c*-supplemented subgroups and p -nilpotency of finite groups
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Wang, Youyu Wei, H. 2020-11-02T11:28:54Z 2020-11-02T11:28:54Z 2007 c*-supplemented subgroups and p -nilpotency of finite groups / Youyu Wang, H. Wei // Український математичний журнал. — 2007. — Т. 59, № 8. — С. С. 1011–1019. — Бібліогр.: 13 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/172464 517.42 en Інститут математики НАН України Український математичний журнал Статті c*-supplemented subgroups and p -nilpotency of finite groups c*-доповнені підгрупи та p-нільпотентність скінченних груп Article published earlier |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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c*-supplemented subgroups and p -nilpotency of finite groups Wang, Youyu Wei, H. Статті |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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c*-supplemented subgroups and p -nilpotency of finite groups |
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Wang, Youyu Wei, H. |
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Wang, Youyu Wei, H. |
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c*-доповнені підгрупи та p-нільпотентність скінченних груп |
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c*-supplemented subgroups and p -nilpotency of finite groups / Youyu Wang, H. Wei // Український математичний журнал. — 2007. — Т. 59, № 8. — С. С. 1011–1019. — Бібліогр.: 13 назв. — англ. |
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UDC 517.42
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 such
that 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 ∩ 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 extend
and 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 pidhrupa
iz 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\potentnyj
zalyßok 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 on
the Sylow p-subgroup, he may obtain his desired result. For example, Wielandt proved
that a group G is p-nilpotent if it has a regular Sylow p-subgroup whose G-normalizer
is 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 some
minimal 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 and
the c-supplementation [5, 6] of subgroups. Then, we give several sufficient conditions for
a 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 ofG. ThenG is p-nilpotent ifNG(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�-
supplemented in P ;
* Project supported by NSFC (10571181), NSF of Guangxi (0447038) and Guangxi Education Department.
c© H. WEI, Y. WANG, 2007
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 1011
1012 H. WEI, Y. WANG
(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
quaternion-free.
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
in P ;
(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
quaternion-free.
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 is
p-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
in NG(P );
(b) Every minimal subgroup of P ∩GNp is c�-supplemented inNG(P ) and, if p = 2,
P is quaternion-free.
Our results improve and extend the following theorems of Guo and Shum [7, 8].
Theorem 1.4 ([7], Main theorem). LetG be a group such thatG is S4-free and let P
be a Sylow p-subgroup ofG, where p is the smallest prime divisor of |G|. If every minimal
subgroup of P ∩GN is c-supplemented inNG(P ) and, when p = 2, P is quaternion-free,
then G is p-nilpotent.
Theorem 1.5 ([8], Main theorem). LetP be a Sylow p-subgroup of a groupG,where
p is a prime divisor of |G| with (|G|, p− 1) = 1. If every minimal subgroup of P ∩GN is
permutable inNG(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) inG ifH permutes with every subgroup ofG. H is c-supplemented inG if there
exists a subgroup K1 of G such that G = HK1 and H ∩ K1 ≤ HG = CoreG(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 subgroupK ofG such thatG = HK andH ∩K is a permutable subgroup
of G. We say thatK 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-supplemented
but not permutable in G. If we take G = 〈a, b|a16 = b4 = 1, ba = a3b〉, 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.
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8
c�-SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS 1013
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 inM ;
(2) Let N ✁ G and N ≤ H. Then H is c�-supplemented in G if and only if H/N is
c�-supplemented in G/N ;
(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
in G.
Lemma 2.2. Let c be an element of a groupG 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−1xc = 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
elementary abelian group.
Lemma 2.4 ([10], Lemma 2.8(1)). LetM be a maximal subgroup of a groupG and
let P be a normal p-subgroup of G such that G = PM, where p a prime. Then P ∩M is
a 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
(|G|, p− 1) = 1. Then:
(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) IfM is a subgroup of G of index p, thenM 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 ∈ Sylp(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 )| must
divide (|G|, p − 1) = 1. Thus NG(P ) = CG(P ), and statement (2) follows by the well-
known Burnside theorem.
(3) We may assume thatMG = 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 some
prime q. It is obvious that G =MN andM ∩N is normal in G. ThereforeM ∩N = 1
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8
1014 H. WEI, Y. WANG
and |N | = |G : M | = p. Now N ≤ Z(G) by statement (1) and, of course,M is normal
in G as desired.
3. Proofs of theorems.
Proof of Theorem 1.1. Let G be a minimal counterexample. Then we have the
following claims:
(1) M is p-nilpotent whenever P ≤M < G.
SinceNM (P ) ≤ NG(P ), NM (P ) is p-nilpotent. Let x be an element ofM\NM (P ).
Then, since P ∩P x∩MNp ≤ P ∩P x∩GNp , every minimal subgroup of P ∩P x∩MNp
is c�-supplemented in P by Lemma 2.1. If G satisfies (a), then every cyclic subgroup of
P ∩ P x ∩MNp with order 4 is c�-supplemented in P. If G satisfies (b), then
[Ω2(P ∩ P x ∩MNp), P ] ≤ Z(P ∩GNp) ∩ (P ∩MNp) ≤ Z(P ∩MNp).
Now we see thatM satisfies the hypotheses of the theorem. The minimality of G implies
thatM is p-nilpotent.
(2) Op′(G) = 1.
If not, we consider G = G/N, where N = Op′(G). Clearly NG(P ) = NG(P )N/N
is p-nilpotent, where P = PN/N. For any xN ∈ G\NG(P ), since G
Np = GNpN/N
and P ∩ P xN = P xn for some n ∈ N, we have
P ∩ P xN ∩GNp = (P ∩ P xn ∩GNpN)N/N = (P ∩ P xn ∩GNp)N/N.
Because xN ∈ G\NG(P ), xn ∈ G\NG(P ). Now let P 0 = P0N/N be a minimal
subgroup of P ∩ P xN ∩ GNp
.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 such
that P = P0K0 and P0 ∩ K0 is a permutable subgroup of P. It follows that PN/N =
= (P0N/N)(K0N/N) and (P0N/N)∩ (K0N/N) = (P0∩K0N)N/N. If P0∩K0N =
= P0 then P0 ≤ P ∩ K0N = K0 and consequently P0 = P0 ∩ K0 is permutable in
P. In this case, P 0 is permutable in P . If P0 ∩ K0N = 1 then P 0 is complemented in
P . Thus P 0 is c�-supplemented in P . Assume that G satisfies (a). Let P 1 = P1N/N
be a cyclic subgroup of P ∩ P xN ∩ GNp 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 = P1K1 and P1∩K1 is permutable in P.We have PN/N = (P1N/N)(K1N/N) and
(P1N/N)∩ (K1N/N) = (P1∩K1N)N/N. If P1∩K1N = 1 then P 1 is complemented
in P . If P1 ∩ K1N = 〈z2〉, since z2 ≤ Φ(P ) and 〈z2〉 is c�-supplemented in P, 〈z2〉
is permutable in P by Lemma 2.1. Furthermore, 〈z2〉N/N is permutable in PN/N and
P 1 is c�-supplemented in P . If P1 ∩K1N = P1 then P1 = P1 ∩K1 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
[
Ω2(P ∩ P xN ∩GNp), P
]
=
[
Ω2(P ∩ P xn ∩GNp), P
]
N/N ≤ Z(P ∩GNp)N/N,
namely
[
Ω2(P ∩ P xN ∩GNp), P
]
≤ Z(P ∩GNp).
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 1015
(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. DenoteH = NG(P1). It is obvious that P1 < P0 ≤ P
for some Sylow p-subgroup P0 ofH. The choice of P1 implies thatNG(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,
P0 ∩ P x
0 ∩HNp ≤ P ∩ P x ∩GNp ,
so every minimal subgroup of P0∩P x
0 ∩HNp is c�-supplemented in P0 by Lemma 2.1. If
(a) is satisfied then every cyclic subgroup of P0∩P x
0 ∩HNp of order 4 is c�-supplemented
in P0. If (b) is satisfied then
[
Ω2(P0 ∩ P x
0 ∩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 the
theorem. The choice of G yields that H is p-nilpotent, which is contrary to the choice
of 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 = PQ, 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 normal
in 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 = PQ 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), a
contradiction. Thus G = PQ and so G is solvable. Now let T/Op(G) be a minimal
normal subgroup of G/Op(G) contained in Opp′(G)/Op(G). Then T = Op(G)(T ∩Q).
If T ∩Q < Q, then PT < G and therefore PT 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 = P0Opp′(G). Then P0 is a Sylow p-subgroup of G0. The maximality of P in G
implies that eitherNG(P0) = G orNG(P0) = P. If the latter holds, thenNG0(P0) = P0.
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 the
theorem. 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 = GNpL, 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) = P
or NG(P0) = G. If NG(P0) = P, then NM (P0) = P0, where M = P0T = P0Q.
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8
1016 H. WEI, Y. WANG
Evidently, P0 ∩ P x
0 ∩ MNp ≤ 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 = GNpQ = 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 cyclic
group. Now applying the Frattini argument we have G = GNpNG(Q). Therefore we
may assume that G = GNpL, where L = 〈a〉[Q] is a non-abelian split extension of a
normal 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,
GNpQ/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 = GNpM. 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 letM0 be a maximal subgroup
of M containing P0. Then M0 = P0(M0 ∩ Q) and GNpM0 < G. By applying (1) we
know that GNpM0 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 thatM 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 thatGNp ∩〈a〉 �= 1 ifGNp has exponent p. Then we can take
an element c inGNp∩〈a〉 such that c is of order p. SinceP is not normal inG,GNp∩〈a〉 <
< 〈a〉. Consequently c ∈ 〈ap〉 ≤ Φ(P ) and 〈c〉 is permutable in P. By (6), (7) and
Lemma 2.2, we see that c is centralized by bothGNp andL, hence c ∈ Z(G). IfG 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 NG(P ) = NG(P )/〈c〉 is
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 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 = P1K1 and
P1 ∩ K1 = 1. In general, we may find minimal subgroups P1, P2, . . . , Pm of GNp and
also subgroups K1,K2, . . . ,Km of P such that P = PiKi and Pi ∩Ki = 1 for each i
and every minimal subgroup ofGNp ∩K1∩. . .∩Km is permutable in P. Furthermore, we
may assume that Pi ≤ K1 ∩ . . .∩Ki−1, i = 2, 3, . . . ,m. HenceforthK1 ∩ . . .∩Ki−1 =
= Pi(K1∩. . .∩Ki) for i = 2, 3, . . . ,m. It is easy to see thatGNp∩Ki is normal in P and
(GNp ∩Ki)〈a〉 is a complement of Pi in P, so we may replaceKi by (GNp ∩Ki)〈a〉 and
further 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
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c�-SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS 1017
xa ∈ (GNp ∩K1 ∩ . . . ∩Km) ∩ 〈x〉〈a〉 = 〈x〉.
This means that a induces a power automorphism of p-power order in the elementary
abelian 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〉| = 2n, 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, GN2
is minimal normal in G. In fact, let N be a minimal normal subgroup of G contained
in GN2 and let H = NL. Since N〈a〉 is maximal but not normal in H, we see that
NH(N〈a〉) = N〈a〉.Noticing thatN〈a〉∩HN2 ≤ N, every minimal subgroup ofN〈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, NQ = N × Q and
so 1 �= N ∩ Z(P ) ≤ Z(G). The choice of N implies that N = N ∩ Z(P ) is of order
2. 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 Gaschütz
Theorem [12] (I, 17.4), N also has a complement to G, say M. It follows that M is a
normal 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. For
any minimal subgroup 〈y〉N/N of (G/N)N2 = GN2/N, by the hypotheses, P = 〈y〉K
and 〈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 and
so 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 a
permutation of the set {c1, c2, . . . , cq}. Noticing that ba = b−1 and (c1)a−1ba = (c1)b−1
,
(c2)a = cq. By using (bi)a = b−i and (c1)a−1bia = (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 and
so N1 is normal in G. The minimal normality of GN2 implies that GN2 = N1, thus we
have 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 5.3.8
of [12], there exists some positive integer h such that |P | = 22h+1. Hence |GN2 | = 22h.
However, 22h−1 = (2h +1)(2h−1) and q = 22h−1, hence h = 1, q = 3 and |P | = 23.
Now we see that L ∼= S3 and GN2Q ∼= A4, therefore G ∼= S4, which is contrary to the
hypothesis on G.
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1018 H. WEI, Y. WANG
(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〉 �= 1
then 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-power
order in the elementary abelian 2-group Φ(GN2), hence [Φ(GN2), a] = 1. In view of
Lemma 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, namely
Φ(GN2) ≤ Z(G). Thus we can also take an element c in Φ(GN2) such that c is of order
2 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 NG(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〉T
and 〈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 and
so 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), hence
Φ(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 latter
holds, then [GN2 , Q] ≤ Φ(GN2). This means that Q stabilizes the chain of subgroups
1 ≤ Φ(GN2) ≤ GN2 . It follows from [13] (Theorem 5.3.2) that [GN2 , Q] = 1 and Q is
normal in G, which is impossible. This completes our proof.
Proof of Theorem 1.3. By applying Theorem 1.1, we only need to prove thatNG(P )
is p-nilpotent.
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. By
results of Itô [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 of
exponent p if p > 2 and of exponent at most 4 if p = 2. On the other hand, the minimality
of H implies that HNp = Hp. Let P0 be a minimal subgroup of Hp and let K0 be
a c�-supplement of P0 in H. Then H = P0K0 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 inH. It follows fromK0 is nilpotent thatHq is normal inH, a contradiction.
If P0 ∩K0 = P0 then P0 is permutable in H. In this case, if P0Hq = H, then Hp = P0
is cyclic and H is p-nilpotent by Lemma 2.6, a contradiction. Hence P0Hq < H and
P0Hq = P0 ×Hq. Thus Ω1(Hp) is centralized byHq. If further CH(Ω1(Hp)) < H then
CH(Ω1(Hp)) is nilpotent normal inH. This leads toHq ✁H, a contradiction. Therefore
Ω1(Hp) ≤ Z(H). If Hp has exponent p, then Hp = Ω1(Hp) and H = Hp × Hq,
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c�-SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS 1019
again a contradiction. Thus p = 2 and H2 has exponent 4. If G satisfies (b) then H2 is
quaternion-free and, by Lemma 2.5, Hq acts trivially on H2, thus Hq is normal in H, a
contradiction. Now assume that G satisfies (a). Let P1 = 〈x〉 be a cyclic subgroup of H2
of order 4. Since P1 is c�-supplemented in H, H = P1K1 with P1 ∩ K1 is permutable
in 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 P1Hq is 2-nilpotent
and consequently Hq is normalized by H2. This final contradiction completes our proof.
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Received 03.05.2006
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