1 Introduction
Throughout this paper, all groups are finite and G always denotes a finite
group. Moreover, p is always supposed to be a prime, ℙ denotes
the set of all primes. A subgroup H of G is said to be 𝔘-subnormal (in the sense of Kegel [11]) or K-𝔘-subnormal in G (see [1, p. 236]) if there exists a chain of
subgroups
H=H0≤H1≤⋯≤Ht=G
such that either Hi-1 is normal in Hi or Hi/(Hi-1)Hi is supersoluble
for all i=1,…,t.
Recall that a subgroup H of G is said to be S-propermutable in G provided there is a subgroup B of G such that G=NG(H)B and H
permutes with all Sylow subgroups P of B, that is, HP=PH. Nice results
obtained in the papers [18, 19] were based on applications of this
concept. In this paper we study the following generalization of the concept
of S-permutability.
Definition 1.1We say that a subgroup H of G is weaklyS-propermutable in G provided there is a K-𝔘-subnormal
subgroup T of G such that HT=G and H∩T≤S≤H for some S-propermutable subgroup S of G.
Note that in the symmetric group S4 of degree 4 a Carter subgroup is
weakly S-propermutable but not S-propermutable in the group.
Our main goal here is to prove the following result.
Theorem 1.2Let X be a normal subgroup of G and P a
Sylow p-subgroup of X, where (p-1,|X|)=1. Suppose that there is a
subgroup 1<D<P such that every subgroup of P of order |D| and
also, in the case when P is non-abelian and |D|=2, every cyclic subgroup
of order 4 of P is weakly S-propermutable in G. Then X is p-nilpotent and every p-chief factor of G below X is cyclic.
The following special case of Theorem 1.2 gives the positive answer to
Question 18.91 (b) in [14].
Corollary 1.3Let X be a normal subgroup of G and P
a Sylow p-subgroup of X, where (p-1,|X|)=1. Suppose that there is a
subgroup 1<D<P such that every subgroup of P of order |D| and also, in
the case when P is non-abelian and |D|=2, every cyclic subgroup of order
4 of P is S-propermutable in G. Then X is p-nilpotent and every p-chief factor of G below X is cyclic.
Theorem 1.2 also allows us to prove the following useful fact.
Corollary 1.4Let X≤E be normal subgroups of G.
Suppose that for every non-cyclic Sylow subgroup P of X, there is a
subgroup 1<D<P such that every subgroup of P of order |D| and
also, in the case when P is non-abelian and |D|=2, every cyclic subgroup
of order 4 of P is weakly S-propermutable in G. If X=E or X=F*(E), then every chief factor of G below E is cyclic.
In this corollary F*(E) denotes the generalized Fitting subgroup of E, that is, the product of all normal quasinilpotent subgroups of E.
The main application of Corollary 1.4 is the following fact which covers
results of many papers.
Corollary 1.5Let F be a solubly saturated
formation (see [4, C̱hapter IV]) containing all supersoluble groups
and let X≤E be normal subgroups of G such that G/E∈F.
Suppose that for every non-cyclic Sylow subgroup P of X there is a
subgroup 1<D<P such that every subgroup of P of order |D| and also, in
the case when P is non-abelian and |D|=2, every cyclic subgroup of order
4 of P is weakly S-propermutable in G. If X=E or X=F∗(E),
then G∈F.
One can find in the literature a lot of special cases of Corollary 1.5.
Recall, for example, that a subgroup H of G is called 𝑆𝑆-quasinormal [13] in G if H permutes with all Sylow subgroups
of some supplement of H to G. It is clear that every 𝑆𝑆-quasinormal
subgroup and every S-propermutable subgroup are weakly S-propermutable.
Therefore Corollary 1.5 covers the main result in [17], [13, Theorem 1.5], [12, Theorems 3.2 and 3.3] and [19, Theorem B].
All unexplained notation and terminology are standard. The reader is
referred to[1], [4], or [6] if necessary.
2 Basic lemmas
Lemma 2.1Lemma 2.1 (see [18, Lemma 2.4])
Let H and B be
subgroups of G. Suppose that G=NG(H)B and HVb=VbH for some subgroup V
of B and for all b∈B. Then HVx=VxH for all x∈G.
Lemma 2.2Lemma 2.2 (see [18, Lemma 2.5])
Suppose that for
subgroups A and B of G we have AB=BA and G=NG(A)B. Then:
If A permutes with all Sylow p-subgroups of B, then A
permutes with all Sylow p-subgroups of AG∩B.
Lemma 2.3Lemma 2.3 (see Kegel [10])
Let A and B be
subgroups of G such that G≠AB and ABx=BxA for all x∈G.
Then G has a proper normal subgroup N such that either A≤N or B≤N.
Lemma 2.4Let H, K and N be subgroups of G. If N
is subnormal in G, HK equals KH and π(H)∩π(K)=∅, then N∩HK=(N∩H)(N∩K).
Proof.
Let D=N∩HK. Then D is subnormal in HK, so D=(D∩H)(D∩K). But (D∩H)(D∩K)≤(N∩H)(N∩K)≤D, so N∩HK=(N∩H)(N∩K). The lemma is proved.
∎
Lemma 2.5Lemma 2.5 (see [7, Theorem A*])
Let H be a
Hall π-subgroup of G. Suppose that G=HT for some subgroup T of G, and q
a prime. If H permutes with every Sylow p-subgroup of T for all primes
p≠q, then T contains a complement of H in G and any two
complements of H in G are conjugate.
We say H is completelyS-propermutable in G if H is S-propermutable in any subgroup of G containing H.
In order to use inductive reasoning more actively in the proofs, it is
convenient to work with the following generalizations of S-propermutable
and weakly S-propermutable subgroups.
Definition 2.6We say that a subgroup H of G is:
A generalizedS-propermutable subgroup of G if H is
completely S-propermutable in HG and H permutes with some Sylow p-subgroup of G for any prime p dividing |G|,
P-embedded in G if there is a K-𝔘-subnormal
subgroup T of G such that HT=G and H∩T≤S≤H for some
generalized S-propermutable subgroup S of G.
If H is P-embedded in G, then we use Σ(H) to denote the set of
all triples (H,S,T), where T is a K-𝔘-subnormal subgroup
of G and S is a generalized S-propermutable subgroup of G such that HT=G and H∩T≤S≤H.
Lemma 2.7Let H, L, K and N be subgroups of G.
Suppose that H is S-propermutable (completely S-propermutable,
respectively) in G, L is a generalized S-propermutable subgroup of G, K is P-embedded in G and N is normal in G.
The quotient HN/N is S-propermutable (completely S-propermutable, respectively) in G/N.
If H≤E≤G and H is S-propermutable (completely S-propermutable, respectively) in E, then NH is S-propermutable
(completely S-propermutable, respectively) in NE.
If H is S-propermutable (completely S-propermutable,
respectively) in HG, then the quotient HN/N is S-propermutable (completely S-propermutable, respectively) in (NH/N)G/N.
The group H permutes with some Sylow p-subgroup of G for every prime
p dividing |G|.
|G:NG(L∩N)| is a
π
-number, where π=π(N)∪π(L).
The quotient NL/N is a generalized S-propermutable subgroup of G/N.
If L≤P, where P is a p-subgroup of G, then L is generalized S-propermutable in PN.
Suppose that L is a p-group, LG=G and Q is a Sylow q-subgroup of G, where q≠p. Suppose that either L is not a Sylow
subgroup of G or G is not soluble, then QG≠G.
Suppose that L is a p-group, LG=G and D is a normal
subgroup of G. If L∩D is a Sylow p-subgroup of D, then L∩D
is completely S-propermutable in D.
Suppose that L is a p-group and N is a minimal subnormal
subgroup of LLG. If L∩N≠1, then N is a p-group.
If N≤K or (|N|,|K|)=1, then KN/N is P-embedded in G/N.
If K≤N, then K is P-embedded in N.
Proof.
(1) First note that if H is S-propermutable in G, then
HN/N is S-propermutable in G/N by Lemma 2.1 in [18].
Now let us assume that H is completely S-propermutable in G. Let HN/N≤E/N≤G/N. Then H is S-propermutable in E and so, in view of the
previous paragraph, HN/N is S-propermutable in E/N. Hence HN/N is
completely S-propermutable in G/N.
(2) Assume that H is S-propermutable in E and let B be a subgroup of
E such that E=NE(H)B and H permutes with all Sylow subgroups of B. Then NH permutes with all Sylow subgroups of B. On the other hand,
since NE(H)≤NNE(H)≤NNE(NH), we have NE=NNE(NH)B.
Hence NH is S-propermutable in NE.
Now assume that H is completely S-propermutable in E and let NH≤W≤NE. Then W=N(W∩E), where H≤W∩E. Hence NH is S-propermutable in W.
(3) Since H is (completely) S-propermutable in HG, it follows that NH is
(completely) S-propermutable in NHG by assertion (2). On the other
hand, we have (NH/N)G/N=(NH)G/N=NHG/N. Therefore assertion (3) follows
from assertion (1).
(4) By [9, Chapter VI, Section 4.6], there are Sylow p-subgroups P1, P2
and P of NG(H), B and G, respectively, such that P=P1P2.
Hence H permutes with P.
(5) Let p be a prime such that p∉π. Then by (4), there is a
Sylow p-subgroup P of G such that LP=PL is a subgroup of G. Hence LP∩N=L∩N is a normal subgroup of HP. Thus p does not divide |G:NG(L∩N)|.
(6) This follows from assertion (3) and the fact that for any Sylow p-subgroup P/N of G/N there is a Sylow p-subgroup P1 of G such
that P1N/N=P/N.
(7) Let E=PN. It is clear that HE≤HG∩E. Hence H is
completely S-propermutable in HE. On the other hand, for any prime q≠p dividing |E| and for a Sylow q-subgroup Q of G such that HQ=QH we have HQ∩E=HQ∩PN=H(Q∩PN)=H(Q∩N)=(Q∩N)H, where
Q∩N is a Sylow q-subgroup of E. Thus H is generalized S-propermutable in E.
(8) Since LG=G, it follows that L is completely S-propermutable in G by
hypothesis. Hence there is a subgroup B of G such that LB=G and L
permutes with all Sylow subgroups of B. Hence L permutes with Qx
for all x∈G by Lemma 2.1. Note also that LQ≠G. Indeed, if L is
a Sylow p-subgroup of G, then G is not soluble by hypothesis and so LQ≠G by the Burnside paqb-theorem. Hence QG≠G by Lemma 2.4.
(9) Since LG=G, Lemma 2.3 implies that there is a subgroup B of G
such that G=LB=PB and L permutes with all Sylow subgroups of B. Hence L permutes with every Sylow q-subgroup of G for all primes q≠p by
Lemma 2.1. Let R be a Sylow q-subgroup of D∩B, where q≠p.
Then for some Sylow q-subgroup Bq of B we have R=D∩Bq. On
the other hand, LBq=BqL. Hence we have D∩LBq=(D∩L)(D∩Bq)=DpR=RDp by Lemma 2.4. Therefore, by Lemma 2.5, D has a Hall p′-subgroup E. Then D=DpE and Dp permutes with all
Sylow subgroups of E. Hence D∩H=Dp is completely S-propermutable in D.
(10) By Lemma 2.2 (2), we have that L is generalized S-propermutable in LD, where
D=LG. Since L∩N≠1, LD∩N≠1 and hence N≤LD,
since N is a minimal subnormal subgroup of G. Hence in the case when LD≠G the assertion is true by induction.
Now assume that LD=G. Then for some subgroup B of G we have G=LB
and L permutes with all Sylow subgroups of B. Suppose that N is a
simple non-abelian group. Let q≠p be a prime dividing |N| and Nq
a Sylow q-subgroup of N. Then for some Sylow q-subgroup Q of G we
have Nq=N∩Q, LQ≠N and L permutes with Qx for all x∈G by Lemma 2.1. Since LQx=QxL, we have N∩LQx=(N∩L)(N∩Qx)=(N∩L)(N∩Q)x=(N∩L)(Nq)x=(Nq)x(N∩L)
by Lemma 2.4. Therefore N is not simple by Lemma 2.3, a contradiction.
Hence we have (10).
(11) Let (K,S,T)∈Σ(K). Clearly, (KN/K)(TN/N)=G/N. Moreover, TN/N is K-𝔘-subnormal in G/N by [1, Theorem 6.1.6 (3)] and SN/N is generalized S-propermutable in G/N by assertion (6).
Note also that KN∩NT≤NS. Indeed, if N≤K, then KN∩NT=N(KN∩T)=N(K∩T)≤NS. On the other hand, if (|N|,|K|)=1, then
(|N|,|G:T|)=1 and so N≤T, which as above implies that KN∩NT≤NS. Hence (KN/K)∩(TN/N)≤SN/N, so KN/N is P-embedded in G/N.
(12) Note that N=K(T∩N), where, by [1, Theorem 6.1.7 (2)], T∩N is K-𝔘-subnormal in N. Also, S is generalized
propermutable in N by assertion (7). Finally, K∩(T∩N)=(K∩T)∩N≤S∩N=S, so K is P-embedded in N.
∎
Lemma 2.8Lemma 2.8 (see Yi and Skiba [19, Theorem C])
If a Sylow p-subgroup of G is S-propermutable in G, then G is p-soluble.
We use Rp(G) to denote the largest p-soluble normal subgroup of G.
Lemma 2.9Let P be a Sylow p-subgroup of G with |P|=pn. Suppose that for some 1≤k≤n, every subgroup of P of
order pk is generalized S-propermutable in G. Then Rp(G)≠1.
Proof.
Assume that this lemma is false and let G be a
counterexample of minimal order. Suppose that for some subgroup H of P
of order pk we have HG≠G. By Lemma 2.7 (7) the hypothesis holds
for HG, so Rp(HG)≠1 by the choice of G. Since Rp(HG) is a characteristic subgroup of HG, it is normal in G.
Hence we have 1<Rp(HG)≤Rp(G). But this contradicts the choice of G.
Hence for every subgroup H of P of order pk we have HG=G,
which in view of Lemma 2.7 (8) implies that G has a proper non-identity
normal subgroup D. By Lemma 2.7 (6), the hypothesis holds for G/D. Hence
the choice of G implies that p divides |D|. Let Dp=P∩D be a
Sylow p-subgroup of D and |Dp|=pa. If k≤a, the hypothesis
holds for D and so Rp(D)≠1 by the choice of G. Thus Rp(D)≤Rp(G)≠1, a contradiction. Hence a<k and so for some
subgroup H of P of order pk we have Dp=H∩D=P∩D. Since HG=G, Lemma 2.7 (9) implies that Dp is completely S-propermutable
in D. Hence 1<Rp(D)≤Rp(G) by Lemma 2.8. This contradiction
completes the proof of the lemma.
∎
Lemma 2.10If a Sylow p-subgroup P of G is generalized
S-propermutable in G, then G is p-soluble.
Proof.
The hypothesis holds for PG by Lemma 2.7 (7). Thus in
the case, when PG≠G, PG is p-soluble by induction and so G
is p-soluble. Now assume that PG=G. Then P is completely S-propermutable in G by hypothesis. Hence G is p-soluble by Lemma 2.8.
∎
Lemma 2.11If every maximal subgroup of a Sylow 2-subgroup
P of G is generalized S-propermutable in G and |P|=8, then G is
2-nilpotent.
Proof.
Suppose that this lemma is false and let G be a
counterexample of minimal order.
(1) O2′(N)=1 for every normal subgroup N of G.
Suppose that for some normal subgroup N of G we have O2′(N)≠1. Since O2′(N) is a characteristic subgroup of N, it is normal in G. On the other hand, by Lemma 2.7 (6), the hypothesis
holds for G/O2′(N). Hence G/O2′(N) is 2-nilpotent by the choice of G and so G is 2-nilpotent, a contradiction.
(2) O2(G)≠1.
By Lemma 2.9, R2(G)≠1. On the other hand, by (1), O2′(R2(G))=1. Hence 1<O2(R2(G))≤O2(G).
(3) G is soluble.
If O2(G)=P, this assertion follows from the Feit–Thompson Odd theorem. Let O2(G)<P. In view of (2), for some minimal normal subgroup N of G we
have N≤O2(G). By Lemma 2.7 (6), the hypothesis holds for G/N, so P/N is 2-nilpotent by the choice of G. Thus again we get that G is
soluble.
(4) If N is a minimal normal subgroup of G, then |N|=4 and N=CG(N).
In view of (1) and (3), N≤P. It is also clear that G/N is 2-nilpotent. Hence N is the only minimal normal subgroup of G, N≰Φ(G) and N≰Z∞(G). Thus N=CG(N) by [4, A, 15.2]. It follows that |N|>2. Assume that N=P and let H be a maximal
subgroup of N. Then H permutes for some Sylow q-subgroup Q of G
for any prime q dividing |G|. Hence H=HQ∩P is normal in HQ.
Therefore H is normal in G, which implies that H=1. This contradiction
shows that |N|=4.
(5) |G|=24.
Indeed, in view of (4), G/N=G/CG(N) is isomorphic to some subgroup of GL(2,2).
Final contradiction. In view of (4), P is not abelian. Hence P
has a cyclic subgroup C4 of order 4. By hypothesis, there is a Sylow
3-subgroup Q of G such that M=C4Q=QC4. It is clear that M=NG(Q) and N≰M. Hence 1<M∩N<N, where M∩N is
normal in G, contrary to the minimality of N. The lemma is proved.
∎
Lemma 2.12Lemma 2.12 (see [5, Chapter 5, Section 3.11])
Let P be a p-group and let D be a Thompson critical subgroup of P. Then D is of class
at most 2 and D/Z(D) is elementary abelian.
Let P be a p-group. If P is not a non-abelian 2-group we use Ω(P) to denote the subgroup Ω1(P). Otherwise, Ω(P)=Ω2(P).
Lemma 2.13Let P be a p-group of class at most 2.
Suppose that exp(P/Z(P)) divides p.
If p>2, then 𝑒𝑥𝑝(Ω(P))=p.
If P is a non-abelian
2
-group, then exp(Ω(P))=4.
See [3, Lemma 2.12].
The symbol Z𝔘(G) denotes the largest normal subgroup of G
such that each chief factor of G below Z𝔘(G) is cyclic.
Lemma 2.14Lemma 2.14 (see [3, Lemma 2.12])
Let P be a
normal p-subgroup of G and let D be a Thompson critical subgroup of P. If D≤ZU(G) or Ω≤ZU(G), then P≤ZU(G).
Lemma 2.15Let X be a normal subgroup of G and P a
Sylow p-subgroup of X. Suppose that (|X|,p-1)=1. Suppose that every
cyclic subgroup of P of prime order and of order 4 (if P is a
non-abelian 2-group) are generalized S-propermutable in G. Then X is p-nilpotent and every chief factor of G between X and Op′(X) is cyclic.
Proof.
Suppose that this lemma is false and let G be a
counterexample with |G|+|X| minimal. Let Z=Z𝔘(G).
(1) Op′(X)=1. (See (1) in the proof of Lemma 2.11.)
The hypothesis holds for (X,X) by Lemma 2.7 (7). Suppose X≠G. Then the choice of G implies that X is p-nilpotent, so Claim
(1) implies that X=G.
(2) Op(X)≤Z∩Z∞(X).
Since (p-1,|X|)=1, it is enough to prove that Op(X)≤Z. Suppose
that this assertion is false. Then Op(X)≠1.
(a) G has a normal subgroup R≤Op(X) such that Op(X)/R
is a non-cyclic chief factor of G, R≤Z and V≤R for any normal
subgroup V≠P of G contained in P.
Let Op(X)/R be a chief factor of G. Then the hypothesis holds for (G,R). Therefore R≤Z and so Op(X)/R is not cyclic by the choice
of (G,X). Now let V≠Op(X) be any normal subgroup of G contained
in Op(X). Then V≤Z. If V≰R, then from the G-isomorphism Op(X)/R=VR/R≃V/V∩R we have Op(X)≤Z,
which contradicts our choice of (G,X). Hence V≤R.
(b) If D is a Thompson critical subgroup of Op(X), then we have Ω(Op(X))=Op(X)=D.
Indeed, suppose that Ω(Op(X))<Op(E). Then, in view of (a), we have that Ω(Op(X))≤Z𝔘(G). Hence Op(X)≤Z𝔘(G) by Lemma 2.14, a contradiction. Hence Ω(Op(X))=Op(X).
Similarly we get that Op(X)=D.
The final contradiction for assertion (2).
Let H/R be any minimal subgroup of (Op(X)/R)∩Z(Gp/R), where Gp is a Sylow p-subgroup of G. Let x∈H∖R and L=〈x〉. Then H/R=LR/R. On the other hand, by (b) and Lemmas
2.12 and 2.13, |L|=p or |L|=4. Hence L is generalized S-propermutable in G by hypothesis, so H/R is generalized S-propermutable in G/R by Lemma 2.7 (6). Hence by Lemma 2.7 (5), |(G/R):NG/R(H/R)| is a power of p, so H/R is normal in G, which
implies that Op(X)/R=H/R is cyclic. This contradiction completes the
proof of (2).
(3) Op(X)≠X.
Indeed, in view of (2) we have Op(X)≤Z. Hence Op(X)≠X by
the choice of (G,X).
Final contradiction.
By (3), there is a chief factor V/Op(X) of G such that V≤X. If V/Op(X) is a p′-group, then V is p-nilpotent since Op(X)≤Z∞(X) by (2). Therefore 1<Op′(V)≤Op′(X)=1, a contradiction. Hence p divides |V/Op(X)| and
V/Op(X) is non-abelian. Moreover, since (|X|,p-1)=1, it follows that p=2. Let W/Op(X) be a minimal normal subgroup of V/Op(X). By
[9, Chapter IV, Section 5.4], W/Op(X) has a 2-closed Schmidt subgroup L/Op(X). Let A be a minimal supplement to Op(X) in L. Then we have that A∩Op(X)≤Φ(A). Thus, in view of assertion (2), A is a Schmidt group.
Hence by [15, Chapter V, Section 24.2], A=A2⋊Aq, where A2 is the
Sylow 2-subgroup of A and Aq is a Sylow q-subgroup of A (q≠2), A2 is of exponent 2 or exponent 4 (if A2 is non-abelian) and
if Φ=Φ(A2), then A2/Φ is a non-cyclic chief factor of A. Without loss of generality, we may assume that A2≤P. Therefore
there is a cyclic subgroup H of order 2 or order 4 such that H≰Φ. Hence H≰O2(X) and H is generalized S-propermutable in G.
Thus O2(X)H/O2(X) is a non-identity generalized S-propermutable
subgroup of W/O2(X) by Lemma 2.7 (7). Hence, by Lemma 2.7 (8), for some
non-identity Sylow subgroup Q of W/O2(X) we have QW/O2(X)≠W/O2(X), which contradicts the minimality of W/O2(X). This
contradiction completes the proof of the lemma.
∎
Lemma 2.16Let X be a normal subgroup of G and P a
Sylow 2-subgroup of X. Suppose that every subgroup of P of order 4 is
generalized S-propermutable in G. Then X is 2-nilpotent and every
chief factor of G between X and O2′(X) is cyclic.
Proof.
Suppose that this lemma is false and let G be a
counterexample with |G|+|X| minimal.
(1) Op′(N)=1 for each normal subgroup N of G.
(See (1) in the proof of Lemma 2.11.)
(2) X is not a 2-group.
Assume that X=P. Let Z≤N be subgroups of P such that N is a
minimal normal subgroup of G, |Z|=2 and Z is contained in the center
of a Sylow 2-subgroup of G.
(a) There is a subgroup B of P of order 2 such that |G:NG(B)| is not a power of 2. (This follows from Lemma 2.15 and the
choice of G.)
(b) |N|≤4. (In view of Lemma 2.7 (5), the minimality of N
implies that |N|≤4.)
(c) There is a subgroup A of P of order 2 such that |G:NG(A)| is a power of 2.
First suppose also that P has a cyclic subgroup C4=〈a〉
of order 4. In view of Lemma 2.7 (5), |G:NG(C4)| is a power of 2.
Hence |G:NG(〈a2〉)| is a power of 2. Finally, assume
that P is elementary and let H=Z×B, where B is a subgroup of P
of order 2 not contained in N. Then |G:NG(H)| is a power of 2 by
Lemma 2.7 (5), so |G:NG(Z)|=|G:NG(H∩N)| is a power of 2.
(d) Z=N.
It is enough to show that |G:NG(Z)| is a power of 2. Assume that this
is false. Then A≠Zx for all x∈G, so H=AZ is a group of order
4. If A≰N, then Z=H∩N and hence |G:NG(Z)| is a power of 2. This contradiction shows that A≤N, so N=Z×A by Claim (b).
But in this case we have |G:NG(Z)|=2, a contradiction. Hence we have
(d).
(e) |G:NG(V)| is a power of 2 for each subgroup V≠Z of P
of order 2.
Let H=ZV. Then |NG(H):NG(V)∩NG(H)|=2. On the other hand, by
Lemma 2.7 (5), |G:NG(H)| is a power of 2. Thus |G:NG(V)| is a
power of 2. By Claim (e), |G:NG(V)| is a power of 2 for each
subgroup V of P of order 2. Hence every subgroup of P of prime order
or order 4 is generalized S-propermutable in G. Therefore by Lemma
2.15, X=P≤Z∞(G). This contradiction shows that X≠P.
(3) X=G. Hence |P|>8.
Assume that X<G. By Lemma 2.7 (7) the hypothesis holds for X, so X is p-nilpotent by the choice of G. Hence we have X=P by (1), contrary to (2).
Thus X=G, so |P|>8 by Lemma 2.11 and the choice of G.
(4) For every proper normal subgroup N of G we have NP<G.
Indeed, assume that NP=G and let NP≤M, where |G:M|=2. Then, in
view of Claim (3) and Lemmas 2.7 (7) and 2.11, the hypothesis holds for M,
so M is 2-nilpotent by the choice of G. Hence M≤P by (1), a
contradiction.
(5) For every subgroup H of P of order 4 we have HG≤O2(G).
Suppose that HG=G. Let Q be a Sylow p-subgroup of G, where p
is an odd prime dividing |G|. Then, by Lemma 2.7 (8), QG<G. Hence QGP<G by (4). By Lemma 2.7 (7) the hypothesis holds for QGP.
Therefore QG is 2-nilpotent, contrary to Claim (1). Hence HG<G,
so HGP<G by (4). On the other hand, the hypothesis holds for HGP
by Lemma 2.7 (7) and so HG≤O2(G) by (1).
Final contradiction. Since G is not 2-nilpotent, it has a 2-closed Schmidt subgroup A=A2⋊Aq, where A2 is the Sylow 2-subgroup of A of exponent dividing 4 and Aq is a Sylow q-subgroup of A for some prime q≠2 by [9, Chapter IV, Section 5.4]. Moreover,
A2/Φ(A2) is a non-cyclic chief factor of A, Z∞(A)∩A2=Φ(A2), and Φ(A2)=1 if A2 is abelian by
[15, V, 24.2]. Without loss of generality, we may assume that A2≤P. Hence there is a subgroup H of order 4 such that H≰Φ(A2) and H∩Φ(A2)≠1. Hence Φ(A2)H/Φ(A2) is a subgroup of A2/Φ(A2) of order 2. Let D=HG.
By (5), D≤O2(G). If H=D, then we have A2/Φ(A2)=Φ(A2)H/Φ(A2). This contradiction shows that H<D. Thus the
hypothesis holds for (G,D), so D≤Z∞(G). But then H≤A∩Z∞(G)≤Z∞(A)∩A2=Φ(A2). This
contradiction completes the proof of the lemma.
∎
Lemma 2.17Lemma 2.17 (see [3, Lemma 1.6])
Let F be a
solubly saturated formation containing all supersoluble groups and E a
normal subgroups of G with G/E∈F. If every chief factor of
G below E is cyclic, then G∈F.
In fact, the following lemma is well known.
Lemma 2.18The following statements hold.
If G/Φ(G) is p-supersoluble, then G is p-supersoluble [9, Chapter IV, Section 8.6].
Let N and R be minimal normal subgroups of G. If G/N and
G/R are p-supersoluble, then G is p-supersoluble.
Let A=G/Op′(G). Then G is p-supersoluble if
and only if A/Op(A) is an abelian group of exponent dividing p-1,
p
is the largest prime dividing |A| and F(A)=Op(A) is a normal Sylow
subgroup of A.
3 Proof of Theorem 1.2 and Corollaries 1.4 and 1.5
First we prove the following result.
Theorem 3.1Let X be a normal subgroup of G and P a
Sylow p-subgroup of X, where (p-1,|X|)=1. Suppose that there is a
subgroup 1<D<P such that every subgroup of P of order |D| and
also, in the case when P is non-abelian and |D|=2, every cyclic subgroup
of order 4 of P are P-embedded in G. Then X is p-nilpotent and
every p-chief factor of G below X is cyclic.
Proof.
Suppose that this theorem is false and let G be a
counterexample with |G|+|X| minimal. Let Gp be a Sylow p-subgroup
of G containing P.
(1) Op′(N)=1 for any normal subgroup N of G. (See
(1) in the proof of Lemma 2.11.)
(2) X=G or X=P.
Suppose that X<G. By Lemma 2.7 (7), the hypothesis holds for (X,X). Hence X is p-nilpotent by the choice of G, so X=P by Claim (1).
(3) Op(X)≠1.
Suppose that this is false. Let R be a minimal normal subgroup of G
contained in X and Z a minimal normal subgroup of R. Then Z is a
simple non-abelian group and p divides |Z| by Claim (1). Let Cp be
a subgroup of Z of order p. Let H be a subgroup of P of order pk containing Cp and (H,S,T)∈Σ(H).
Suppose that R≰T. Then T∩R<R. In view of [1, 6.1.7 (2)], T∩R is K-𝔘-subnormal in R. On the other hand, |R:T∩R|=|RT:T| is a power of p since G=HT. Hence Op(R)≠R
since (|X|,p-1)=1, so L is a p-group. This contradiction shows that R≤T, so 1<Cp≤H∩Z=S∩Z. Then we have Z≤(SG)G and
hence, by Lemma 2.7 (10), Z is a p-group. Thus Z≤Op(X). This
contradiction shows that we have (3).
(4) k<n-1.
Suppose that k=n-1. By Claim (3), we have Op(X)≠1. Let R be a minimal
normal subgroup of G contained in Op(X). The hypothesis holds on G/R
by Lemma 2.7 (6), so X/R is p-nilpotent and every p-chief factor of G
below X is cyclic by the choice of G. Therefore X is p-soluble,
hence R is the only minimal normal subgroup of G contained in X.
Moreover, |R|>p and, in view of [8, Corollary 1.6], R≰Φ(G).
Now we show that X≠P. Suppose that X=P. Then Φ(X)=1, otherwise
R≤Φ(X)≤Φ(G)∩X. Thus X=P is an elementary abelian p-group. Let W be a maximal subgroup of R such that W is normal in Gp. Then W≠1. Let B be a complement of R in E and H=WB.
Then H is a maximal subgroup of E. Hence H is P-embedded in G. Let
(H,S,T)∈Σ(H). Then T∩X is a non-identity normal subgroup
of G. Hence R≤T, which implies that W≤T∩H=T∩S. Then
W=R∩H=R∩S, and so |G:NG(W)|=|G:NG(S∩R)| is a power of p
by Lemma 2.7 (5). Hence W is normal in G and consequently |R|=p, a
contradiction. Therefore X≠P.
Suppose that X<G. By Lemma 2.7 (7), the hypothesis holds for (X,X). Hence X is p-nilpotent by the choice of G, so X=P by Claim (1), a
contradiction. Hence X=G is p-soluble. Then R=CG(R)=Op(G) and G=R⋊M for some p-nilpotent maximal subgroup M of G. Let W be
a maximal subgroup of R such that W is normal in P. Let Mp be a
Sylow p-subgroup of M and V=WMp. Without loss of generality we may
assume that Mp≤P. Hence V is P-embedded in G. Let (V,S,T)∈Σ(V). Then, since (p-1,|G|)=1, T is subnormal in G.
Hence R≤T. Now arguing, as above, one can show that W is normal in G. Hence |N|=p. This contradiction completes the proof of assertion (4).
(5) All subgroups H of P of order pk are generalized S-propermutable in G.
Indeed, by hypothesis H is P-embedded in G. Let (H,S,T)∈Σ(H). Suppose that H is not generalized S-propermutable in G. Then T≠G and hence there is a maximal subgroup M of G such that T≤M
and G/MG is supersoluble, so |G:M|=p. Moreover, since G=HT=HM=XM, |X:X∩M|=p and so X∩M is normal in X (since (|X|,p-1)=1) and
also in M. Hence X∩M is normal in G. By Claim (4), the hypothesis
holds for (G,X∩M), so the choice of G implies that X∩M is p-nilpotent and every p-chief factor of G below X∩M is cyclic. But
then X is p-nilpotent and every p-chief factor of G below X is
cyclic. This contradiction shows that we have (5).
(6) There is a minimal normal subgroup N of G such that N<P
and the hypothesis does not hold for (G/N,X/N).
Assume that this is false. Let N be a minimal normal subgroup of G
contained in X. Then X/N is p-nilpotent and every chief factor of G/N
between X/N and Op′(X/N) is cyclic by the choice of (G,X). Therefore we have N≰Φ(X) by [8, Corollary 1.6].
Suppose that there is a minimal normal subgroup R≠N of G contained
in R. Then X/N and X/R are p-supersoluble. Hence X≃X/N∩R
is p-supersoluble by Lemma 2.18. Assertion (1) implies that Op′(X)=1. Hence, by Lemma 2.18, X is supersoluble group such that every
chief factor of G between X and N and also every chief factor of G
between X and R are cyclic. But then from the G-isomorphism NR/R≃N we get that X≤Z𝔘(G), a contradiction.
Therefore N is the unique minimal normal subgroup of G contained in X,
which in view of (2) implies that N≤Op(X) and |N|>p.
Now we show that N has a complement V in P. Suppose that N≤Φ(G). Then X<G and so, by Claim (2), so X=P. Since N≰Φ(X),
it follows that P is elementary, so N has a complement in P. Now
suppose that N≤Φ(G). Then there is a maximal subgroup M of G
such that G=N⋊M and so we have that P=P∩NM=N(P∩M)=N⋊(P∩M).
Finally, there are two subgroups L and F such that 1<L<N, L is normal in Gp, F≤V and H=L⋊F is of order pk. Then H is P-embedded in G. Let (H,S,T)∈Σ(H). Then N≤T. Indeed, if X=G, then T is subnormal in G since (|G|,p-1)=1. But |G:T| is a
power of p, so in this case we have N≤T. Now suppose that X≠G,
then Claim (2) implies that X=P. Hence X is elementary, so T∩P is
normal in G since TH=G=TP. Also, P=P∩HT=H(P∩T), so P∩T≠1. Hence N≤T. It follows that L=N∩H=N∩S is a p-number by Lemma 2.7 (5). But then |G:NG(L)|=1 since L is normal in Gp. Thus L is normal in G, which contradicts the minimality of N.
Hence we have (6).
(7) k>1. (This follows from Lemma 2.15 and Claim (5).)
(8) |N|≤pk for every minimal normal subgroup N of G.
Suppose that this is false and let H be a subgroup of N of order pk
such that H is normal Gp. Let (H,S,T)∈Σ(H). Then N≤T. Indeed, we have G=HT, so N=H(N∩T), which implies that N∩T≠1. But N∩T is normal in G. Hence N≤T, so H=N∩H=N∩S=S is generalized S-propermutable in G. Hence H is normal
in G by Lemma 2.7 (5), which contradicts the minimality of N.
(9) pk=4.
In view of Claim (6), it is enough to show that if pk≠4, then the
hypothesis holds on G/N for every minimal normal subgroup N of G. If
either p>2 and |N|<pk or p=2 and |N|<2k-1, then it is
clear. Now let either p>2 and |N|=pk or p=2 and |N|∈{2k,2k-1}.
Suppose that |N|=pk. Then, in view of (4), N is non-cyclic and so
every subgroup of G containing N is not cyclic. Let N≤K≤P,
where |K:N|=p. Since K is non-cyclic, it has a maximal subgroup L≠N. By hypothesis, L is generalized S-propermutable in G and hence K/N=NL/N is generalized S-propermutable in G/N by Lemma 2.7 (6). Thus if
either P/N is abelian or p>2, the hypothesis is true for G/N.
Next suppose that P/N is a non-abelian 2-group. Let N≤V≤P,
where |V:N|=4, V/N is cyclic and |V:K|=2. Since V/N is not
elementary, it follows that N≰Φ(V). Hence for some maximal subgroup of K1 of
V we have V=K1N. Suppose that K1 is cyclic. Then |K1∩N|=2 and 2=|V:K1|=|K1N:K1|=|N:K1∩N|. This implies that |N|=4=pk, a contradiction. Hence K1 is not cyclic. Let S and R
be two distinct maximal subgroup of K1. Then K1=SR. If SN≤K
and RN≤K, we have K1=SR≤K, which implies that V=K1N≤KN≤K, a contradiction. Since N/N<K/N<V/N, where K/N is a maximal
subgroup of V/N, we have that V/N=K1N/N=SRN/N=(SN/N)(RN/N). But since
V/N is cyclic, either V/N=SN/N or V/N=RN/N. Without loss of
generality, we may assume that NS=V. Since S is a maximal subgroup of K1 and K1 is a maximal subgroup of V, |S|=|N|=pk. Thus V/N=NS/N is generalized S-propermutable in G by Lemma 2.7 (6). Therefore
every subgroup of P/N of order p and every cyclic subgroup of P/N of
order 4 (if P/N is a non-abelian 2-group ) are generalized S-propermutable in G/N.
Next suppose that 2k-1=|N| and P/N is non-abelian. If |N|>2, then N is not cyclic and as above one can show that every cyclic subgroup of P/N of order 4 is generalized S-propermutable in G/N. Finally, if |N|=2, then pk=4. This contradiction shows that we have (9).
Final contradiction. In view of (9), pk=4. Hence X is p-nilpotent and every chief factor of G between X and Op′(X)
is cyclic by Lemma 2.16. This contradiction completes the proof of the
result.
∎
Proof of Theorem 1.2.
In view of Lemmas 2.2 and 2.7 (5), every
weakly S-propermutable subgroup is P-embedded. Hence, in fact, Theorem
1.2 is a corollary of Theorem 3.1.
∎
Proof of Corollary 1.4.
First we show that X≤Z𝔘(G). Suppose that this is false and let G be a counterexample with |G|+|X| minimal. Let p be the smallest prime dividing |X| and P a
Sylow p-subgroup of X. Then, by Theorem 1.2, X is p-nilpotent. Let V be the Hall 2′-subgroup of X. It is clear that V is
normal in G and the hypothesis holds for (G,V) and, by Lemma 2.7 (11),
for (G/V,X/V), so the choice of G implies that in the case, when V≠1, V≤Z𝔘(G) and X/V≤Z𝔘(G/V). Hence
X≤Z𝔘(G), a contradiction. Therefore V=1, that is, X=P. But then X≤Z𝔘(G) by Theorem 1.2. Finally, if X=F*(E), then E≤Z𝔘(G) by [16, Theorem B].
∎
Proof of Corollary 1.5.
This corollary directly follows from Corollary 1.4
and Lemma 2.17.
∎
