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
S4-free is
p-nilpotent if
NG(
P ) is
p-nilpotent and, for all
x ∈ G\NG(
P ), every minimal subgroup of
P ∩ P x ∩ GNpis
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)
Pis 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 [
S4-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 S4
-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 S4
-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 S4
-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 order4
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
K1 of
G such that
G =
HK1 and
H ∩ K1
≤ HG = Core
G(
H) [5,6]; in this case, if we denote
K =
HGK1
, 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 =
A4
,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|a16 =
b4 = 1
, ba =
a3
b , then
b2(
aibj) == (
aibj)9+2((
−1)
j−1)
b2
. Hence
b2 is permutable in
G. However,
b2 is not
c-supple-mented in
G as
b2 is in Φ(
G) and not normal in
G.
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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 or4 (
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
Nbe 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 Npis
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 =
P0
N/N be a minimal
∩ G p. We may assume that
P0 =
y , where
y is an element of
P ∩ P xn ∩ GNp of order
p. By the hypotheses, there exists a subgroup
K0 of
P suchthat
P =
P0
K0 and
P0
∩ K0 is a permutable subgroup of
P. It follows that
P N/N == (
P0
N/N )(
K0
N/N ) and (
P0
N/N )
∩ (
K0
N/N ) = (
P0
∩ K0
N )
N/N. If
P0
∩ K0
N ==
P0 then
P0
≤ P ∩ K0
N =
K0 and consequently
P0 =
P0
∩ K0 is permutable in
P. In this case,
P 0 is permutable in
P . If
P0
∩ K0
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 =
P1
N/N
∩ G p of order 4
. We may assume that
P1 =
z ,
where
z is an element of
P ∩ P xn ∩ GNp of order 4
. Since
P1 is
c -supplemented in
P,P =
P1
K1 and
P1
∩ K1 is permutable in
P. We have
P N/N = (
P1
N/N )(
K1
N/N ) and(
P1
N/N )
∩ (
K1
N/N ) = (
P1
∩ K1
N )
N/N. If
P1
∩ K1
N = 1 then
P 1 is complementedin
P . If
P1
∩ K1
N =
z2
, since
z2
≤ Φ(
P ) and
z2 is
c -supplemented in
P, z2is permutable in
P by Lemma 2.1. Furthermore,
z2
N/N is permutable in
P N/N and
P 1 is
c -supplemented in
P . If
P1
∩ K1
N =
P1 then
P1 =
P1
∩ K1 is permutable in
Pand
P 1 is permutable in
P . In a ward,
P 1 is
c -supplemented in
P . Now assume that
Gsatisfies (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.
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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
P1 of
P such that
Op(
G)
< P1
< P and
NG(
P1) is not
p-nilpotent but
NG(
P2) is
p-nilpotent whenever
P1
< P2
≤ P. Denote
H =
NG(
P1)
. It is obvious that
P1
< P0
≤ Pfor some Sylow
p-subgroup
P0 of
H. The choice of
P1 implies that
NG(
P0) is
p-nilpotent,hence
NH (
P0) is also
p-nilpotent. Take
x ∈ H\NH(
P0)
. Since
P0 =
P ∩ H, we have
x ∈ G\NG(
P )
. Again,
so every minimal subgroup of
P0
∩P x ∩
HNp is
c -supplemented in
P0 by Lemma 2.1. If
(a) is satisfied then every cyclic subgroup of
P0
∩P x ∩
HNp of order 4 is
c -supplemented
in
P0
. If (b) is satisfied then
HNp )
, P0
≤ Z(
P ∩ GNp)
∩ (
P0
∩ HNp)
≤ Z(
P0
∩ HNp)
.
If (c) is satisfied then
P0 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
P1
. 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
G0 =
P Q is a subgroup of
G [13] (Theorem 6.3.5). If
G0
< G, then, by (1),
G0 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
P0 be a maximal subgroup of
P containing
Op(
G) and
let
G0 =
P0
Opp (
G)
. Then
P0 is a Sylow
p-subgroup of
G0
. The maximality of
P in
Gimplies that either
NG(
P0) =
G or
NG(
P0) =
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
G0 satisfies the hypotheses of thetheorem. Thereby
G0 is
p-nilpotent and
Q ≤ CG(
Op(
G))
≤ Op(
G)
, a contradiction.
Thus
NG(
P0) =
G and
P0 =
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
P0 be a maximal
subgroup of
P containing
GNp . Then, by the maximality of
P, either
NG(
P0) =
Por
NG(
P0) =
G. If
NG(
P0) =
P, then
NM (
P0) =
P0
, where
M =
P0
T =
P0
Q.
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M Np ≤ GNp for all
x ∈ M \NM (
P0)
, 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(
P0) =
G and
P0
≤ Op(
G)
. This infers from (5) that
Op(
G) =
P0 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)
| =
pand
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
P0 =
a (
M ∩ GNp) and let
M0 be a maximal subgroupof
M containing
P0
. Then
M0 =
P0(
M0
∩ Q) and
GNpM0
< G. By applying (1) weknow that
GNp M0 is
p-nilpotent, therefore
M0
∩ Q ≤ CG(
Op(
G))
≤ Op(
G)
.
It follows that
M0
∩ Q = 1 and so
P0 is maximal in
M. In this case, if
P0
✁ M, then
a =
P0
∩ L ✁ L, which is contrary to (6). Hence
NM (
P0) =
P0 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
Kin
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
yis 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
P1 be a minimal subgroup of
GNp not permutable
in
P. Then, by the hypotheses, there is a subgroup
K1 of
P such that
P =
P1
K1 and
P1
∩ K1 = 1
. In general, we may find minimal subgroups
P1
, P2
, . . . , Pm of
GNp andalso subgroups
K1
, K2
, . . . , Km of
P such that
P =
PiKi and
Pi ∩ Ki = 1 for each
iand every minimal subgroup of
GNp ∩K1
∩. . .∩Km is permutable in
P. Furthermore, wemay assume that
Pi ≤ K1
∩ . . . ∩ Ki−1
, i = 2
, 3
, . . . , m . Henceforth
K1
∩ . . . ∩ Ki−1 ==
Pi(
K1
∩. . .∩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,
K1
∩. . .∩Km = (
GNp ∩K1
∩. . .∩Km)
a .
Since, for any
x ∈ GNp ∩ K1
∩ . . . ∩ 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 ∩ K1
∩ . . . ∩ Km)
∩ x a =
x .
This means that
a induces a power automorphism of
p-power order in the elementaryabelian
p-group
GNp ∩ K1
∩ . . . ∩ Km. Hence [
GNp ∩ K1
∩ . . . ∩ Km, a] = 1 and
K1
∩ . . . ∩ Km is abelian.
Now we claim that
p is even. If it is not the case, then, by [13] (Theorem 6.5.2),
K1
∩ . . . ∩ Km ≤ Op(
G)
. Consequently,
P =
GNp(
K1
∩ . . . ∩ Km)
≤ Op(
G)
, a con-tradiction. We proceed now to consider the following two cases:
Case 1. | a | = 2
n, n > 1
.
Since
K1
∩ . . . ∩ Km is an abelian normal subgroup of
P and
a ∈ K1
∩ . . . ∩ Km,
Φ(
K1
∩ . . . ∩ Km) =
a2
P and so Ω1(
a2 ) =
c ≤ Z(
P )
, where
c =
a2
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,
GN2is minimal normal in
G. In fact, let
N be a minimal normal subgroup of
G containedin
GN2 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 ∩HN2
≤ N, every minimal subgroup of
N a ∩
∩ HN2 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 ≤ GN2
≤ 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 )
N2 =
GN2
/N, by the hypotheses,
P =
y Kand
y ∩ K is permutable in
P, where
y ∈ GN2
. 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
GN2 must be a minimal normal subgroup of
G; of course,
GN2 is an elementary abelian 2-group. Since
GN2
∩ NG(
Q)
✁ NG(
Q)
, we know that
GN2
∩ NG(
Q) = 1 and so
b acts fixed-point-freely on
GN2
. We may assume that
N1 ==
{1
, c1
, c2
, . . . , cq} is a subgroup of
GN2 with
c1
∈ Z(
P ) and
b = (
c1
, c2
, . . . , cq) is apermutation of the set
{c1
, c2
, . . . , cq}. Noticing that
ba =
b−1 and (
c1)
a−1
ba = (
c1)
b−1
,(
c2)
a =
cq. By using (
bi)
a =
b−i and (
c1)
a−1
bia = (
c1)
b−i, we see that (
ci+1)
a ==
cq−i+1 for
i = 1
, 2
, . . . , (
q + 1)
/2
. Hence
N1 is normalized by both
GN2 and
L andso
N1 is normal in
G. The minimal normality of
GN2 implies that
GN2 =
N1
, thus wehave
Z(
P ) =
{1
, c1
}. Since
GN2
∩ K1
∩ . . . ∩ Km is centralized by both
GN2 and
a , we have 1
< GN2
∩ K1
∩ . . . ∩ 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
|GN2
| = 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 ∼
=
A4
, therefore
G ∼
=
S4
, 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
GN2 is 4
. By applying
Lemma 2.3,
Z(
GN2 ) = Φ(
GN2 ) is an elementary abelian 2-group. If Φ(
GN2 )
∩ a = 1then there exists an element
c in Φ(
GN2 )
∩ a such that
c is of order 2
. Since Φ(
GN2)
∩
a < a , we have
c ∈ a2
≤ Z(
L)
. But
c is also centralized by
GN2 by Lemma 2.2,
so
c ∈ Z(
G)
. If Φ(
GN2)
∩ a = 1 then
a induces a power automorphism of 2-powerorder in the elementary abelian 2-group Φ(
GN2 )
, hence [Φ(
GN2 )
, a] = 1
. In view ofLemma 2.2, Φ(
GN2 ) is also centralized by
GN2
, hence Φ(
GN2 )
≤ Z(
P )
. Furthermore,by the Frattini argument,
G =
NG(Φ(
GN2)) =
CG(Φ(
GN2))
NG(
P )
.
Noticing that
NG(
P ) =
P and
P ≤ CG(Φ(
GN2))
, we get
CG(Φ(
GN2)) =
G, namelyΦ(
GN2 )
≤ Z(
G)
. Thus we can also take an element
c in Φ(
GN2) 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 ∈ GN2
,since
y is
c -supplemented in
P, there exists a subgroup
T of
P such that
P =
y Tand
y ∩ T is permutable in
P. However,
y2
∈ Φ(
GN2)
, hence
y2 is permutable in
P and
y2
T forms a group. Because
|P :
y2
T | ≤ 2
, N ≤ y2
T. It follows that
P/N = (
y N/N )(
y2
T /N ) and
y N/N ∩ y2
T /N =
y2 (
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(
GN2 )
Q is 2-nilpotent and [Φ(
GN2 )
, Q] = 1
. Write
K =
CG(
GN2
/Φ(
GN2))
. 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 [
GN2
, Q]
≤ Φ(
GN2)
. This means that
Q stabilizes the chain of subgroups1
≤ Φ(
GN2)
≤ GN2
. It follows from [13] (Theorem 5.3.2) that [
GN2
, 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
P0 be a minimal subgroup of
Hp and let
K0 bea
c -supplement of
P0 in
H. Then
H =
P0
K0 and
P0
∩ K0 is permutable in
H. If
P0
∩ K0 = 1 then
K0 is maximal in
H of index
p. By applying Lemma 2.6 we see that
K0 is normal in
H. It follows from
K0 is nilpotent that
Hq is normal in
H, a contradiction.
If
P0
∩ K0 =
P0 then
P0 is permutable in
H. In this case, if
P0
Hq =
H, then
Hp =
P0is cyclic and
H is
p-nilpotent by Lemma 2.6, a contradiction. Hence
P0
Hq < H and
P0
Hq =
P0
× 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
H2 has exponent 4
. If
G satisfies (b) then
H2 isquaternion-free and, by Lemma 2.5,
Hq acts trivially on
H2
, thus
Hq is normal in
H, acontradiction. Now assume that
G satisfies (a). Let
P1 =
x be a cyclic subgroup of
H2of order 4
. Since
P1 is
c -supplemented in
H, H =
P1
K1 with
P1
∩ K1 is permutablein
H. If
|H :
K1
| = 4 then
|H :
K1
x2
| = 2
, hence
K1
x2
✁ H and so
Hq ✁ H,a contradiction. If
|H :
K1
| = 2 then
K1
✁ H. We still get a contradiction. Therefore
K1 =
H and
P1 is permutable in
H. Now Lemma 2.6 implies that
P1
Hq is 2-nilpotentand consequently
Hq is normalized by
H2
. This final contradiction completes our proof.
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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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