*
(2) Let

*P ∈ *Syl

*p*(

*G*) and

*|P | *=

*pn. *Since

*P *is cyclic,

*|*Aut(

*P *)

*| *=

*pn−*1(

*p − *1)

*.*
Again,

*NG*(

*P *)

*/CG*(

*P *) is isomorphic to a subgroup of Aut(

*P *)

*, *so

*|NG*(

*P *)

*/CG*(

*P *)

*| *mustdivide (

*|G|, p − *1) = 1

*. *Thus

*NG*(

*P *) =

*CG*(

*P *)

*, *and statement (2) follows by the well-known Burnside theorem.

(3) We may assume that

*MG *= 1 by induction. As everyone knows the result is true
in the case where

*p *= 2

*. *So assume that

*p > *2 and consequently

*G *is of odd order as(

*|G|, p − *1) = 1

*. *Now we know that

*G *is solvable by the Odd Order Theorem. Let

*N*be a minimal normal subgroup of

*G. *Then

*N *is an elementary abelian

*q*-group for someprime

*q. *It is obvious that

*G *=

*M N *and

*M ∩ N *is normal in

*G. *Therefore

*M ∩ N *= 1

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
and

*|N | *=

*|G *:

*M | *=

*p. *Now

*N ≤ Z*(

*G*) by statement (1) and, of course,

*M *is normalin

*G *as desired.

**3. Proofs of theorems.**

*Proof of Theorem ***1.1. **Let

*G *be a minimal counterexample. Then we have the

(1)

*M *is

*p*-nilpotent whenever

*P ≤ M < G.*

Since

*NM *(

*P *)

*≤ NG*(

*P *)

*, NM *(

*P *) is

*p*-nilpotent. Let

*x *be an element of

*M \NM *(

*P *)

*.*
Then, since

*P ∩ P x ∩ M Np ≤ P ∩ P x ∩ GNp, *every minimal subgroup of

*P ∩ P x ∩ M Np*is

*c *-supplemented in

*P *by Lemma 2.1. If

*G *satisfies (a), then every cyclic subgroup of

*P ∩ P x ∩ M Np *with order 4 is

*c *-supplemented in

*P. *If

*G *satisfies (b), then
[Ω2(

*P ∩ P x ∩ M Np*)

*, P *]

*≤ Z*(

*P ∩ GNp*)

*∩ *(

*P ∩ M Np*)

*≤ Z*(

*P ∩ M Np*)

*.*
Now we see that

*M *satisfies the hypotheses of the theorem. The minimality of

*G *impliesthat

*M *is

*p*-nilpotent.

(2)

*Op *(

*G*) = 1

*.*

If not, we consider

*G *=

*G/N, *where

*N *=

*Op *(

*G*)

*. *Clearly

*N *(

*P *) =

*N*
is

*p*-nilpotent, where

*P *=

*P N/N. *For any

*xN ∈ G\N *(

*P *)

*, *since

*G p *=

*GNpN/N*
and

*P ∩ P xN *=

*P xn *for some

*n ∈ N, *we have

*∩ G p *= (

*P ∩ P xn ∩ GNpN*)

*N/N *= (

*P ∩ P xn ∩ GNp*)

*N/N.*
Because

*xN ∈ G\N *(

*P *)

*, xn ∈ G\N*
*G*(

*P *)

*. *Now let

*P *0 =

*P*0

*N/N *be a minimal

*∩ G p. *We may assume that

*P*0 =

*y , *where

*y *is an element of

*P ∩ P xn ∩ GNp *of order

*p. *By the hypotheses, there exists a subgroup

*K*0 of

*P *suchthat

*P *=

*P*0

*K*0 and

*P*0

*∩ K*0 is a permutable subgroup of

*P. *It follows that

*P N/N *== (

*P*0

*N/N *)(

*K*0

*N/N *) and (

*P*0

*N/N *)

*∩ *(

*K*0

*N/N *) = (

*P*0

*∩ K*0

*N *)

*N/N. *If

*P*0

*∩ K*0

*N *==

*P*0 then

*P*0

*≤ P ∩ K*0

*N *=

*K*0 and consequently

*P*0 =

*P*0

*∩ K*0 is permutable in

*P. *In this case,

*P *0 is permutable in

*P . *If

*P*0

*∩ K*0

*N *= 1 then

*P *0 is complemented in

*P . *Thus

*P *0 is

*c *-supplemented in

*P . *Assume that

*G *satisfies (a). Let

*P *1 =

*P*1

*N/N*
*∩ G p *of order 4

*. *We may assume that

*P*1 =

*z ,*
where

*z *is an element of

*P ∩ P xn ∩ GNp *of order 4

*. *Since

*P*1 is

*c *-supplemented in

*P,P *=

*P*1

*K*1 and

*P*1

*∩ K*1 is permutable in

*P. *We have

*P N/N *= (

*P*1

*N/N *)(

*K*1

*N/N *) and(

*P*1

*N/N *)

*∩ *(

*K*1

*N/N *) = (

*P*1

*∩ K*1

*N *)

*N/N. *If

*P*1

*∩ K*1

*N *= 1 then

*P *1 is complementedin

*P . *If

*P*1

*∩ K*1

*N *=

*z*2

*, *since

*z*2

*≤ *Φ(

*P *) and

*z*2 is

*c *-supplemented in

*P, z*2is permutable in

*P *by Lemma 2.1. Furthermore,

*z*2

*N/N *is permutable in

*P N/N *and

*P *1 is

*c *-supplemented in

*P . *If

*P*1

*∩ K*1

*N *=

*P*1 then

*P*1 =

*P*1

*∩ K*1 is permutable in

*P*and

*P *1 is permutable in

*P . *In a ward,

*P *1 is

*c *-supplemented in

*P . *Now assume that

*G*satisfies (b), then

*∩ G *)

*, P *= Ω2(

*P ∩ P xn ∩ GNp*)

*, P N/N ≤ Z*(

*P ∩ GNp*)

*N/N,*
*∩ G *)

*, P ≤ Z*(

*P ∩ G *)

*.*
If

*G *satisfies (c) then

*P ∼*
=

*P *is quaternion-free. Therefore

*G *=

*G/N *satisfies the
hypotheses of the theorem. The choice of

*G *implies that

*G *is

*p*-nilpotent and so is

*G,*a contradiction.

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
*c *-SUPPLEMENTED SUBGROUPS AND

*p*-NILPOTENCY OF FINITE GROUPS
(3)

*G/Op*(

*G*) is

*p*-nilpotent and

*CG*(

*Op*(

*G*))

*≤ Op*(

*G*)

*.*

Suppose that

*G/Op*(

*G*) is not

*p*-nilpotent. Then, by Frobenius’ theorem (refer [12],
Theorem 10.3.2), there exists a subgroup of

*P *properly containing

*Op*(

*G*) such that its

*G*-normalizer is not

*p*-nilpotent. Since

*NG*(

*P *) is

*p*-nilpotent, we may choice a subgroup

*P*1 of

*P *such that

*Op*(

*G*)

*< P*1

*< P *and

*NG*(

*P*1) is not

*p*-nilpotent but

*NG*(

*P*2) is

*p*-nilpotent whenever

*P*1

*< P*2

*≤ P. *Denote

*H *=

*NG*(

*P*1)

*. *It is obvious that

*P*1

*< P*0

*≤ P*for some Sylow

*p*-subgroup

*P*0 of

*H. *The choice of

*P*1 implies that

*NG*(

*P*0) is

*p*-nilpotent,hence

*NH *(

*P*0) is also

*p*-nilpotent. Take

*x ∈ H\NH*(

*P*0)

*. *Since

*P*0 =

*P ∩ H, *we have

*x ∈ G\NG*(

*P *)

*. *Again,
so every minimal subgroup of

*P*0

*∩P x ∩*
*HNp *is

*c *-supplemented in

*P*0 by Lemma 2.1. If
(a) is satisfied then every cyclic subgroup of

*P*0

*∩P x ∩*
*HNp *of order 4 is

*c *-supplemented
in

*P*0

*. *If (b) is satisfied then

*HNp *)

*, P*0

*≤ Z*(

*P ∩ GNp*)

*∩ *(

*P*0

*∩ HNp*)

*≤ Z*(

*P*0

*∩ HNp*)

*.*
If (c) is satisfied then

*P*0 is quaternion-free. Therefore

*H *satisfies the hypotheses of thetheorem. The choice of

*G *yields that

*H *is

*p*-nilpotent, which is contrary to the choiceof

*P*1

*. *Thereby

*G/Op*(

*G*) is

*p*-nilpotent and

*G *is

*p*-solvable with

*Op *(

*G*) = 1

*. *Conse-quently, we obtain

*CG*(

*Op*(

*G*))

*≤ Op*(

*G*) (refer [13], Theorem 6.3.2).

(4)

*G *=

*P Q, *where

*Q *is an elementary abelian Sylow

*q*-subgroup of

*G *for a
prime

*q *=

*p. *Moreover,

*P *is maximal in

*G *and

*QOp*(

*G*)

*/Op*(

*G*) is minimal normalin

*G/Op*(

*G*)

*.*
For any prime divisor

*q *of

*|G| *with

*q *=

*p, *since

*G *is

*p*-solvable, there exists a Sylow

*q*-subgroup

*Q *of

*G *such that

*G*0 =

*P Q *is a subgroup of

*G *[13] (Theorem 6.3.5). If

*G*0

*< G, *then, by (1),

*G*0 is

*p*-nilpotent. This leads to

*Q ≤ CG*(

*Op*(

*G*))

*≤ Op*(

*G*)

*, *acontradiction. Thus

*G *=

*P Q *and so

*G *is solvable. Now let

*T /Op*(

*G*) be a minimalnormal subgroup of

*G/Op*(

*G*) contained in

*Opp *(

*G*)

*/Op*(

*G*)

*. *Then

*T *=

*Op*(

*G*)(

*T ∩ Q*)

*.*

If

*T ∩ Q < Q, *then

*P T < G *and therefore

*P T *is

*p*-nilpotent by (1). It follows that
1

*< T ∩ Q ≤ CG*(

*Op*(

*G*))

*≤ Op*(

*G*)

*,*
which is impossible. Hence

*T *=

*Opp *(

*G*) and

*QOp*(

*G*)

*/Op*(

*G*) is an elementary abelian

*q*-group complementing

*P/Op*(

*G*)

*. *This yields that

*P *is maximal in

*G.*
(5)

*|P *:

*Op*(

*G*)

*| *=

*p.*

Clearly,

*Op*(

*G*)

*< P. *Let

*P*0 be a maximal subgroup of

*P *containing

*Op*(

*G*) and
let

*G*0 =

*P*0

*Opp *(

*G*)

*. *Then

*P*0 is a Sylow

*p*-subgroup of

*G*0

*. *The maximality of

*P *in

*G*implies that either

*NG*(

*P*0) =

*G *or

*NG*(

*P*0) =

*P. *If the latter holds, then

*NG *(

*P*
On the other hand, in view of (3), we have

*GNp ≤ Op*(

*G*)

*, *hence

*P ∩ P x ∩ GNp *==

*GNp *for every

*x ∈ G. *Now it is easy to see that

*G*0 satisfies the hypotheses of thetheorem. Thereby

*G*0 is

*p*-nilpotent and

*Q ≤ CG*(

*Op*(

*G*))

*≤ Op*(

*G*)

*, *a contradiction.

Thus

*NG*(

*P*0) =

*G *and

*P*0 =

*Op*(

*G*)

*. *This proves (5).

(6)

*G *=

*GNp L, *where

*L *=

*a *[

*Q*] is a non-abelian split extension of

*Q *by a cyclic

*p*-subgroup

*a , ap ∈ Z*(

*L*) and the action of

*a *(by conjugate) on

*Q *is irreducible.

From (3) we see that

*GNp ≤ Op*(

*G*)

*. *Clearly,

*T *=

*GNpQ ✁ G. *Let

*P*0 be a maximal
subgroup of

*P *containing

*GNp . *Then, by the maximality of

*P, *either

*NG*(

*P*0) =

*P*or

*NG*(

*P*0) =

*G. *If

*NG*(

*P*0) =

*P, *then

*NM *(

*P*0) =

*P*0

*, *where

*M *=

*P*0

*T *=

*P*0

*Q.*
*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
*M Np ≤ GNp *for all

*x ∈ M \NM *(

*P*0)

*, *hence

*M *satisfies the
hypotheses of the theorem. By the minimality of

*G, M *is

*p*-nilpotent. It follows that

*T *=

*GNp Q *=

*GNp × Q *and so

*Q ✁ G, *a contradiction. Thereby

*NG*(

*P*0) =

*G *and

*P*0

*≤ Op*(

*G*)

*. *This infers from (5) that

*Op*(

*G*) =

*P*0 and hence

*P/GNp *is a cyclicgroup. Now applying the Frattini argument we have

*G *=

*GNp NG*(

*Q*)

*. *Therefore wemay assume that

*G *=

*GNp L, *where

*L *=

*a *[

*Q*] is a non-abelian split extension of anormal Sylow

*q*-subgroup

*Q *by a cyclic

*p*-group

*a . *Noticing that

*|P *:

*Op*(

*G*)

*| *=

*p*and

*Op*(

*G*)

*∩ NG*(

*Q*)

*✁ NG*(

*Q*)

*, *we have

*ap ∈ Z*(

*L*)

*. *Also since

*P *is maximal in

*G,GNp Q/GNp *is minimal normal in

*G/GNp *and consequently

*a *acts irreducibly on

*Q.*
(7)

*GNp *has exponent

*p *if

*p > *2 and exponent at most 4 if

*p *= 2

*.*

By Lemma 2.3 it will suffice to show that there exists a

*p*-nilpotent maximal subgroup

*M *of

*G *such that

*G *=

*GNp M. *In fact, let

*M *be a maximal subgroup of

*G *containing

*L. *Then

*M *=

*L*(

*M ∩ GNp*) and

*G *=

*GNpM. *By Lemma 2.4,

*M ∩ GNp ✁ G, *hence

*M *= (

*a *(

*M ∩ GNp*))

*Q. *Write

*P*0 =

*a *(

*M ∩ GNp*) and let

*M*0 be a maximal subgroupof

*M *containing

*P*0

*. *Then

*M*0 =

*P*0(

*M*0

*∩ Q*) and

*GNpM*0

*< G. *By applying (1) weknow that

*GNp M*0 is

*p*-nilpotent, therefore

*M*0

*∩ Q ≤ CG*(

*Op*(

*G*))

*≤ Op*(

*G*)

*.*
It follows that

*M*0

*∩ Q *= 1 and so

*P*0 is maximal in

*M. *In this case, if

*P*0

*✁ M, *then

*a *=

*P*0

*∩ L ✁ L, *which is contrary to (6). Hence

*NM *(

*P*0) =

*P*0 and

*M *satisfies the
hypotheses of the theorem. The choice of

*G *implies that

*M *is

*p*-nilpotent, as desired.

Without losing generality, we assume in the following that

*P *=

*GNp a .*

(8) If

*GNp *has exponent

*p, *then

*GNp ∩ a *= 1

*.*

Assume on the contrary that

*GNp ∩ a *= 1 if

*GNp *has exponent

*p. *Then we can take
an element

*c *in

*GNp ∩ a *such that

*c *is of order

*p. *Since

*P *is not normal in

*G, GNp∩ a << a . *Consequently

*c ∈ ap ≤ *Φ(

*P *) and

*c *is permutable in

*P. *By (6), (7) andLemma 2.2, we see that

*c *is centralized by both

*GNp *and

*L, *hence

*c ∈ Z*(

*G*)

*. *If

*G *satisfies(c) then, since

*GNp ≤ GN , c *= 1 by Lemma 2.5, a contradiction. If

*G *satisfies (a) or(b), we consider the factor group

*G *=

*G/ c . *It is obvious that

*N *(

*P *) =

*N*
*p*-nilpotent, where

*P *=

*P/ c . *Now let

*y c / c *be a minimal subgroup of

*GNp / c ,*where

*y ∈ GNp. *Since

*y *is of order

*p, *by the hypotheses,

*y *has a

*c *-supplement

*K*in

*P. *If

*y ∩ K *= 1 then

*K *is a maximal subgroup of

*P *and

*c ≤ K. *It follows that

*P/ c *= (

*y c / c *)(

*K/ c *) with

*y c / c ∩ K/ c *= 1

*. *If

*y ∩ K *=

*y *then

*y*is permutable in

*P *and hence

*y c / c *is permutable in

*P/ c . *That is

*y c / c *is

*c *-supplemented in

*P/ c , *therefore

*G *satisfies (a) or (b). The choice of

*G *implies that

*G/ c *is

*p*-nilpotent and so

*G *is

*p*-nilpotent, a contradiction.

(9) The exponent of

*GNp *is not

*p.*

If not,

*GNp *has exponent

*p. *Let

*P*1 be a minimal subgroup of

*GNp *not permutable
in

*P. *Then, by the hypotheses, there is a subgroup

*K*1 of

*P *such that

*P *=

*P*1

*K*1 and

*P*1

*∩ K*1 = 1

*. *In general, we may find minimal subgroups

*P*1

*, P*2

*, . . . , Pm *of

*GNp *andalso subgroups

*K*1

*, K*2

*, . . . , Km *of

*P *such that

*P *=

*PiKi *and

*Pi ∩ Ki *= 1 for each

*i*and every minimal subgroup of

*GNp ∩K*1

*∩. . .∩Km *is permutable in

*P. *Furthermore, wemay assume that

*Pi ≤ K*1

*∩ . . . ∩ Ki−*1

*, i *= 2

*, *3

*, . . . , m . *Henceforth

*K*1

*∩ . . . ∩ Ki−*1 ==

*Pi*(

*K*1

*∩. . .∩Ki*) for

*i *= 2

*, *3

*, . . . , m . *It is easy to see that

*GNp ∩Ki *is normal in

*P *and(

*GNp ∩ Ki*)

*a *is a complement of

*Pi *in

*P, *so we may replace

*Ki *by (

*GNp ∩ Ki*)

*a *andfurther assume that

*a ≤ Ki *for each

*i. *Now,

*K*1

*∩. . .∩Km *= (

*GNp ∩K*1

*∩. . .∩Km*)

*a .*

Since, for any

*x ∈ GNp ∩ K*1

*∩ . . . ∩ Km, x a *=

*a x , *we have

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
*c *-SUPPLEMENTED SUBGROUPS AND

*p*-NILPOTENCY OF FINITE GROUPS

*xa ∈ *(

*GNp ∩ K*1

*∩ . . . ∩ Km*)

*∩ x a *=

*x .*
This means that

*a *induces a power automorphism of

*p*-power order in the elementaryabelian

*p*-group

*GNp ∩ K*1

*∩ . . . ∩ Km. *Hence [

*GNp ∩ K*1

*∩ . . . ∩ Km, a*] = 1 and

*K*1

*∩ . . . ∩ Km *is abelian.

Now we claim that

*p *is even. If it is not the case, then, by [13] (Theorem 6.5.2),

*K*1

*∩ . . . ∩ Km ≤ Op*(

*G*)

*. *Consequently,

*P *=

*GNp*(

*K*1

*∩ . . . ∩ Km*)

*≤ Op*(

*G*)

*, *a con-tradiction. We proceed now to consider the following two cases:

**Case 1. ***| a | *= 2

*n, n > *1

*.*

Since

*K*1

*∩ . . . ∩ Km *is an abelian normal subgroup of

*P *and

*a ∈ K*1

*∩ . . . ∩ Km,*
Φ(

*K*1

*∩ . . . ∩ Km*) =

*a*2

*P *and so Ω1(

*a*2 ) =

*c ≤ Z*(

*P *)

*, *where

*c *=

*a*2

*n−*1

*. *Again,

*c ∈ Z*(

*L*) by (6), so

*c ∈ Z*(

*G*)

*. *If

*G *satisfies (c) then we obtain

*c *= 1 by Lemma 2.5,which is absurd. If

*G *satisfies (a) or (b), then, with the same arguments to those used in(8), we may prove that

*G/ c *satisfies the hypotheses of the theorem. The minimality of

*G *implies that

*G/ c *is 2-nilpotent and therefore

*G *is also 2-nilpotent, a contradiction.

**Case 2. ***| a | *= 2

*.*

Since

*a *acts irreducibly on

*Q, a *is an involutive automorphism of

*Q*; consequently,

*Q *is a cyclic subgroup of order

*q *and

*ba *=

*b−*1

*, *where

*Q *=

*b . *In this case,

*GN*2is minimal normal in

*G. *In fact, let

*N *be a minimal normal subgroup of

*G *containedin

*GN*2 and let

*H *=

*N L. *Since

*N a *is maximal but not normal in

*H, *we see that

*NH *(

*N a *) =

*N a . *Noticing that

*N a ∩HN*2

*≤ N, *every minimal subgroup of

*N a ∩*
*∩ HN*2 is

*c *-supplemented in

*NH*(

*N a *) =

*N a *by Lemma 2.1. If further

*H < G,*then the choice of

*G *implies that

*H *is 2-nilpotent. Consequently,

*N Q *=

*N × Q *andso 1 =

*N ∩ Z*(

*P *)

*≤ Z*(

*G*)

*. *The choice of

*N *implies that

*N *=

*N ∩ Z*(

*P *) is of order2

*. *This is contrary to Lemma 2.5 if

*G *satisfies (c). Now assume that

*G *satisfies (a) or(b). In this case, if

*N ≤ *Φ(

*P *)

*, *then

*N *has a complement to

*P. *By applying Gasch¨utzTheorem [12] (I, 17.4),

*N *also has a complement to

*G, *say

*M. *It follows that

*M *is anormal subgroup of

*G. *Furthermore,

*G/M *is cyclic of order 2 and so

*N ≤ GN*2

*≤ M,*a contradiction. Hence

*N ≤ *Φ(

*P *)

*. *Now we go to consider the factor group

*G/N. *Forany minimal subgroup

*y N/N *of (

*G/N *)

*N*2 =

*GN*2

*/N, *by the hypotheses,

*P *=

*y K*and

*y ∩ K *is permutable in

*P, *where

*y ∈ GN*2

*. *Since

*N ≤ K, *we have

*P/N *== (

*y N/N *)(

*K/N *) and (

*y N/N *)

*∩ *(

*K/N *) = (

*y ∩ K*)

*N/N *is permutable in

*P/N,*so

*y N/N *is

*c *-supplemented in

*P/N. *This yields at once that

*G/N *is 2-nilpotent andso is

*G. *Hence

*H *=

*G *and

*GN*2 must be a minimal normal subgroup of

*G*; of course,

*GN*2 is an elementary abelian 2-group. Since

*GN*2

*∩ NG*(

*Q*)

*✁ NG*(

*Q*)

*, *we know that

*GN*2

*∩ NG*(

*Q*) = 1 and so

*b *acts fixed-point-freely on

*GN*2

*. *We may assume that

*N*1 ==

*{*1

*, c*1

*, c*2

*, . . . , cq} *is a subgroup of

*GN*2 with

*c*1

*∈ Z*(

*P *) and

*b *= (

*c*1

*, c*2

*, . . . , cq*) is apermutation of the set

*{c*1

*, c*2

*, . . . , cq}. *Noticing that

*ba *=

*b−*1 and (

*c*1)

*a−*1

*ba *= (

*c*1)

*b−*1

*,*(

*c*2)

*a *=

*cq. *By using (

*bi*)

*a *=

*b−i *and (

*c*1)

*a−*1

*bia *= (

*c*1)

*b−i, *we see that (

*ci*+1)

*a *==

*cq−i*+1 for

*i *= 1

*, *2

*, . . . , *(

*q *+ 1)

*/*2

*. *Hence

*N*1 is normalized by both

*GN*2 and

*L *andso

*N*1 is normal in

*G. *The minimal normality of

*GN*2 implies that

*GN*2 =

*N*1

*, *thus wehave

*Z*(

*P *) =

*{*1

*, c*1

*}. *Since

*GN*2

*∩ K*1

*∩ . . . ∩ Km *is centralized by both

*GN*2 and

*a , *we have 1

*< GN*2

*∩ K*1

*∩ . . . ∩ Km ≤ Z*(

*P *)

*. *In view of

*P *is not abelian, we get
Φ(

*P *) =

*P *=

*Z*(

*P *)

*, *namely

*P *is an extra-special 2-group. By applying Theorem 5.3.8of [12], there exists some positive integer

*h *such that

*|P | *= 22

*h*+1

*. *Hence

*|GN*2

*| *= 22

*h.*

However, 22

*h − *1 = (2

*h *+ 1)(2

*h − *1) and

*q *= 22

*h − *1

*, *hence

*h *= 1

*, q *= 3 and

*|P | *= 23

*.*

Now we see that

*L ∼*
=

*A*4

*, *therefore

*G ∼*
=

*S*4

*, *which is contrary to the

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
(10) The final contradiction.

From (7) and (9) we see that

*p *= 2 and the exponent of

*GN*2 is 4

*. *By applying
Lemma 2.3,

*Z*(

*GN*2 ) = Φ(

*GN*2 ) is an elementary abelian 2-group. If Φ(

*GN*2 )

*∩ a *= 1then there exists an element

*c *in Φ(

*GN*2 )

*∩ a *such that

*c *is of order 2

*. *Since Φ(

*GN*2)

*∩*
*a < a , *we have

*c ∈ a*2

*≤ Z*(

*L*)

*. *But

*c *is also centralized by

*GN*2 by Lemma 2.2,
so

*c ∈ Z*(

*G*)

*. *If Φ(

*GN*2)

*∩ a *= 1 then

*a *induces a power automorphism of 2-powerorder in the elementary abelian 2-group Φ(

*GN*2 )

*, *hence [Φ(

*GN*2 )

*, a*] = 1

*. *In view ofLemma 2.2, Φ(

*GN*2 ) is also centralized by

*GN*2

*, *hence Φ(

*GN*2 )

*≤ Z*(

*P *)

*. *Furthermore,by the Frattini argument,

*G *=

*NG*(Φ(

*GN*2)) =

*CG*(Φ(

*GN*2))

*NG*(

*P *)

*.*
Noticing that

*NG*(

*P *) =

*P *and

*P ≤ CG*(Φ(

*GN*2))

*, *we get

*CG*(Φ(

*GN*2)) =

*G, *namelyΦ(

*GN*2 )

*≤ Z*(

*G*)

*. *Thus we can also take an element

*c *in Φ(

*GN*2) such that

*c *is of order2 and

*c ∈ Z*(

*G*)

*. *This is contrary to Lemma 2.5 if

*G *satisfies (c). Now assume that

*G *satisfies (a). Denote

*N *=

*c *and consider

*G *=

*G/N. *It is clear that

*N *(

*P *) =
=

*NG*(

*P *)

*/N *is 2-nilpotent because

*NG*(

*P *) is, where

*P *=

*P/N. *For any

*y ∈ GN*2

*,*since

*y *is

*c *-supplemented in

*P, *there exists a subgroup

*T *of

*P *such that

*P *=

*y T*and

*y ∩ T *is permutable in

*P. *However,

*y*2

*∈ *Φ(

*GN*2)

*, *hence

*y*2 is permutable in

*P *and

*y*2

*T *forms a group. Because

*|P *:

*y*2

*T | ≤ *2

*, N ≤ y*2

*T. *It follows that

*P/N *= (

*y N/N *)(

*y*2

*T /N *) and

*y N/N ∩ y*2

*T /N *=

*y*2 (

*y ∩ T *)

*N/N*
is permutable in

*P/N. *This shows that

*G *satisfies (a). Thereby

*G *is 2-nilpotent andso is

*G, *a contradiction. Finally we assume that

*G *satisfies (b). Let

*M *be a max-imal subgroup of

*G *containing

*L. *Then

*M *is 2-nilpotent by the proof of (7), henceΦ(

*GN*2 )

*Q *is 2-nilpotent and [Φ(

*GN*2 )

*, Q*] = 1

*. *Write

*K *=

*CG*(

*GN*2

*/*Φ(

*GN*2))

*. *Then,by the hypotheses,

*P ≤ K ✁ G. *The maximality of

*P *yields that

*P *=

*K *or

*K *=

*G.*

If the former holds, then

*G *=

*NG*(

*P *) is 2-nilpotent, a contradiction. If the latterholds, then [

*GN*2

*, Q*]

*≤ *Φ(

*GN*2)

*. *This means that

*Q *stabilizes the chain of subgroups1

*≤ *Φ(

*GN*2)

*≤ GN*2

*. *It follows from [13] (Theorem 5.3.2) that [

*GN*2

*, Q*] = 1 and

*Q *isnormal in

*G, *which is impossible. This completes our proof.

**Proof of Theorem ****1.3. **By applying Theorem 1.1, we only need to prove that

*NG*(

*P *)

If

*NG*(

*P *) is not

*p*-nilpotent, then

*NG*(

*P *) has a minimal non-

*p*-nilpotent subgroup
(that is, every proper subgroup of a group is

*p*-nilpotent but itself is not

*p*-nilpotent)

*H. *Byresults of Itˆo [2] (IV, 5.4) and Schmidt [2] (III, 5.2),

*H *has a normal Sylow

*p*-subgroup

*Hp *and a cyclic Sylow

*q*-subgroup

*Hq *such that

*H *= [

*Hp*]

*Hq. *Moreover,

*Hp *is ofexponent

*p *if

*p > *2 and of exponent at most 4 if

*p *= 2

*. *On the other hand, the minimalityof

*H *implies that

*HNp *=

*Hp. *Let

*P*0 be a minimal subgroup of

*Hp *and let

*K*0 bea

*c *-supplement of

*P*0 in

*H. *Then

*H *=

*P*0

*K*0 and

*P*0

*∩ K*0 is permutable in

*H. *If

*P*0

*∩ K*0 = 1 then

*K*0 is maximal in

*H *of index

*p. *By applying Lemma 2.6 we see that

*K*0 is normal in

*H. *It follows from

*K*0 is nilpotent that

*Hq *is normal in

*H, *a contradiction.

If

*P*0

*∩ K*0 =

*P*0 then

*P*0 is permutable in

*H. *In this case, if

*P*0

*Hq *=

*H, *then

*Hp *=

*P*0is cyclic and

*H *is

*p*-nilpotent by Lemma 2.6, a contradiction. Hence

*P*0

*Hq < H *and

*P*0

*Hq *=

*P*0

*× Hq. *Thus Ω1(

*Hp*) is centralized by

*Hq. *If further

*CH*(Ω1(

*Hp*))

*< H *then

*CH *(Ω1(

*Hp*)) is nilpotent normal in

*H. *This leads to

*Hq ✁ H, *a contradiction. ThereforeΩ1(

*Hp*)

*≤ Z*(

*H*)

*. *If

*Hp *has exponent

*p, *then

*Hp *= Ω1(

*Hp*) and

*H *=

*Hp × Hq,*
*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
*c *-SUPPLEMENTED SUBGROUPS AND

*p*-NILPOTENCY OF FINITE GROUPS
again a contradiction. Thus

*p *= 2 and

*H*2 has exponent 4

*. *If

*G *satisfies (b) then

*H*2 isquaternion-free and, by Lemma 2.5,

*Hq *acts trivially on

*H*2

*, *thus

*Hq *is normal in

*H, *acontradiction. Now assume that

*G *satisfies (a). Let

*P*1 =

*x *be a cyclic subgroup of

*H*2of order 4

*. *Since

*P*1 is

*c *-supplemented in

*H, H *=

*P*1

*K*1 with

*P*1

*∩ K*1 is permutablein

*H. *If

*|H *:

*K*1

*| *= 4 then

*|H *:

*K*1

*x*2

*| *= 2

*, *hence

*K*1

*x*2

*✁ H *and so

*Hq ✁ H,*a contradiction. If

*|H *:

*K*1

*| *= 2 then

*K*1

*✁ H. *We still get a contradiction. Therefore

*K*1 =

*H *and

*P*1 is permutable in

*H. *Now Lemma 2.6 implies that

*P*1

*Hq *is 2-nilpotentand consequently

*Hq *is normalized by

*H*2

*. *This final contradiction completes our proof.

*Doerk H., Hawkes T. *Finite solvable groups. – Berlin; New York, 1992.

*Huppert B. *Endliche Gruppen I. – New York: Springer, 1967.

*Ballester-Bolinches A., Esteban-Romero R. *Sylow permutable subnormal subgroups of finite groups //

J. Algebra. – 2002. –

**251**. – P. 727 – 738.

*Guo X., Shum K. P. p*-Nilpotence of finite groups and minimal subgroups // Ibid. – 2003. –

**270**. – P. 459 –

470.

*Wang Y. *Finite groups with some subgroups of Sylow subgroups

*c*-supplemented // Ibid. – 2000. –

**224**. –

P. 467 – 478.

*Ballester-Bolinches A., Wang Y., Guo X. C*-supplemented subgroups of finite groups // Glasgow Math. J.

– 2000. –

**42**. – P. 383 – 389.

*Guo X., Shum K. P. *On

*p*-nilpotence and minimal subgroups of finite groups // Sci. China. Ser. A. – 2003.

–

**46**. – P. 176 – 186.

*Guo X., Shum K. P. *Permutability of minimal subgroups and

*p*-nilpotency of finite groups // Isr. J. Math.

– 2003. –

**136**. – P. 145 – 155.

*Asaad M., Ballester-Bolinches A., Pedraza-Aguilera M. C. *A note on minimal subgroups of finite groups

// Communs Algebra. – 1996. –

**24**. – P. 2771 – 2776.

*Wang Y., Wei H., Li Y. *A generalization of Kramer’s theorem and its applications // Bull. Austral. Math.

Soc. – 2002. –

**65**. – P. 467 – 475.

*Dornhoff L. M *-groups and 2-groups // Math. Z. – 1967. –

**100**. – P. 226 – 256.

*RobinsonD. J. S. *A course in the theory of groups. – New York: Springer, 1980.

*Gorenstein D. *Finite groups. – New York: Chelsea, 1980.

*ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8*
Source: http://www.imath.kiev.ua/~umzh/Archiv/2007/08/UMZh_2007_08_1011.pdf

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