Q -permutable subgroups of finite groups
A subgroup $H$ of a group $G$ is called $Q$-permutable in $G$ if there exists a subgroup $B$ of $G$ such that (1) $G = HB$ and (2) if $H_1$ is a maximal subgroup of $H$ containing $H_{QG}$, then $H_1B = BH_1 < G$, where $H_{QG}$ is the largest permutable subgroup of $G$ contained in $H$. In...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508803432185856 |
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| author | Miao, L. Pu, Zh. Мяо, Л. Пу, Ж. |
| author_facet | Miao, L. Pu, Zh. Мяо, Л. Пу, Ж. |
| author_sort | Miao, L. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
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| datestamp_date | 2020-03-18T19:37:24Z |
| description | A subgroup $H$ of a group $G$ is called $Q$-permutable in $G$ if there exists a subgroup $B$ of $G$ such that (1) $G = HB$ and (2) if $H_1$
is a maximal subgroup of $H$ containing $H_{QG}$, then $H_1B = BH_1 < G$, where $H_{QG}$ is the largest permutable subgroup of $G$ contained in $H$. In this paper we prove that:
Let $F$ be a saturated formation containing $U$ and $G$ be a group with a normal subgroup $H$ such that $G/H \in F$. If every maximal subgroup of every noncyclic Sylow subgroup of $F∗(H)$ having no supersolvable supplement in $G$
is $Q$-permutable in $G$, then $G \in F$. |
| first_indexed | 2026-03-24T02:31:01Z |
| format | Article |
| fulltext |
UDC 512.5
Zh. Pu (School Math. and Statistics, Hexi Univ., China),
L. Miao (School Math. Sci., Yangzhou Univ., China)
Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS*
Q-ПЕРЕСТАВНI ПIДГРУПИ СКiНЧЕННИХ ГРУП
A subgroup H of a group G is called Q-permutable in G if there exists a subgroup B of G such that (1)
G = HB and (2) if H1 is a maximal subgroup of H containing HQG, then H1B = BH1 < G, where
HQG is the largest permutable subgroup of G contained in H. In this paper we prove that: Let F be a
saturated formation containing U and G be a group with a normal subgroup H such that G/H ∈ F . If every
maximal subgroup of every noncyclic Sylow subgroup of F ∗(H) having no supersolvable supplement in G
is Q-permutable in G, then G ∈ F .
Пiдгрупу H групи G називають Q-переставною в G, якщо iснує пiдгрупа B групи G така, що: 1) G =
= HB та 2) якщо H1 — максимальна пiдгрупа H, що мiстить HQG, то H1B = BH1 < G, де
HQG є найбiльшою переставною пiдгрупою G, що мiститься в H. У цiй роботi доведено наступне
твердження. Нехай F — насичена формацiя, що мiстить U , а G — група з нормальною пiдгрупою
H такою, що G/H ∈ F . Якщо кожна максимальна пiдгрупа кожної нециклiчної силовської пiдгрупи
F ∗(H), що не має надрозв’язного доповнення в G, є Q-переставною в G, то G ∈ F .
1. Introduction. All groups considered in this paper are finite. Our terminology and
notation are standard (see [2, 6, 12]). In what follows, U denotes the formation of all
supersolvable groups.
It has been of interest to use the supplementation of subgroups to characterize the
structure of a group. In this context, Hall and Kegel proved some interesting results
for solvable groups (see [5, 8, 9]). Recently, by considering some special supplemented
subgroups, Wang introduced the concept of c-normal [14] and Ballester – Bolinches,
Guo and Wang introduced the notion of c-supplemented subgroups [1]. More recently,
A. N. Skiba introduced the concept of weakly s-permutable subgroups [13] and Miao
and Lempken introduced the definition ofM-supplemented subgroups [10]. They used
certain types of supplement to study conditions for solvability and supersolvability of
finite groups.
In these paper, we continue this work and introduce the concept of Q-permutable
subgroups.
Definition 1.1. A subgroup H of a group G is called Q-permutable in G if there
exists a subgroup B of G such that (1) G = HB and (2) if H1 is a maximal subgroup
of H containing HQG, then H1B = BH1 < G, where HQG is the largest permutable
subgroup of G contained in H.
Recall that a subgroup H is calledM-supplemented in G, if there exists a subgroup
B of G such that G = HB and H1B is a proper subgroup of G for every maximal
subgroup H1 of H. Moreover, a subgroup H is called weakly s-permutable in G if there
exists a subnormal subgroup K of G such that G = HK and H ∩ K ≤ HsG where
HsG is the largest s-permutable subgroup of G contained in H.
The following examples indicate that the Q-permutability of subgroups cannot be
deduced from Skiba’s result nor from other related results.
Example 1.1. Let G = S4 be the symmetric group of degree 4 and H = 〈(1234)〉
be a cyclic subgroup of order 4. Then G = HA4 where A4 is the alternating group of
*This research is supported by the grant of NSFC (Grant #10901133)and sponsored by Qing Lan Project
and Natural science fund for colleges and universities in Jiangsu Province.
c© ZH. PU, L. MIAO, 2011
1534 ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 11
Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS 1535
degree 4. Clearly, since A4 E G, we have A4 permutes all maximal subgroups of H
and hence H is Q-permutable in G. On the other hand, we have HsG = 1. Otherwise,
if H is s-permutable in G, then H is normal in G, a contradiction. If HsG = 〈(13)(24)〉
is s-permutable in G, then we also have 〈(13)(24)〉 is normal in G, a contradiction.
Therefore we know that H is not weakly s-permutable in G.
Example 1.2. Let G = S4 be the symmetrical group of degree 4 and H be a Sylow
2-subgroup of G. Clearly, H is Q-permutable in G and G = HA4. Furthermore, H is
notM-supplemented in G.
2. Preliminaries. For the sake of convenience, we first list here some known results
which will be useful in the sequel.
Lemma 2.1. Let G be a group. Then:
(1) If H is Q-permutable in G, H ≤M ≤ G, then H is Q-permutable in M.
(2) Let N E G and N ≤ H. Then H is Q-permutable in G if and only if H/N is
Q-permutable in G/N.
(3) Let π be a set of primes. Let K be a normal π′-subgroup and H be a π-subgroup
of G. If H is Q-permutable in G, then HK/K is Q-permutable in G/K.
(4) Let R be a solvable minimal normal subgroup of G and R1 be a maximal
subgroup of R. If R1 is Q-permutable in G, then R is a cyclic group of prime order.
(5) Let P be a p-subgroup of G where p is a prime divisor of |G|. If P is Q-
permutable in G, then there exists a subgroup B of G such that |G : TB| = p for any
maximal subgroup T of P containing PQG.
Proof. (1) If H is Q-permutable in G, then there exists a subgroup B of G such
that G = HB and TB < G for any maximal subgroup T of H with HQG ≤ T.
Since H ≤ M ≤ G, we have HQG ≤ HQM . Thus we may set L = M ∩ B. Clearly,
L = M ∩B ≤M and M = M ∩HB = H(M ∩B) = HL. Since TB < G for every
maximal subgroup T of H with HQM ≤ T, we easily see that TL = T (M ∩ B) =
= M ∩ TB is a proper subgroup of M.
(2) This follows easily from the definition of Q-permutable subgroups.
(3) If H is Q-permutable in G, then there exists a subgroup B of G such that
G = HB and H1B < G for any maximal subgroup H1 of H containing HQG. Clearly,
(HK/K)(BK/K) = G/K. For any maximal subgroup T/K of HK/K containing
(HK/K)Q(G/K), since K is a normal π′-subgroup and H is a π-subgroup of G,
we have T = T1K where T1 is a maximal subgroup of H containing HQG. There-
fore (T1K/K)(BK/K) = T1BK/K = (BK/K)(T1K/K) < G/K. Otherwise, if
T1BK = G, then |G : T1B| = |K : K ∩ T1B| is a π′-number, on the other hand,
|G : T1B| = |HB : T1B| is a π-number, which is a contradiction.
Conversely, if HK/K is Q-permutable in G/K by the subgroup B/K, we easily
verify that H is Q-permutable in G by B.
(4) If R1 is permutable in G, then R = R1 since the minimal normal subgroup
of G is also a minimal permutable subgroup of G [11]), a contradiction. On the other
hand, if (R1)QG ≤ R1, then there exists a subgroup B of G such that G = R1B and
TB = BT < G for any maximal subgroup T of R1 with (R1)QG ≤ T. If R ∩B = R,
then B = G, a contradiction. If R ∩B = 1, then R is a cyclic subgroup of prime order.
(5) If P is Q-permutable in G, then there exists a subgroup B of G such that
G = PB and TB = BT < G for any maximal subgroup T of P with PQG ≤ T. Since
|P : T | = p, we get |G| = |PB| = p|T ||B|/|P ∩B| = (p/|(P ∩B) : (T ∩B)|) · |TB|.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 11
1536 ZH. PU, L. MIAO
As p is a prime and TB < G, we conclude that P ∩ B = T ∩ B and |G : TB| = p.
Now the claim follows.
Lemma 2.2 ([4], Theorem 1.8.17). LetN be a nontrivial solvable normal subgroup
of a group G. If N ∩ Φ(G) = 1, then the Fitting subgroup F (N) of N is the direct
product of minimal normal subgroups of G which are contained in N.
Lemma 2.3 ([16], Theorem 4.6). If H is a subgroup of G with |G : H| = p, where
p is the smallest prime divisor of |G|, then H E G.
Lemma 2.4 ([3], main theorem). Suppose a finite group G has a Hall π-subgroup
where π is a set of primes not containing 2. Then all Hall π-subgroups of G are conju-
gate.
Lemma 2.5. Let G be a finite group and P be a Sylow p-subgroup of G where p
is the smallest prime divisor of |G|. If every maximal subgroup of P has a p-nilpotent
supplement or is Q-permutable in G, then G/Op(G) is solvable p-nilpotent.
Proof. Assume that the claim is false and choose G to be a counterexample of
smallest order. Furthermore we have,
(1) Op(G) 6= 1.
If 1 < Op(G) ≤ P, then G/Op(G) satisfies the hypotheses and the minimal choice
of G implies that G/Op(G) ∼= (G/Op(G))/Op(G/Op(G)) is p-nilpotent, a contradic-
tion.
(2) Op(G) = 1.
Let P1 be a maximal subgroup of P. If |P | = p, then G is p-nilpotent by Burnside
p-nilpotent Theorem, a contradiction. So we may assume |P | > p2. By hypotheses, if
P1 has a p-nilpotent supplement in G, then there exists a subgroup K of G such that
G = P1K and K is p-nilpotent. Therefore we have Kp′ E K where Kp′ is a Hall
p′-subgroup of K and of course of G. Hence G = P1NG(Kp′ ). If P ∩NG(Kp′ ) = P,
then Kp′ E G, a contradiction. If P ∩NG(Kp′ ) = L where L is a maximal subgroup of
P, then |G : NG(Kp′ )| = |P : P ∩NG(Kp′ )| = |P : L| = p and hence NG(Kp′ ) E G
by Lemma 2.3, a contradiction. So we may assume P ∩ NG(Kp′ ) ≤ L2 < L1 where
L1 is a maximal subgroup of P and L2 is a maximal subgroup of L1. If L1 has a
p-nilpotent supplement in G, then there exists a p-nilpotent subgroup H such that G =
= L1H. With the similar discussion we have G = L1NG(Hp′ ) where Hp′ is the Hall
p′-subgroup of H and of course of G. By Lemma 2.4, there exists an element x of P
such that NG(Kp′ ) = (NG(Hp′ ))x. Therefore G = L1NG(Hp′ ) = (L1NG(Hp′ ))x =
= L1NG(Kp′ ). Furthermore, P = P ∩ L1NG(Kp′ ) = L1(P ∩ NG(Kp′ )) = L1, a
contradiction. Hence L1 is Q-permutable in G, there exists a subgroup B of G such
that G = L1B and TB < G for any maximal subgroup T of L1 with (L1)QG ≤ T.
Moreover, (L1)QG ≤ Op(G) = 1 and hence L1 is M-supplemented in G in this case.
Therefore L2B < G and |G : L2B| = p by Lemma 2.1(5). Since p is the smallest prime
divisor of |G|, Lemma 2.3 implies that L2B E G. We have G = L1B = PB = PL2B
and P ∩ L2B = L2(P ∩ B) is the Sylow p-subgroup of L2B. Clearly, L2(P ∩ B)
is the maximal subgroup of P. By hypotheses if L2(P ∩ B) is Q-permutable in G,
then L2(P ∩ B) is M-supplemented in G and hence M-supplemented in L2B by
[10] (Lemma 2.1(1)). So L2B is p-nilpotent by [10] (Lemma 2.11). Therefore G is
p-nilpotent, a contradiction.
So we may assume L2(P ∩B) has a p-nilpotent supplement in G. With the similar
discussion as above, there exists a p-nilpotent subgroup S such that G = L2(P ∩
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 11
Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS 1537
∩ B)S = L2(P ∩ B)NG(Sp′ ) where Sp′ is a Hall p′-subgroup of S and also of G.
By Lemma 2.4, there exists an element g of P such that NG(Kp′ ) = (NG(Sp′ ))g.
Therefore G = L2(P ∩B)NG(Sp′ ) = (L2(P ∩B)NG(Sp′ ))g = L2(P ∩B)NG(Kp′ ).
Furthermore, P = P ∩L2(P ∩B)NG(Kp′ ) = L2(P ∩B)(P ∩NG(Kp′ )) = L2(P ∩B),
a contradiction.
Therefore G/Op(G) is p-nilpotent.
Lemma 2.6 ([7], Chapter X). Let G be a group and N a subgroup of G. The gen-
eralized Fitting subgroup F ∗(G) of G is the unique maximal normal quasinilpotent
subgroup of G. Then:
(1) if N is normal in G, then F ∗(N) ≤ F ∗(G);
(2) F ∗(G) 6= 1 if G 6= 1; in fact, F ∗(G)/F (G) = Soc (F (G)CG(F (G))/F (G);
(3) F ∗(F ∗(G)) = F ∗(G) ≥ F (G); if F ∗(G) is solvable, then F ∗(G) = F (G);
(4) CG(F ∗(G)) ≤ F (G);
(5) let P E G and P ≤ Op(G); then F ∗(G/Φ(P )) = F ∗(G)/Φ(P );
(6) if K is a subgroup of G contained in Z(G), then F ∗(G/K) = F ∗(G)/K.
Lemma 2.7 ([10], Lemma 2.7). Let G be a finite group with normal subgroups H
and L and let p ∈ π(G). Then the following hold:
1) If L ≤ Φ(G), then F (G/L) = F (G)/L.
2) If L ≤ H ∩ Φ(G), then F (H/L) = F (H)/L.
3) If H is a p-group and L ≤ Φ(H), then F ∗(G/L) = F ∗(G)/L.
4) If L ≤ Φ(G) with |L| = p, then F ∗(G/L) = F ∗(G)/L.
5) If L ≤ H ∩ Φ(G) with |L| = p, then F ∗(H/L) = F ∗(H)/L.
Lemma 2.8 ([15], Theorem 3.1). Let F be a saturated formation containing U , G
a group with a solvable normal subgroup H such that G/H ∈ F . If for every maximal
subgroup M of G, either F (H) ≤ M or F (H) ∩M is a maximal subgroup of F (H),
then G ∈ F . The converse also holds, in the case where F = U .
3. Main results.
Theorem 3.1. Let G be a group and H a normal subgroup of G with G/H ∈ U .
If every maximal subgroup of every noncyclic Sylow subgroup of H has a supersolvable
supplement or is Q-permutable in G, then G is supersolvable.
Proof. Assume that the theorem is false and let G be a counterexample with minimal
order. Then we have following claims:
(1) G is solvable.
By hypotheses and Lemma 2.5, H/Or(H) is solvable r-nilpotent where r is the
smallest prime divisor of |H| and hence G is solvable. Let L be a minimal normal
subgroup of G contained in H. Clearly, L is an elementary abelian p-group for some
prime divisor p of |G|.
(2) G/L ∈ U and L is the unique minimal normal subgroup of G contained in H
such that H ∩ Φ(G) = 1. Furthermore, L = F (H) = CH(L).
Firstly, we check that (G/L,H/L) satisfies the hypotheses for (G,H). We know
that H/L E G/L and (G/L)/(H/L) ∼= G/H is supersolvable. Let Q = QL/L be
a Sylow q-subgroup of H/L. We may assume that Q is a Sylow q-subgroup of H.
If p = q, we may assume that L ≤ P, where P is a Sylow p-subgroup of H. If
L ≤ P = Q, then every maximal subgroup of P/L is of the form P1/L where P1 is
a maximal subgroup of P. If P1/L has no supersolvable supplement in G/L, then P1
has no supersolvable supplement in G, by hypotheses, P1 is Q-permutable in G and
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 11
1538 ZH. PU, L. MIAO
hence P1/L is Q-permutable in G/L by Lemma 2.1(3). Now we assume that p 6= q.
Let Q1 be a maximal subgroup of a Sylow q-subgroup of H. Without loss of generality,
we may assume that Q1 = Q1L/L where Q1 is a maximal subgroup of a Sylow
q-subgroup of H. Clearly, if Q1L/L has no supersolvable supplement in G/L, then
Q1L/L is Q-permutable in G/L by Lemma 2.1(3). Hence G/L satisfies the hypotheses
of the theorem. The minimal choice of G implies that G/L ∈ U . Since U is a saturated
formation, we know that L is the unique minimal normal subgroup of G which is
contained in H and L � Φ(G). By Lemma 2.2 we have F (H) = L. The solvability of
H implies that L ≤ CH(L) = CH(F (H)) ≤ F (H) and so CH(L) = L = F (H).
(3) L is a Sylow subgroup of H.
Let q be the largest prime divisor of |H| and Q be a Sylow q-subgroup of H.
Since G/L is supersolvable, we have H/L is supersolvable. Consequently, LQ/L char
H/L E G/L and hence LQ E G. If p = q, then L ≤ Q E G. Therefore Q ≤ F (H) =
= L and L is a Sylow q-subgroup of H as desired.
Now we assume p < q. Let P be a Sylow p-subgroup of H. Clearly, P is not cyclic.
Otherwise, G/L ∈ U implies that G ∈ U . Then L ≤ P and PQ = PLQ is a subgroup
of H. Note that every maximal subgroup of noncyclic Sylow subgroup of PQ having no
supersolvable supplement in PQ is Q-permutable in PQ by Lemma 2.1(1). Therefore
PQ satisfies the hypotheses for G. If PQ < G, the minimal choice of G implies that
PQ is supersolvable; in particular, Q E PQ. Hence LQ = L×Q and Q ≤ CH(L) ≤ L,
a contradiction.
Now we may assume that G = PQ = H and L < P. Since G/L ∈ U , LQ E G. By
the Frattini argument, G = LNG(Q). Note that L ∩ NG(Q) is normalized by NG(Q)
and L. Since L is the unique minimal normal subgroup of G, we have L∩NG(Q) = 1.
Let P2 be a Sylow p-subgroup of NG(Q). Then LP2 is a Sylow p-subgroup of G.
Choose a maximal subgroup P1 of LP2 such that P2 ≤ P1. Clearly, L
P1 and hence
(P1)G = 1. If P1 is Q-permutable in G, then there exists a subgroup B of G such
that G = P1B and TB < G for any maximal subgroup T of P1 containing (P1)QG.
Furthermore, (P1)QG = 1. Otherwise, if (P1)QG 6= 1, then we have (P1)QG = L
since 1 < (P1)QG ≤ Op(G) = L and a minimal normal subgroup of G and also is
the minimal permutable subgroup of G by [11], contrary to L � P1. We may choose a
maximal subgroup T of P1 with P2 ≤ T. Otherwise, P2 = P1, then we have |L| = p
and hence G/L ∈ U implies that G is supersolvable, a contradiction. By Lemma 2.1(5),
|G : TB| = p. Therefore L ≤ TB or L∩TB = 1. If L∩TB = 1, then |G : TB| = |L| =
= p, a contradiction. Therefore L ≤ TB and hence LP2 ≤ TB, contrary to |G : TB| =
= p.
So P1 has a supersolvable supplement in G, that is, there exists a supersolvable
subgroup K of G such that G = P1K. In fact, K has a normal p-complement Q1 which
is also a Sylow q-subgroup of G. By Sylow’s theorem, there exists an element g ∈ L
such that Qg
1 = Q. Since P1 E LP2, we have that G = P1K = (P1K)g = P1K
g.
Since Kg ∼= K has a normal Sylow q-subgroup and Q = Qg
1 ≤ Kg, it follows that
Kg ≤ NG(Q). Since LP2 = LP2 ∩ G = LP2 ∩ P1K
g = P1(LP2 ∩ Kg), we have
that LP2 ∩Kg
P2. Otherwise, LP2 ≤ P1P2 = P1, a contradiction. Therefore P2 is
a proper subgroup of P3 =< P2, LP2 ∩Kg > . On the other hand, since both P2 and
Kg are contained in NG(Q), P3 is a p-subgroup of NG(Q) which contains a Sylow
p-subgroup P2 of NG(Q) as a proper subgroup, which is a contradiction.
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Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS 1539
(4) G ∈ U .
Let L1 be a maximal subgroup of L. If L1 has a supersolvable supplement in G, then
there exists a supersolvable subgroup K of G such that G = L1K. Since L is a minimal
normal subgroup of G, we have L ∩K ∈ {1, L}. If L ∩K = L, then G = L1K = K,
a contradiction. If L ∩ K = 1, then |L| = p, also a contradiction. Thus we have that
L1 is Q-permutable in G. In this case we know that L is a cyclic subgroup by Lemma
2.1(4), a contradiction.
The final contradiction completes our proof.
Corollary 3.1. Let G be a group. If every maximal subgroup of every noncyclic
Sylow subgroup of G having no supersolvable supplement in G is Q-permutable in G,
then G is supersolvable .
Theorem 3.2. Let F be a saturated formation containing U . Suppose that G is a
finite group with a normal subgroup H such that G/H ∈ F . If every maximal subgroup
of every noncyclic Sylow subgroup of H having no supersolvable supplement in G is
Q-permutable in G, then G ∈ F .
Proof. Assume that the claim is false and choose G to be a counterexample of
minimal order. Since the pair (H,H) satisfies the hypotheses for the pair (G,H) with
H/H ∈ U , H is supersolvable by Theorem 3.1.
Now let p be the largest prime in π(H) and P ∈ Sylp(H); so P = Op(H) E G. Let
L be a minimal normal subgroup of G contained in P. Using similar arguments as for the
proof of Claim (2) of Theorem 3.1 we easily establish that G/L ∈ F and that L is the
unique minimal normal subgroup of G contained in H; moreover, L = F (H) = CH(L)
is noncyclic and H ∩ Φ(G) = 1.
Clearly, Ω1(Z(P )) E G and so L ≤ Ω1(Z(P )); hence P ≤ CH(L) = L and thus
L = P ∈ Sylp(H). The same arguments as the last step of the proof of Theorem 3.1
now yield a contradiction.
Theorem 3.3. Let F be a saturated formation containing U and G be a group
with a solvable normal subgroup H such that G/H ∈ F . If every maximal subgroup of
every noncyclic Sylow subgroup of F (H) having no supersolvable supplement in G is
Q-permutable in G, then G ∈ F .
Proof. Assume that the assertion is false and choose G to be a counterexample of
minimal order.
By Lemma 2.8, we may pick a maximal subgroup M of G not containing F (H).
Actually, since F (H) �M, there at least exists a prime p of π(|H|) with Op(H) �
� M. Then G = Op(H)M and Op(H) ∩M E G. If |Op(H)| = p, then |G : M | = p
and hence G ∈ F by Lemma 2.8, a contradiction. Let Mp be a Sylow p-subgroup of M.
Then we know that Gp = Op(H)Mp is a Sylow p-subgroup of G. Now, let P1 be a max-
imal subgroup ofGp containingMp and set P2 = P1∩Op(H). Then P1 = P2Mp.More-
over, P2 ∩Mp = Op(H)∩Mp, so |Op(H) : P2| = |Op(H)Mp : P2Mp| = |Gp : P1| =
= p, that is, P2 is a maximal subgroup of Op(H). Hence P2(Op(H)∩M) is a subgroup
of Op(H). By the maximality of P2 in Op(H), we have P2(Op(H) ∩M) = P2 or
Op(H).
1) If P2(Op(H)∩M) = Op(H), then G = Op(H)M = P2M. Notice that Op(H)∩
∩M = P2 ∩M. So Op(H) = P2, a contradiction.
2) P2(Op(H) ∩ M) = P2, that is, Op(H) ∩ M ≤ P2. Clearly, Op(H) ∩ M E
E G, so Op(H) ∩M ≤ (P2)G. On the other hand, if P2 has a supersolvable supple-
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1540 ZH. PU, L. MIAO
ment in G, then there exists a supersolvable subgroup N of G such that G = P2N.
Set K = (P2)GN, then G = P2N = P2K and K/K ∩ (P2)G = K/(P2)G =
= (P2)GN/(P2)G ∼= N/N ∩ (P2)G ∈ U ⊆ F .
Now, we consider the following cases.
a) K < G. Suppose that K1 is a maximal subgroup of G containing K. Then
Op(H)∩K1 E G, which implies that (Op(H)∩K1)M is a subgroup of G. If (Op(H)∩
∩K1)M = G = Op(H)M, then Op(H) ∩K1 = Op(H) since (Op(H) ∩K1) ∩M =
= Op(H) ∩M. This implies that Op(H) ≤ K1, and hence G = Op(H)K1 = K1,
which is contrary to the above hypotheses on K1. Thus (Op(H) ∩ K1)M = M and
hence Op(H) ∩ K1 ≤ M. Furthermore, P2 ∩ K ≤ Op(H) ∩ K ≤ Op(H) ∩ M ≤
≤ (P2)G ≤ P2 ∩K, that is, Op(H) ∩K = Op(H) ∩M = P2 ∩K. This is contrary to
G = P2K = Op(H)K.
b) K = (P2)GN = G. In this case, if (P2)G = 1, then N = G ∈ F , a contradiction.
So we may assume that (P2)G 6= 1. Thus (P2)GM = M or G. If (P2)GM = G,
then G = (P2)GM = Op(H)M = P2M. Note that Op(H) ∩ M = P2 ∩ M, so
Op(H) = P2, a contradiction. Therefore (P2)GM = M. It follows from (P2)G ≤
≤ Op(H) ∩M ≤ (P2)G that Op(H) ∩M = (P2)G. By hypotheses, G/(P2)G ∈ U
implies that |G/(P2)G : M/(P2)G| = |G : M | = p. This is contrary to the choice of M.
So we may assume that P2 is Q-permutable in G. There exists a subgroup B of G
such that P2B = G and TB < G for any maximal subgroup T containing (P2)QG.
Assume P2 is permutable in G. The maximality of M in G implies P2M = M or G. If
P2M = G, then G = Op(H)M = P2M and hence Op(H) = P2 since Op(H) ∩M =
= P2∩M, a contradiction. Thus Op(H)∩M = P2∩M = P2 and hence |F (H) : F (H)∩
∩M | = |G : M | = |Op(H) : Op(H) ∩M | = p, a contradiction.
Finally we may assume (P2)QG < P2. For any maximal subgroup T of P2 con-
taining (P2)QG, we have |G : TB| = p by Lemma 2.1(5). Clearly, TB is a maximal
subgroup of G. Then Op(H) ∩ TB E G, which implies that (Op(H) ∩ TB)M is a
subgroup of G. If (Op(H) ∩ TB)M = G = Op(H)M, then Op(H) ∩ TB = Op(H)
since (Op(H) ∩ TB) ∩M = Op(H) ∩M. This leads to Op(H) ≤ TB, and hence
G = Op(H)TB = TB, which is contrary to the above hypotheses on TB. Thus
Op(H) ∩ TB ≤ M. Furthermore, P2 ∩ TB ≤ Op(H) ∩ TB ≤ Op(H) ∩ M ≤
≤ (P2)QG ≤ P2 ∩ TB, from this, Op(H) ∩ TB = Op(H) ∩M = P2 ∩ TB. This
is contrary to G = P2B = Op(H)B.
The final contradiction completes our proof.
Corollary 3.2. Let G be a group with a solvable normal subgroup H such that
G/H ∈ U . If every maximal subgroup of every noncyclic Sylow subgroup of F (H)
having no supersolvable supplement in G is Q-permutable in G, then G ∈ U .
Theorem 3.4. Let F be a saturated formation containing U and G be a group
with a normal subgroup H such that G/H ∈ F . If every maximal subgroup of every
noncyclic Sylow subgroup of F ∗(H) having no supersolvable supplement in G is Q-
permutable in G, then G ∈ F .
Proof. Suppose that the theorem is false and choose G to be a counterexample of
minimal order; so in particular, H 6= 1. We consider the following two cases.
Case 1. F = U .
By Corollary 3.1 we easily verify that F ∗(H) is supersolvable and hence F (H) =
= F ∗(H) 6= 1 by Lemma 2.6(3). Since H satisfies the hypotheses of the theorem,
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Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS 1541
the minimal choice of G implies that H is supersolvable if H < G. Then G ∈ U by
Corollary 3.2, a contradiction. Thus we have
(1) H = G, F ∗(G) = F (G) 6= 1.
Let S be a proper normal subgroup of G containing F ∗(G). By Lemma 2.6(1),
F ∗(G) = F ∗(F ∗(G)) ≤ F ∗(S) ≤ F ∗(G), so F ∗(S) = F ∗(G). By hypotheses and
Lemma 2.1(1), every maximal subgroup of every noncyclic Sylow subgroup of F ∗(S)
having no supersolvable supplement in S is Q-permutable in S. Hence S is supersolv-
able by the minimal choice of G and we get
(2) Every proper normal subgroup of G containing F ∗(G) is supersolvable.
Suppose now that Φ(Op(G)) 6= 1 for some p ∈ π(F (G)). By Lemma 2.6(5) we
have F ∗(G/Φ(Op(G))) = F ∗(G)/Φ(Op(G)). Using Lemma 2.1(2) we observe that
the pair (G/Φ(Op(G)), F ∗(G)/Φ(Op(G))) satisfies the hypotheses of the theorem. The
minimal choice of G then implies G/Φ(Op(G)) ∈ U . Since U is a saturated formation,
we get G ∈ U , a contradiction. Thus we have
(3) If p ∈ π(F (G)), then Φ(Op(G)) = 1 and Op(G) is elementary abelian; in
particular, F ∗(G) = F (G) is abelian and CG(F (G)) = F (G).
If L is a minimal normal subgroup of G contained in F (G) and |L| = p where
p ∈ π(F (G)), then set C = CG(L). Clearly, F (G) ≤ C E G. If C < G, then C is
solvable by (2). On the other hand, since G/C is cyclic, then we have G is solvable, a
contradiction. So we may assume C = G. Now we have L ≤ Z(G). Then we consider
subgroup G/L. By Lemma 2.6(6), we have F ∗(G/L) = F ∗(G)/L = F (G)/L. In fact,
G/L satisfies the condition of the theorem by Lemma 2.1. Therefore the minimal choice
of G implies that G/L ∈ U and hence G is supersolvable, a contradiction. This proves
(4) There is no normal subgroup of prime order in G contained in F (G).
If every Sylow subgroup of F (G) is cyclic,then F (G) = H1 × . . .×Hr where Hi,
i = 1, . . . , r, is the cyclic Sylow subgroup of F (G) and hence G/CG(Hi) is abelian for
any i ∈ {1, . . . , r}. Moreover, we have G/
⋂r
i=1
CG(Hi) = G/CG(F (G)) is abelian
and hence G/F (G) is abelian since CG(F (G)) = CG(F ∗(G)) ≤ F (G). Therefore G
is solvable, a contradiction. This proves that
(5) There exists noncyclic Sylow subgroup Op(G) of F (G) for some prime p ∈
∈ π(F (G)).
Let P1 be a maximal subgroup of Op(G). If P1 has a supersolvable supplement
in G, then there exists a supersolvable subgroup K of G such that G = P1K =
= Op(G)K. Clearly, G/Op(G) ∼= K/K ∩ Op(G) is supersolvable and hence G is
solvable, a contradiction. So we obtain that
(6) Every maximal subgroup of every noncyclic Sylow subgroup of F (G) has no
supersolvable supplement in G.
Set R = Op(G)∩Φ(G). If R = 1, then by Lemma 2.2, Op(G) is the direct product
of some minimal normal subgroup of G. So we may assume that Op(G) = R1×. . .×Rt,
where Ri is a minimal normal subgroup of G, i = 1, 2, . . . , t. Consider the maximal
subgroup P1 of P, where P1 has the following form:
P1 = R1 × . . .×Ri−1 ×Ri
∗ ×Ri+1 × . . .×Rt.
Where Ri
∗ is a maximal subgroup of Ri for some i. By hypotheses and (6), P1 is Q-
permutable in G. Let T denote the normal subgroup R1×. . .×Ri−1×Ri+1×. . .×Rt of
G, then P1 = Ri
∗T. Let T denote the normal subgroup R1×. . .×Ri−1×Ri+1×. . .×Rt
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1542 ZH. PU, L. MIAO
of G, then P1 = Ri
∗T. Clearly, T ≤ (P1)QG and P/T is the minimal normal subgroup
of G/T. Since (P1)QG/T is a permutable subgroup of G/T, by [11] a minimal normal
subgroup of G is also the minimal permutable subgroup of G, we have (P1)QG = T. By
hypotheses P1 is Q-permutable in G, there exists a subgroup B of G such that G = P1B
and SB = BS < G for any maximal subgroup S of P1 containing (P1)QG = T. By
Lemma 2.1(5) we have |G : SB| = p. Since Ri is the minimal normal subgroup of G,
we have Ri ∩ SB ∈ {1, Ri}. Clearly, if Ri ≤ SB, then we have SB = RiSB = G, a
contradiction. Hence we have Ri � SB, we know that |Ri| = p, contrary to (4). This
contradiction leads to
(7) R = Op(G) ∩ Φ(G) 6= 1.
LetQ be a Sylow q-subgroup of F (G), and let L be a minimal normal subgroup ofG
contained in R, where q 6= p. Then Q is elementary abelian by (3). By the definition of a
generalized Fitting subgroup, F ∗(G/L) = F (G/L)E(G/L) and [F (G/L), E(G/L)] =
= 1, where E(G/L) is the layer of G/L. Since L ≤ Φ(G), F (G/L) = F (G)/L. Now
set E/L = E(G/L). Since Q is normal in G and [F (G)/L,E/L] = 1, [Q,E] ≤ Q ∩
∩ L = 1, i.e., [Q,E] = 1. Therefore F (G)E ≤ CG(Q). If CG(Q) < G, then CG(Q)
is supersolvable by (2). Thus E(G/L) = E/L is supersolvable. The semisimplicity of
E(G/L)/Z(E(G/L)) implies that E(G/L) = Z(E(G/L)). So E(G/L) ≤ F (G/L)
and F ∗(G/L) = F (G)/L, with the same argument in (3), we have that G/L satisfies
the hypotheses of the theorem. By the minimal choice of G, G/L is supersolvable
and so is G, a contradiction. If CG(Q) = G, then Q ≤ Z(G). By Lemma 2.6(6),
F ∗(G/Q) = F ∗(G)/Q = F (G)/Q. Similarly, G/Q is supersolvable and so is G by
Corollary 3.2, a contradiction. This verifies
(8) F (G) = Op(G).
If R = Op(G), then by hypotheses every maximal subgroup P1 of Op(G) is Q-
permutable in G. That is, there exists a subgroup B of G such that G = P1B and
TB < G for any maximal subgroup T of P1 containing (P1)QG. Then G = P1B = B
since P1 ≤ Φ(G), a contradiction. Hence R 6= Op(G). Now Φ(G/R) = 1. Then
by Lemma 2.2, Op(G)/R = (H1/R) × . . . × (Hm/R), where Hi/R, i = 1, . . . ,m,
are minimal normal in G/R. With the same argument as in (7), we know that Hi/R,
i = 1, . . . ,m, are all of order p because all maximal subgroups of Op(G)/R are Q-
permutable in G/R by Lemma 2.1(2). Again, since Op(G) is an elementary abelian
p-group, Hi is of the form 〈xi〉R, i = 1, . . . ,m. This proves
(9) Op(G) = 〈x1〉 × . . .× 〈xm〉 ×R where 〈xi〉 6= 1 and 〈xi〉R E G, i = 1, . . . ,m.
Now let L be a minimal normal subgroup of G contained in R and set G := G/L.
Clearly, F (G) = F (G)/L = Op(G)/L, because L ≤ Φ(G). If F ∗(G) = F (G), then
we easily verify that G/L satisfies the hypotheses of the theorem, and thus G/L ∈
∈ U by the minimal choice of G. Since L ≤ Φ(G) and U is saturated, we get G ∈
∈ U , a contradiction. Therefore F ∗(G) = F (G)E(G) > F (G) and so there exists
a perfect normal subgroup E in G such that EL/L = E(G). Clearly, Op(G)E is a
nonsolvable normal subgroup of G; hence, by (2) and (1), G = Op(G)E. In particular,
G = Op(G)E(G) = F ∗(G) with [Op(G), E(G)] = [F (G), E(G)] = 1 and hence
[Op(G), E] ≤ L. Since L is minimal normal in G and since [Op(G), E] E G as well as
CG(Op(G)) = Op(G), we get [Op(G), E] = L. Therefore we have the following:
(10) G = Op(G)E with L = [Op(G), E] ≤ Op(G) ∩ E , G = Op(G)E = F ∗(G)
and Op(G) ≤ Z(G).
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Q-PERMUTABLE SUBGROUPS OF FINITE GROUPS 1543
Now assume that M is a minimal normal subgroup of G contained in Op(G) with
M 6= L. Then M = ML/L is a minimal normal subgroup of G contained in Z(G). As
M ∩ L = 1, we get |M | = |M | = p, contrary to (4). This proves
(11) L is the unique minimal normal subgroup of G contained in Op(G).
Now let T be a complement of L in Op(G) and set P1 := TL1 where L1 is a
maximal subgroup of L. Then P1 is a maximal subgroup of Op(G) and so, by hy-
potheses, is Q-permutable in G with G = P1B = PB and SB = BS < G for
any maximal subgroup S containing (P1)QG. Obviously, (P1)QG is normalized by
Op(G)Op(G) = Op(G)Op(E) = Op(G)E = G. Since (P1)QG does not contain L,
we conclude (P1)QG = 1 by (11).
By Lemma 2.1(5), |G : SB| = p. Clearly, SB is a maximal subgroup of G, and
L ∩ SB ∈ {1, L}. If L ∩ SB = 1, then |L| = p, this is contrary to (4). So we have
L ≤ SB for any maximal subgroup S of P1. Furthermore, if L ∩ P1 = 1, then we also
have |L| = p, a contradiction. So we get L ∩ P1 6= 1. We claim that L ∩ P1 ≤ S for
any maximal subgroup S of P1. Otherwise, there exists a maximal subgroup S of P1
such that L∩P1 � S. So we consider SB = (L∩P1)SB = P1B = G, a contradiction.
Based on the discussion as above, we have 1 < L∩P1 ≤ Φ(P1) ≤ Φ(Op(G)), contrary
to (3), thereby completing the proof for Case I .
Case 2. F 6= U .
By case 1, H is supersolvable. Particularly, H is solvable and hence F ∗(H) =
= F (H). Therefore G ∈ F by Corollary 3.2.
The final contradiction completes our proof.
Corollary 3.3. Let G be a group with a normal subgroup H such that G/H ∈ U .
If every maximal subgroup of every noncyclic Sylow subgroup of F ∗(H) having no
supersolvable supplement in G is Q-permutable in G, then G ∈ U .
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Received 22.02.11
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| id | umjimathkievua-article-2823 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:31:01Z |
| publishDate | 2011 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/20/d29b682d6be84a2e6c36ce3a1e6ddc20.pdf |
| spelling | umjimathkievua-article-28232020-03-18T19:37:24Z Q -permutable subgroups of finite groups Q-переставнi пiдгрупи скiнченних груп Miao, L. Pu, Zh. Мяо, Л. Пу, Ж. A subgroup $H$ of a group $G$ is called $Q$-permutable in $G$ if there exists a subgroup $B$ of $G$ such that (1) $G = HB$ and (2) if $H_1$ is a maximal subgroup of $H$ containing $H_{QG}$, then $H_1B = BH_1 < G$, where $H_{QG}$ is the largest permutable subgroup of $G$ contained in $H$. In this paper we prove that: Let $F$ be a saturated formation containing $U$ and $G$ be a group with a normal subgroup $H$ such that $G/H \in F$. If every maximal subgroup of every noncyclic Sylow subgroup of $F∗(H)$ having no supersolvable supplement in $G$ is $Q$-permutable in $G$, then $G \in F$. Пiдгрупу $H$ групи $G$ називають $Q$-переставною в $G$, якщо iснує пiдгрупа $B$ групи $G$ така, що: 1) $G = HB$ та 2) якщо $H_1$ — максимальна пiдгрупа $H$, що мiстить $H_{QG}$, то $H_1B = BH_1 < G$, де $H_{QG}$ є найбiльшою переставною пiдгрупою $G$, що мiститься в $H$. У цiй роботi доведено наступне твердження. Нехай $F$ — насичена формацiя, що мiстить $U$, а $G$ — група з нормальною пiдгрупою $H$ такою, що $G/H \in F$. Якщо кожна максимальна пiдгрупа кожної нециклiчної силовської пiдгрупи $F∗(H)$, що не має надрозв’язного доповнення в $G$, є $Q$-переставною в $G$, то $G \in F$. Institute of Mathematics, NAS of Ukraine 2011-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2823 Ukrains’kyi Matematychnyi Zhurnal; Vol. 63 No. 11 (2011); 1534-1543 Український математичний журнал; Том 63 № 11 (2011); 1534-1543 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2823/2399 https://umj.imath.kiev.ua/index.php/umj/article/view/2823/2400 Copyright (c) 2011 Miao L.; Pu Zh. |
| spellingShingle | Miao, L. Pu, Zh. Мяо, Л. Пу, Ж. Q -permutable subgroups of finite groups |
| title | Q -permutable subgroups of finite groups |
| title_alt | Q-переставнi пiдгрупи скiнченних груп |
| title_full | Q -permutable subgroups of finite groups |
| title_fullStr | Q -permutable subgroups of finite groups |
| title_full_unstemmed | Q -permutable subgroups of finite groups |
| title_short | Q -permutable subgroups of finite groups |
| title_sort | q -permutable subgroups of finite groups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2823 |
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