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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

TREATING Based on Practice Guideline for the Treatment of Patients With Obsessive-Compulsive Disorder,originally published in July 2007. A guideline watch, summarizing significant developments in the scientific literature since publication of this guideline, may be available in the Psychiatric Practice section of the APA web site at www.psych.org. American Psychiatric Association Steer

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