On weakly S-embedded subgroups and weakly $τ$-embedded subgroups

0.2 MB·English

by  Xiaoyu Chen

τ
On weakly S-embedded subgroups and weakly -embedded
3
subgroups∗
1
0
2
Xiaoyu Chen, Wenbin Guo†
n
a Department of Mathematics, University of Science and Technology of China,
J
Hefei 230026,P. R. China
9
2 E-mail: [email protected], [email protected]

]
R
G
Abstract
.
h
t Let G be a finite group. A subgroup H of G is said to be weakly S-embedded in G if there
a
m existsKEGsuchthatHK isS-quasinormalinGandH∩K ≤HseG,whereHseG isthesubgroup
generated by all those subgroups of H which are S-quasinormally embedded in G. We say that
[
H is weakly τ-embedded in G if there exists K EG such that HK is S-quasinormal in G and

1 H ∩K ≤ HτG, where HτG is the subgroup generated by all those subgroups of H which are
v τ-quasinormal in G. In this paper, we study the properties of the weakly S-embedded subgroups
5
andthe weaklyτ-embedded subgroups,anduse them to determine the structure offinite groups.
6
8
6
.
1 1 Introduction
0
3
1 Throughout this paper, all groups mentioned are finite and G always denotes a finite group. All
:
v unexplained notation and terminology are standard, as in [6,11,20].
i
X Recall that a subgroup H of G is said to be S-quasinormal (S-permutable, or π-quasinormal) in
r G if H permutes with every Sylow subgroupof G. This concept was introduced by Kegel [13]in 1962
a
and has been investigated by many authors. Ballester-Bolinches and Pedraza-Aguilera [1] extended
the notion to S-quasinormally embedding in 1998: a subgroup H of G is said to be S-quasinormally
embedded in G if every Sylow subgroup of H is a Sylow subgroup of some S-quasinormal subgroup of
G. Also, some authors considered in another way. For example, Li et al. [16] introduced the notion
of SS-quasinormality in 2008: a subgroup H of G is said to be SS-quasinormal in G if there exists a
supplementB of H toGsuchthatH permuteswithevery Sylow subgroupofB. Chen[2]introduced
the notion of S-semipermutability in 1987: a subgroup H of G is said to be S-semipermutable in G
if H permutes with every Sylow p-subgroup of G such that (p,|H|) = 1. Besides, V. O. Lukyanenko
and A. N. Skiba [18] introduced the notion of τ-quasinormality in 2008: a subgroup H of G is said
∗ResearchissupportedbyaNNSFgrantofChina(grant#11071229) andResearchFundfortheDoctoralProgram
of Higher Education of China(Grant 20113402110036).
†Corresponding author.
Keywords: finitegroup, weakly S-embeddedsubgroups, weakly τ-embeddedsubgroups.
Mathematics Subject Classification (2000): 20D10, 20D15, 20D20, 20D25.
1
to be τ-quasinormal in G if H permutes with all Sylow q-subgroups Q of G such that (q,|H|) = 1
and (|H|,|QG|) 6= 1.
It is easy to see that every SS-quasinormal subgroup and every S-semipermutable subgroup of G
are both τ-quasinormal in G. In fact, it is clear that S-semipermutability implies τ-quasinormality
by definition. Now assume that H is SS-quasinormal in G. Then G has a subgroup B such that
G = HB and H permutes with every Sylow subgroup of B. Let P be a Sylow p-subgroup of G with
(p,|H|) = 1. Then there exists an element h ∈ H such that Ph ≤ B. Therefore HPh = PhH, and
so HP = PH. Hence H is S-semipermutable and thus τ-quasinormal in G.
In2009, Guo, ShumandA.N.Skiba[9]gave thedefinitionof S-embeddedsubgroups: asubgroup
H of G is said to be S-embedded in G if there exists a normal subgroup K of G such that HK is
S-quasinormal in G and H∩K ≤ H , where H is the subgroup generated by all those subgroups
sG sG
of H which are S-quasinormal in G.
As a continuation of the research of S-quasinormally embedding, weakly S-embedding was intro-
duced by Li et al. [14] in 2011 as follows.
Definition 1.1 A subgroup H of G is said to be weakly S-embedded in G if there exists a normal
subgroup K of G such that HK is S-quasinormal in G and H ∩ K ≤ H , where H is the
seG seG
subgroup generated by all those subgroups of H which are S-quasinormally embedded in G.
Now we introduce the following notion:
Definition 1.2 A subgroup H of G is said to be weakly τ-embedded in G if there exists a normal
subgroupK of GsuchthatHK is S-quasinormalin GandH∩K ≤ H , whereH is thesubgroup
τG τG
generated by all those subgroups of H which are τ-quasinormal in G.
Evidently every τ-quasinormal subgroup and every S-embedded subgroup of G are weakly τ-
embedded in G. Consequently, every S-quasinormal subgroup, every SS-quasinormal subgroup, and
every S-semipermutable subgroup of G are also weakly τ-embedded in G. However, the next two
examples show that the converse does not hold in general.
Example 1.3 Let G= S be the symmetric group of degree 4 and H = h(14)i. Take Q = h(123)i ∈
4
Syl (G). Then clearly QG = A , where A is the alternating group of degree 4. As HQ 6= QH, H
3 4 4
is not τ-quasinormal in G. But since G = HA and H ∩A = 1, H is weakly τ-embedded in G.
4 4
Example 1.4 Let G = A be the alternating group of degree 5 and H = A . Because of the
5 4
simplicity of A , the normal subgroups of G are only 1 and G. Let P be a Sylow 2-subgroup of G
5
containing h(15)i. As PH 6= HP, H is not S-quasinormal in G, and so H is not S-embedded in G.
However, since G = Hh(12345)i, H is τ-quasinormal in G, and thus it is weakly τ-embedded in G.
Moreover, the following examples show that weakly S-embedding and weakly τ-embedding are
independent of each other.
Example 1.5 Let G =A bethealternating group of degree5, H = h(123)i ∈ Syl (G). Sinceevery
5 3
Hallπ-subgroupisS-quasinormallyembeddedinG, H is weakly S-embeddedinG(takeK = G). On
the other hand, let Q = h(12345)i ∈ Syl (G), then QG = G. It is clear that HQ is not a subgroup
5
of G. Hence H is not τ-quasinormal in G. In view of that H is τ-quasinormal in G (see below
τG
Lemma 2.7), we have H is not weakly τ-embedded in G.
2
Example 1.6 Let A = A be the alternating group of degree 5 and B = Inn(A ) ∼= A . Put
5 5 5
G = A⋊B and H = A⋊h(12)(34)i. As π(G) = π(H), H is τ-quasinormal in G, and thus it is
weakly τ-embedded in G. On the other hand, since Z(A ) = 1, B 5 G. Clearly, the subnormal
5
subgroups of G are only 1, A and G. Suppose that H is weakly S-embedded in G. Then there exists
a normal subgroup K of G such that HK is S-quasinormal in G and H ∩K ≤ H . Since H is
seG
not S-quasinormal in G (see below Lemma 2.1(1)), we only can take K = G. This implies that
H = H . For any non-identity subgroup L of H which is S-quasinormally embedded in G and
seG
any non-identity Sylow subgroup D of L, there exists an S-quasinormal subgroup U of G such that
D ∈ Syl(U). In view of that π(L) ⊆ π(|G : L|) and Lemma 2.1(1), we have U = A, and thereby
L ≤ A. It follows that H = H ≤ A, a contradiction. Therefore H is not weakly S-embedded in
seG
G.
The purpose of this paper is to study the structure of finite groups by using the notion of weakly
S-embedding and weakly τ-embedding. In brief, we say a subgroup H of G satisfies (△) in G if H
is weakly S-embedded or weakly τ-embedded in G.
2 Preliminary Lemmas
In this paper, we use N, N and U to denote the classes of finite nilpotent, p-nilpotent and
p
supersolvable groups, respectively. For a non-empty class F of groups, the symbol Z (G) (usually,
F
Z (G) is written as Z (G)) denotes the F-hypercenter of G, that is, the product of all such normal
N ∞
subgroups L of G whose G-chief factors H/K satisfy that (H/K)⋊(G/C (H/K)) ∈ F. Also, the
G
symbol |G| denotes the order of a Sylow p-subgroup of G.
p
Lemma 2.1 [4,13]. Suppose that H is S-quasinormal in G, U ≤ G and N EG. Then:
(1) H is subnormal in G.
(2) H/H is nilpotent.
G
(3) H ∩U is S-quasinormal in U.
(4) HN/N is S-quasinormal in G/N.
(5) If H is a p-subgroup of G, then Op(G) ≤N (H).
G
Lemma 2.2 [21, Corollary 1]. Suppose that A and B are S-quasinormal in G, then hA,Bi and A∩B
are also S-quasinormal in G.
Lemma 2.3. Suppose that H is S-quasinormal in G, P is a Sylow p-subgroup of H, where p is a
prime divisor of |H|. If H = 1, then P is S-quasinormal in G.
G
Proof. This can be easily deduced from Lemma 2.1(2) and [21, Proposition B].
Lemma 2.4 [1, Lemma 1]. Suppose that H is S-quasinormally embedded in G, U ≤ G and N EG.
Then:
(1) If H ≤ U, then H is S-quasinormally embedded in U.
(2) HN is S-quasinormally embedded in G, and HN/N is S-quasinormally embedded in G/N.
The following lemma can be directly obtained from Lemma 2.4.
Lemma 2.5. Suppose that H ≤ U ≤ G and N EG. Then:
3
(1) H ≤ H .
seG seU
(2) H N/N ≤ (HN/N) .
seG se(G/N)
Lemma 2.6. Suppose that H is τ-quasinormal in G, U ≤ G and N EG.
(1) If H ≤ U, then H is τ-quasinormal in U.
(2) If π(HN/N) = π(H), then HN/N is τ-quasinormal in G/N.
(3) If (|H|,|N|) = 1, then HN/N is τ-quasinormal in G/N.
Proof. See [19, Lemma 2.2] for (1) and (3). Now we prove (2). Let Q/N ∈ Syl (G/N) such that
q
(|HN/N|,|(Q/N)(G/N)|) 6= 1 and q ∈/ π(HN/N) = π(H). Then for some Sylow q-subgroup G
q
of G, we have Q = G N. Since (Q/N)(G/N) = QG/N = (G N)G/N = GGN/N ∼= GG/GG ∩ N,
q q q q q
(|H|,|GG|) 6= 1. Hencebythehypothesis,(HN/N)(Q/N) = HG N/N =G HN/N = (Q/N)(HN/N).
q q q
From Lemma 2.6 we directly have:
Lemma 2.7. Suppose that H ≤ U ≤ G and N EG. Then:
(1) If H is a p-subgroup, then H is τ-quasinormal in G and H ≤ H [19, Lemma 2.3(1)].
τG G τG
(2) H ≤ H [19, Lemma 2.3(2)].
τG τU
(3) If H is a p-subgroup, then H N/N ≤ (HN/N) .
τG τ(G/N)
(4) If (|H|,|N|) = 1, then H N/N ≤ (HN/N) .
τG τ(G/N)
Lemma 2.8. Let P be a p-subgroup of G. Then the following statements are equivalent:
(1) P is S-quasinormal in G.
(2) P ≤ O (G) and P is S-quasinormally embedded in G [17, Lemma 2.4].
p
(3) P ≤ O (G) and P is τ-quasinormal in G [19, Lemma 2.2(4)].
p
Lemma 2.9 [14, Lemma 2.4]. Suppose that H is weakly S-embedded in G, U ≤ G and N EG.
(1) If H ≤ U, then H is weakly S-embedded in U.
(2) If N ≤ H, then H/N is weakly S-embedded in G/N.
(3) If (|H|,|N|) = 1, then HN/N is weakly S-embedded in G/N.
Now we give some basic properties of weakly τ-embedded subgroups.
Lemma 2.10. Suppose that H is weakly τ-embedded in G, U ≤ G and N EG.
(1) If H ≤ U, then H is weakly τ-embedded in U.
(2) If H is a p-subgroup and N ≤ H, then H/N is weakly τ-embedded in G/N.
(3) If (|H|,|N|) = 1, then HN/N is weakly τ-embedded in G/N.
Proof. By the hypothesis, there exists a normal subgroup K of G such that HK is S-quasinormal
in G and H ∩K ≤H . Then:
τG
(1) K ∩U EU and H(K ∩U) = HK ∩U is S-quasinormal in U by Lemma 2.1(3). In view of
Lemma 2.7(2), we have H∩(K∩U)= H∩K ≤ H ≤ H . Hence H is weakly τ-embedded in U.
τG τU
(2) KN/N EG/N and (H/N)(KN/N) = HKN/N is S-quasinormal in G/N by Lemma 2.1(4).
Since (H/N)∩(KN/N) = (H∩K)N/N, (H/N)∩(KN/N) ≤ H N/N ≤ (H/N) by Lemma
τG τ(G/N)
2.7(3). Therefore H/N is weakly τ-embedded in G/N.
(3) KN/NEG/N and(HN/N)(KN/N) = HKN/N is S-quasinormalinG/N by Lemma2.1(4).
It is easy to see that (|N∩HK : N∩H|,|N∩HK :N∩K|)= 1. Then N∩HK = (N∩H)(N∩K),
and so HN ∩KN = (H∩K)N by [5, Chapter A, Lemma (1.2)]. It follows from Lemma 2.7(4) that
4
(HN/N)∩(KN/N) = (H∩K)N/N ≤ H N/N ≤ (HN/N) . Consequently, HN/N is weakly
τG τ(G/N)
τ-embedded in G/N.
Lemma 2.11 [6, Lemma 3.4.7]. Let F be a saturated formation. If G 6∈ F, but every proper subgroup
of G belongs to F. Furthermore, G has a normal Sylow p-subgroup G 6= 1, where p is a prime divisor
p
of |G|. Then:
(1) G = GF.
p
(2) F(G) = F (G) = G Φ(G).
p p
Lemma 2.12. Let p be a prime divisor of |G| with (|G|,(p−1)(p2 −1)···(pn −1)) = 1. If H EG
with pn+1 6 ||H| for some integer n ≥ 1 and G/H is p-nilpotent, then G is p-nilpotent. In particular,
if pn+1 6 ||G|, then G is p-nilpotent.
Proof. Assume that the result is false and let (G,H) be a counterexample for which |G|+ |H| is
minimal. For every non-trivial subgroup F of G, (F,H ∩F) satisfies the hypothesis. Then F is p-
nilpotent dueto thechoice of (G,H). It inducesthat Gis a minimalnon-p-nilpotentgroup, andthus
a minimalnonnilpotentgroup by [11, ChapterIV, Theorem 5.4]. By [6, Theorem3.4.11] and Lemma
2.11, G = P ⋊Q with P = GN = GNp ∈ Syl (G). If N is a minimal normal subgroup of G, then
p
(G/N,HN/N) satisfies the hypothesis. Hence G/N is p-nilpotent. This implies that P = GNp = N
is an elementary abelian group. Clearly P ≤ H, and so |P| ≤ pn. Since N (P)/C (P) > Aut(P),
G G
(cid:12)
|N (P)/C (P)| (cid:12) (p − 1)(p2 − 1)···(pn − 1). Consequently, N (P) = C (P). So by Burnside’s
G G (cid:12) G G
Theorem, G is p-nilpotent, a contradiction.
Lemma 2.13 [11, Chapter VI, Lemma 4.10]. Let A and B be subgroups of G such that G 6= AB. If
AgB = BAg for all g ∈ G, then there exists a non-trivial normal subgroup N of G containing either
A or B.
Lemma 2.14. Let p be a prime divisor of |G| with (|G|,p −1) = 1. If G has a Hall p′-subgroup,
then any two Hall p′-subgroups of G are conjugate in G.
Proof. If p = 2, then by [3, Theorem A], any two Hall p′-subgroups are conjugate in G. If p is an odd
prime, then 26 ||G|. By Feit-Thompson’s Theorem, G is soluble. Hence any two Hall p′-subgroups
are conjugate in G.
The next lemma is evident.
Lemma 2.15. Let p be a prime divisor of |G|, H ≤ G and N EG. If |HN/N| ≥ pn+1 for some
p
integer n ≥ 1, then for every T/N ∈ Syl (HN/N), there exists a Sylow p-subgroup P of H such that
p
T/N = PN/N; for every n-maximal subgroup T /N of T/N, there exists an n-maximal subgroup P
n n
of P such that T /N = P N/N and P ∩N = P ∩N.
n n n
Lemma 2.16 [25, Lemma 2.8]. Let M be a maximal subgroup of a group G, P a normal p-subgroup
of G such that G = PM, where p is a prime divisor of |G|. Then P ∩M is normal in G.
Lemma 2.17 [22]. Let P be a Sylow p-subgroup of G, where p is a prime divisor of |G|. If N EG
and N ∩P ⊆ Φ(P), then N is p-nilpotent.
Lemma 2.18 [5, Chapter A, (4.13) Proposition(b)]. Let G = G ×···×G with each G a non-
1 r i
abelian simple group. Then a subgroup S is subnormal in G if and only if S is a (direct) product of
a subset of the factors G .
i
5
Lemma 2.19 [7, Lemma 2.1]. Let F be a non-empty saturated formation. Suppose that A≤ G.
(1) If AEG, then AZ (G)/A ≤ Z (G/A).
F F
(2) If F is S-closed, then Z (G)∩A ≤ Z (A).
F F
(3) If G ∈ F, then Z (G) = G.
F
Lemma 2.20 [11, Chapter III, Theorem 7.2]. Let G be a p-group. Suppose that H and H are
1 2
(normal) subgroups of G such that H ≤ H and |H : H | = ps. Then for any integer 1 ≤ t ≤ s,
1 2 2 1
there exists a (normal) subgroup H of G, which satisfies that H ≤ H ≤ H and |H :H | = pt.
3 1 3 2 3 1
Lemma2.21[17,Lemma2.8]. LetpbeaprimedividingtheorderofG,GisA -freeand(|G|,p−1) =
4
1. Assume that N is a normal subgroup of G such that G/N is p-nilpotent and the order of N is not
divisible by p3. Then G is p-nilpotent.
Lemma 2.22. Let p be a prime divisor of |G| with (|G|,p−1) = 1. If G has cyclic Sylow p-subgroups,
then G is p-nilpotent.
Proof. See the proof of [20, (10.1.9)].
Lemma 2.23 [10, Lemma 2.9]. Let F be a saturated formation containing U (containing N) and G
a group with a normal subgroup E such that G/E ∈ F. If E is cyclic (if E is contained in Z(G),
respectively), then G ∈ F.
Lemma 2.24 [14, Main Theorem]. Let F be a saturated formation containing U. Then G ∈ F if
and only if G has a normal subgroup E such that G/E ∈ F and for any non-cyclic Sylow subgroup P
of the generalized Fitting subgroup F∗(E), every maximal subgroup of P not having a supersolvable
supplement in G or every cyclic subgroup H of P of prime order or order 4 (when P is a non-abelian
2-group and H * Z (G)) without a supersolvable supplement in G is weakly S-embedded in G.
∞
3 Main Results
Note that a subgroup M of G is said to be an n-maximal (n ≥ 1) subgroup of G if G has a
n
subgroup chain: M < M < ··· < M < M = G such that M is a maximal subgroup of M
n n−1 1 0 i i−1
(1 ≤ i ≤ n).
Theorem 3.1. Let G be a group and p a prime divisor of |G| with (|G|,(p−1)(p2−1)···(pn−1)) = 1
for some integer n ≥ 1. Then G is p-nilpotent if and only if there exists a normal subgroup H of G
such that G/H is p-nilpotent and for any Sylow p-subgroup P of H, every n-maximal subgroup of P
not containing P ∩GNp (if exists) either has a p-nilpotent supplement in G or satisfies (△) in G.
Proof. The necessity is obvious. So we need only to prove the sufficiency. Assume that the result is
false and let (G,H) be a counterexample for which |G|+|H| is minimal. We proceed the proof via
the following steps:
(1) |H| ≥ pn+1.
p
By Lemma 2.12, if |H| ≤ pn, then G is p-nilpotent, a contradiction.
p
(2) G is not a non-abelian simple group.
If G is a non-abelian simple group, then GNp = H = G for GNp 6= 1. Since |G| ≥ pn+1, we
p
6
may let P be an n-maximal subgroup of a Sylow p-subgroup P of G. By the hypothesis, P either
n n
satisfies (△) or has a p-nilpotent supplement in G.
(i) Case 1: P satisfies (△) in G.
n
In this case, there exists a normal subgroup K of G such that P K is S-quasinormal in G and
n
P ∩K ≤ (P ) or P ∩K ≤ (P ) . As G is simple, K = 1 or G.
n n seG n n τG
Suppose that K = 1. Then P is S-quasinormal and so subnormal in G by Lemma 2.1(1). Thus
n
P = 1 for G 6= P . It follows that |G| = pn, which contradicts (1). Now assume that K = G. Then
n n p
P = (P ) orP = (P ) . Intheformercase, foranysubgroupD ofP whichisS-quasinormally
n n seG n n τG n
embeddedinG,thereexistsanS-quasinormalsubgroupU ofGsuchthatD ∈ Syl (U). SinceG 6= U,
p
D = U = 1. Therefore P = (P ) = 1, and so |G| = pn, a contradiction. In the latter case, P
n n seG p n
is τ-quasinormal in G by Lemma 2.7(1). As G is not a p-group, we can take a prime q 6= p dividing
|G| and a Sylow q-subgroup Q of G. Suppose that P 6= 1. As QG = G, (p,|QG|) 6= 1. So we
n
have QgP = P Qg for all g ∈ G. If G = P Q, then G is soluble by Burnside’s paqb-Theorem, a
n n n
contradiction. Hence G 6= P Q. By Lemma 2.13, there exists a non-trivial normal subgroup X of G
n
such that either P ≤ X or Q ≤ X, which is impossible. So we obtain that P = 1 and |G| = pn, a
n n p
contradiction again.
(ii) Case 2: P has a p-nilpotent supplement T in G.
n
Let T be the normal p-complement of T. Then G = P T = P N (T ). If N (T ) = G,
p′ n n G p′ G p′
then T E G, and so T = 1 or G. This implies that G is p-nilpotent, a contradiction. Thus
p′ p′
N (T ) < G. Obviously P ∩N (T ) ∈ Syl (N (T )) and P ∩N (T ) < P. Let P be a maximal
G p′ G p′ p G p′ G p′ 2
subgroup of P containing P ∩N (T ) and P an n-maximal subgroup of P contained in P . If
G p′ n2 2
P satisfies (△) in G, the same discussion as above shows that it is impossible. Hence P has a
n2 n2
p-nilpotent supplement E in G. Let E be the normal p-complement of E, then we also have that
p′
G = P N (E ) = P N (E ). SinceT andE areHall p′-subgroupsof G, there exists an element
n2 G p′ 2 G p′ p′ p′
g ∈ P such that T = (E )g by Lemma 2.14. Consequently, G = (P N (E ))g = P N (T ), and
2 p′ p′ 2 G p′ 2 G p′
thereby P = P (P ∩N (T )) = P , a contradiction. This completes the proof of (2).
2 G p′ 2
(3) If 1 6= LEG such that L ≤ H or (|L|,p) = 1, then G/L is p-nilpotent.
If |HL/L| ≤ pn, then G/L is p-nilpotent owing to Lemma 2.12. So we may assume that
p
|HL/L| ≥ pn+1. Let PL/L be a Sylow p-subgroup of HL/L with P ∈ Syl (H) and P L/L an
p p n
n-maximal subgroup of PL/L not containing (PL/L) ∩ (G/L)Np such that P is an n-maximal
n
subgroup of P and P ∩L= P ∩L (see Lemma 2.15). As GNp ≤ H, (|PGNp∩L :P ∩L|,|PGNp∩L :
n
GNp ∩L|) = 1. Thus PL∩GNpL = (P ∩GNp)L by [5, Chapter A, Lemma (1.2)]. If P ∩GNp ≤ P ,
n
then (PL/L) ∩ (G/L)Np = (PL/L) ∩ (GNpL/L) = (P ∩ GNp)L ≤ P L/L, a contradiction. So
n
by the hypothesis, P either has a p-nilpotent supplement T or satisfies (△) in G. In the former
n
case, P L/L has a p-nilpotent supplement TL/L in G/L. In the latter case, there exists a normal
n
subgroup K of G such that P K is S-quasinormal in G and P ∩K ≤ (P ) or P ∩K ≤ (P ) .
n n n seG n n τG
(cid:12) (cid:12)
Note that |L ∩ P K : L ∩ K| = |K(L ∩ P K) : K| (cid:12) |P K : K| and |L ∩ P K : L ∩ P | (cid:12) |L :
n n (cid:12) n n n (cid:12)
L ∩ P | = |L : L ∩ P| = |PL : P|. Since L ≤ H or (|L|,p) = 1, p 6 | |PL : P|. This implies
n
that (|L ∩ P K : L∩K|,|L ∩P K : L∩P |) = 1. Hence P L ∩KL = (P ∩ K)L as above. In
n n n n n
7
view of Lemma 2.5(2) and Lemma 2.7(3), either (P L/L)∩(KL/L) ≤ (P ) L/L ≤ (P L/L)
n n seG n seG
or (P L/L)∩(KL/L) ≤ (P ) L/L ≤ (P L/L) . This shows that P L/L satisfies (△) in G/L.
n n τG n τG n
Therefore (G/L,HL/L) satisfies the hypothesis. By the choice of (G,H), G/L is p-nilpotent.
(4) If P ≤ F < G, then F is p-nilpotent. In particular, if H < G, then H is p-nilpotent.
Obviously P ∈ Syl (H ∩F). Let P be an n-maximal subgroup of P not containing P ∩FNp.
p n
Suppose that P ∩GNp ≤ P . Then P ∩FNp ≤ P ∩(GNp ∩F) ≤ P , which is impossible. Hence
n n
by the hypothesis, P either has a p-nilpotent supplement T or satisfies (△) in G. It induces from
n
Lemma 2.9(1) and Lemma 2.10(1) that P either has a p-nilpotent supplement T∩F or satisfies (△)
n
in F. In view of the conjugacy of the Sylow p-subgroups, (F,H ∩F) satisfies the hypothesis, and so
F is p-nilpotent by the choice of (G,H).
(5) G has a unique minimal normal subgroup N = GNp contained in H such that N (cid:2) Φ(G) and
|N| ≥ pn+1.
p
It follows directly from (3) and Lemma 2.12.
(6) O (G) = 1.
p′
If not, by (3), G/O (G) is p-nilpotent. Consequently, G is p-nilpotent, a contradiction.
p′
(7) O (H) =1.
p
If not, then:
(i) N = O (H)≤ O (G), and G= N ⋊M for some maximal subgroup M of G.
p p
Since O (H) char H E G, N ≤ O (H), and thereby N is abelian. In view of N (cid:2) Φ(G),
p p
there exists a maximal subgroup M of G such that G = N ⋊ M = O (H)M. It follows from
p
Lemma 2.16 that O (H)∩M EG. The uniqueness of N yields O (H)∩M = 1, which implies that
p p
O (H) = N(O (H)∩M) = N ≤ O (G).
p p p
(ii) Let G be a Sylow p-subgroup of G which satisfies that P = G ∩ H. Then there exist a
p p
maximal subgroup P of P and an n-maximal subgroup P of P contained in P such that P EG
1 n 1 1 p
and P = NP = NP . Furthermore, N (cid:2) P and N ∩P is an n-maximal subgroup of N.
n 1 1 n
Clearly G = N(G ∩M). As |G :G ∩M|= |G M : M|= |G : M|= |N| ≥ pn+1, thereexist an
p p p p p
n-maximalsubgroup(G ) andamaximalsubgroup(G ) of G suchthatG ∩M ≤ (G ) ≤ (G ) .
p n p 1 p p p n p 1
This implies that G = N(G ) = N(G ) . Let P = (G ) ∩H and P = (G ) ∩H. Obviously
p p n p 1 n p n 1 p 1
P EG and P ∩M ≤ P ≤ P . Thus P = N(P ∩M) = NP = NP . Moreover, since |P : P | =
1 p n 1 n 1 n
|G ∩H : (G ) ∩H| = |G : (G ) | = pn and |P : P | = |G ∩H : (G ) ∩H| = |G : (G ) | = p,
p p n p p n 1 p p 1 p p 1
P is an n-maximal subgroup of P and P is a maximal subgroup of P. It is evident that N (cid:2) P .
n 1 1
As |N :N ∩P |= |P :P | = pn, N ∩P is an n-maximal subgroup of N.
n n n
(iii) P satisfies (△) in G, that is, there exists a normal subgroup K of G such that P K is
n n
S-quasinormal in G and P ∩K ≤ (P ) or P ∩K ≤ (P ) .
n n seG n n τG
Assume that P does not satisfy (△) in G, then P has a p-nilpotent supplement T in G by the
n n
hypothesis. LetTp′ bethenormalp-complement ofT, thenG = PnT = PnNG(Tp′). SinceM ∼= G/N
is p-nilpotent, M has a normal p-complement M such that M ≤ N (M ) ≤ G. If N (M ) = G,
p′ G p′ G p′
then M EG, and thus M =1 by (6). This shows that G is a p-group, a contradiction. Therefore
p′ p′
N (M ) = M. By Lemma 2.14, there exists an element g ∈ P such that M = (T )g. Then
G p′ n p′ p′
8
G = (P N (T ))g = P N (M ) = P M, and thereby P = P (P ∩ M) = P , a contradiction.
n G p′ n G p′ n n n
Hence (iii) holds.
(iv) P ∩K = 1.
n
First suppose that P ∩ K ≤ (P ) . Let D ,D ,···D be all subgroups of P which are
n n seG 1 2 s n
S-quasinormally embedded in G, and so (P ) = hD ,D ,···D i. By definition, there exist S-
n seG 1 2 s
quasinormal subgroups U ,U ,···U of G with D ∈ Syl (U ) (1≤i≤s). Assume that (U ) 6= 1.
1 2 s i p i i G
Let N be a minimal normal subgroup of G contained in (U ) . If N (cid:2) H, then N ∩ H = 1.
1 i G 1 1
Let N /H E G/H contained in HN /H, then N E G, and therefore N ∩ N = 1 or N . Thus
2 1 2 1 2 1
N = H(N ∩N ) = H or HN . So HN /H is a minimal normal subgroup of G/H. Since G/H is p-
2 1 2 1 1
nilpotent, it is also p-soluble. It follows from (6) that N ∼= HN /H is a p-subgroup . Consequently,
1 1
N ≤ D ≤ H,acontradiction. HenceN ≤ H,thenby(5), N = N ≤U ≤ P ,alsoacontradiction.
1 i 1 1 i n
Thus (U ) = 1 (1≤i≤s). By Lemma 2.2 and Lemma 2.3, D is S-quasinormal in G and so is
i G i
(P ) . Therefore, (P ) ≤ O (H) = N and Op(G) ≤ N ((P ) ) by Lemma 2.1(1) and Lemma
n seG n seG p G n seG
2.1(5). If (P ) 6= 1, then N ≤ ((P ) )G = ((P ) )Op(G)Gp = ((P ) )Gp ≤ (P ∩N)Gp ≤
n seG n seG n seG n seG n
(P ∩N)Gp = P ∩N ≤ N. This implies that N = P ∩ N and N ≤ P , a contradiction. Thus
1 1 1 1
(P ) = 1, and so P ∩K = 1.
n seG n
Now consider that P ∩ K ≤ (P ) . If H ∩ K 6= 1, then N ≤ H ∩ K ≤ K. Thereupon
n n τG
N ∩ P ≤ P ∩ K ≤ (P ) , and so N ∩ P = N ∩ (P ) . Take an arbitrary prime q 6= p
n n n τG n n τG
dividing |G| and a Sylow q-subgroup Q of G. Suppose that (p,|QG|) = 1, then QG ≤ O (G), which
p′
contradicts (6). By Lemma 2.7(1), (P ) is τ-quasinormal in G, and so (P ) Q is a subgroup of
n τG n τG
G. Note that N ∩P is an n-maximal subgroup of N. Since |N : N ∩(P ) Q| = |N(P ) Q :
n n τG n τG
(P ) Q| = |N(P ) : (P ) | = |N : N ∩ P | = pn, N ∩ (P ) Q = N ∩ P . It follows that
n τG n τG n τG n n τG n
Q ≤ N (N ∩ (P ) Q) = N (N ∩ P ), and thus Op(G) ≤ N (N ∩ P ). If N ∩ P 6= 1, then
G n τG G n G n n
N ≤ (N ∩P )G = (N ∩P )Gp ≤ (N ∩P )Gp = N ∩P . This implies that N ≤ P , a contradiction.
n n 1 1 1
Hence N ∩P = 1. Then |N| = pn, also a contradiction. Therefore, H ∩K = 1, and so P ∩K = 1.
n n
The proof of (iv) is completed.
(v) H = G.
If H < G, then H is p-nilpotent by (4). Let H be the normal p-complement of H. Then
p′
H EG. Hence H = 1 by (6), which implies that H = N = P. Since P (H ∩K) = P K ∩H
p′ p′ n n
is S-quasinormal in G by Lemma 2.2, we have Op(G) ≤ N (P (H ∩K)). If P (H ∩K) 6= 1, then
G n n
H = (P (H ∩K))G = (P (H ∩K))Gp ≤ (P (H ∩K))Gp = P (H ∩K). Therefore, H = P (H ∩K)
n n 1 1 1
and H ∩ K 6= 1. It induces that H ∩ K = H by the minimality of H. Then H ≤ K, and so
P = P ∩K = 1 by (iv). Consequently, |H| = pn, which contradicts (1). Hence P (H ∩K) = 1.
n n n
We also obtain that P = 1, a contradiction as above.
n
(vi) Final contradiction of (7).
By (iv) and (v), |K| = |K : P ∩K| = |P K :P | ≤ pn. If K 6= 1, then N ≤ K, and so G/K
p n p n n p
is p-nilpotent. It follows from Lemma 2.12 that G is p-nilpotent, a contradiction. We may, therefore,
assume that K = 1. Then P is S-quasinormal in G. If P 6= 1, then N ≤ P G ≤ P as above, a
n n n 1
contradiction. Hence P = 1, and thus |G| = pn, the final contradiction completes the proof of (7).
n p
9
(8) H = G.
If not, then H is p-nilpotent by (4). Let H be the normal p-complement of H. Then H EG,
p′ p′
which induces that H = O (H) = 1 by (6) and (7). So G is p-nilpotent, a contradiction.
p
(9) O (G) = 1.
p
It follows directly from (7) and (8).
(10) N is not p-soluble.
If N is p-soluble, then O (N) 6= 1 or O (N) 6= 1, which contradicts (6) or (9).
p′ p
(11) Let P be a Sylow p-subgroup of G, then G = PN and P ∩ N (cid:2) Φ(P). In addition,
N = Op(G).
If PN < G, then PN is p-nilpotent by (4) and so is N, contrary to (10). Hence G = PN.
If P ∩ N ≤ Φ(P), then N is p-nilpotent by Lemma 2.17, which also contradicts (10). Clearly
1 6= Op(G) ≤ N, and so N = Op(G).
(12) Final contradiction.
By (11), there exists a maximal subgroup P of P such that P = (P ∩N)P . It is obvious that
1 1
P ∩GNp = P ∩N (cid:2) P . Let P be an n-maximal subgroup of P contained in P . By the hypothesis,
1 n 1
P either satisfies (△) or has a p-nilpotent supplement T in G.
n
(i) Case 1: P satisfies (△) in G.
n
In this case, there exists a normal subgroup K of G such that P K is S-quasinormal in G and
n
P ∩K ≤ (P ) or P ∩K ≤ (P ) . If K = 1, then P is S-quasinormal in G. It induces from
n n seG n n τG n
Lemma 2.1(1) and (9) that P ≤ O (G) = 1, and so |G| ≤ pn, which contradicts (1). Therefore we
n p p
may suppose that K 6= 1. Then N ≤ K, and so P ∩N ≤ (P ) or P ∩N ≤ (P ) .
n n seG n n τG
First assume that P ∩ N ≤ (P ) . With the similar argument as above, let D ,D ,···D
n n seG 1 2 s
be all subgroups of P which are S-quasinormally embedded in G. Then there exist S-quasinormal
n
subgroups U ,U ,···U of G with D ∈ Syl (U ) (1≤i≤s). If (U ) 6= 1, then N ≤ (U ) ≤ U .
1 2 s i p i i G i G i
Thus D ∩ N ∈ Syl (N). Since P ∩ N ∈ Syl (N), D ∩ N = P ∩ N. It follows that P ∩ N ≤
i p p i
D ≤ P ≤ P , a contradiction. Therefore (U ) = 1, and so D is S-quasinormal in G by Lemma
i n 1 i G i
2.3. As (P ) = hD ,D ,···D i, (P ) ≤ O (G) = 1, which induces that P ∩N = 1. Hence
n seG 1 2 s n seG p n
|N| = |P ∩N| = |P ∩N :P ∩N| = |P (P ∩N): P |≤ |P :P |= pn. It follows from Lemma 2.12
p n n n n
that G is p-nilpotent, a contradiction.
Now consider that P ∩N ≤ (P ) . Let q 6= p be an arbitrary prime divisor of |G| and Q ∈
n n τG
Syl (G). Clearly Q ≤ Op(G) = N and (p,|QG|) 6= 1. Then by Lemma 2.7(1), (P ) Q = Q(P ) .
q n τG n τG
It follows that (P ∩N)Q = ((P ) ∩N)Q = (P ) Q∩N = Q(P ) ∩N = Q((P ) ∩N) =
n n τG n τG n τG n τG
Q(P ∩N). Therefore, (P ∩N)Qn = Qn(P ∩N)for all n ∈N. If N = (P ∩N)Q, thenN is soluble
n n n n
by Burnside’s paqb-Theorem, which contradicts (10). So there exists a non-trivial normal subgroup
X of N such that either P ≤ X or Q ≤ X by Lemma 2.13. Since N is characteristically simple
n
group of G, N ∼= A ×A ×···×A , where A ∼= A (1≤i≤t) is a simple group. Obviously A is
1 2 t i
non-abelian owing to (10). Without loss of generality, we may assume that X ∼= A ×A ×···×A
1 2 k
(k < t) by Lemma 2.18. Since Q ∈ Syl (N), |Q| = |N| = (|A| )t > (|A| )k = |X| . This implies
q q q q q
that P ∩ N ≤ X. Note that |P ∩ N : P ∩ N| = |P (P ∩ N) : P | ≤ |P : P | = pn, then we
n n n n n
10