On Nearly ℳ-Supplemented Subgroups of Finite Groups
A subgroup H is called nearly ℳ-supplemented in a finite group G if there exists a normal subgroup K of G such that HK ⊴ G and TK < HK for every maximal subgroup T of H. We obtain some new results on supersoluble groups and their formation by using nearly ℳ-supplemented subgroups and study th...
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Institute of Mathematics, NAS of Ukraine
2014
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508044686786560 |
|---|---|
| author | Guo, J. Miao, L. Zhang, Jiajia Го, Й. Мяо, Л. Чжан, Їайія |
| author_facet | Guo, J. Miao, L. Zhang, Jiajia Го, Й. Мяо, Л. Чжан, Їайія |
| author_sort | Guo, J. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T10:24:29Z |
| description | A subgroup H is called nearly ℳ-supplemented in a finite group G if there exists a normal subgroup K of G such that HK ⊴ G and TK < HK for every maximal subgroup T of H. We obtain some new results on supersoluble groups and their formation by using nearly ℳ-supplemented subgroups and study the structure of finite groups. |
| first_indexed | 2026-03-24T02:18:57Z |
| format | Article |
| fulltext |
UDC 512.5
J. Guo, J. Zhang (College Math. and Statistics, Yi li Normal Univ., Xinjiang, China),
L. Miao (School Math. Sci., Yangzhou Univ., China)
ON NEARLY M-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS*
ПРО МАЙЖЕ M-ДОПОВНЕНI ПIДГРУПИ СКIНЧЕННИХ ГРУП
A subgroup H is called nearly M-supplemented in a finite group G if there exists a normal subgroup K of G such that
HK �G and TK < HK for every maximal subgroup T of H. We obtain some new results on supersoluble groups and
formation by using nearlyM-supplemented subgroups and investigate the structure of finite groups.
Пiдгрупу H називаємо майже M-доповненою в скiнченнiй групi G, якщо iснує нормальна пiдгрупа K групи G
така, що HK � G i TK < HK для кожної максимальної пiдгрупи T групи H. Отримано деякi новi результати
про суперрозв’язнi групи та їх утворення за допомогою майже M-доповнених пiдгруп та дослiджено структуру
скiнченних груп.
1. Introduction. It is well-known that supplemented subgroups play an important role in the theory
of finite groups. For instance, Hall [4] proved that a group G is soluble if and only if every Sylow
subgroup of G is complemented in G. Srinivasan [10] proved that a finite group is supersoluble if
every maximal subgroup of every Sylow subgroup is normal. Later on, by considering some special
supplemented subgroups (c-supplemented subgroups), Wang [12] proved that G is soluble if and
only if every Sylow subgroup of G is c-supplemented in G. Recently, Miao and Lempken [6] consid-
eredM-supplemented subgroups of finite groups G and obtained some characterization of saturated
formations containing all supersoluble groups. More recently, Wang and Guo [13] introduced the
concept of nearly s-normal subgroups and obtained some interesting results.
Now, we introduce the following concept of nearlyM-supplemented subgroups.
Definition 1.1. A subgroup H is called nearly M-supplemented in group G, if there exists a
normal subgroup K of G such that HK �G and TK < HK for every maximal subgroup T of H.
The following examples indicate that the nearlyM-supplementation of subgroups can neither be
deduced fromM-supplementation of subgroup nor from nearly s-normality of subgroup.
Example 1.1. Let G = S4. Since A4 is normal in G, clearly, A4 is nearly M-supplemented in
G, but A4 is notM-supplemented in G.
Example 1.2. Let G = S4 and H = 〈(1234)〉 be a cyclic subgroup of order 4. Then G = HA4,
where A4 is the alternating group of degree 4. Clearly, since A4 E G, we have H is nearly M-
supplemented in G, but H is not nearly s-normal in G. Otherwise, there exists a normal subgroup
K of G such that HK � G and H ∩ K 6 HsG, we have HsG = 1. Otherwise, if HsG = H is
s-permutable in G, then H is normal in G, a contradiction. If HsG = 〈(13)(24)〉 is s-permutable
in G, then 〈(13)(24)〉 is normal in G, a contradiction. But H ∩K 6= 1. Therefore H is not nearly
s-normal in G.
All the groups in this paper are finite. Most of the notation is standard and can be found in [1]
and [9].
2. Preliminaries. For the sake of convenience, we first list here some known results which will
be useful in the sequel.
* This research is supported by the grant of NSFC (Grant #11271016).
c© J. GUO, J. ZHANG, L. MIAO, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 1 63
64 J. GUO, J. ZHANG, L. MIAO
Lemma 2.1. Let G be a group. Then:
(1) If H is nearlyM-supplemented in G, H ≤ K ≤ G, then H is nearlyM-supplemented in K.
(2) Let N E G and N ≤ H. If H is nearly M-supplemented in G, then H/N is nearly
M-supplemented in G/N.
(3) Let π be a set of primes. Let N be a normal π′-subgroup and let H be a π-subgroup of G.
If H is nearlyM-supplemented in G, then HN/N is nearlyM-supplemented in G/N.
(4) Let R be a soluble minimal normal subgroup of a group G. If there exists a maximal subgroup
R1 of R such that R1 is nearlyM-supplemented in G, then R is a cyclic group of prime order.
Proof. (1) – (4) follow from the definition of nearlyM-supplemented subgroups.
Lemma 2.2 ([7], Lemma 2.6). If H is a subgroup of a group G with |G : H| = p, where p is a
prime divisor of |G| and (|G|, p− 1) = 1, then H E G.
Lemma 2.3 ([8], Lemma 2.7). Let G be a finite group and P 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 and only if P is M-
supplemented in G.
Lemma 2.4 ([2], Theorem 1.8.17). Let N be a nontrivial soluble 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 2.5 ([14], 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 any 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 .
Lemma 2.6 ([3], Main theorem). Suppose that 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 conjugate.
Lemma 2.7. Let F be a formation and G be a group. Suppose that a subgroup H of G has a
F-supplement in G. Then:
(1) If N E G, then HN/N has a F-supplement in G/N.
(2) If H ≤ K ≤ G, then H has a F-supplement in K.
Lemma 2.8 ([8], Lemma 2.9). Let G be a finite group and P be a Sylow p-subgroup of G,
where p is the smallest prime divisor of |G|. Then G is p-nilpotent if and only if every maximal
subgroup of P having no p-nilpotent supplement in G isM-supplemented in G.
Lemma 2.9. Let G be a group and N a subgroup of G. The generalized 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) If P E G with P ≤ Op(G), then F ∗(G/Φ(P )) = F ∗(G)/Φ(P ).
(6) If K ≤ Z(G), then F ∗(G/K) = F ∗(G)/K.
Lemma 2.10 ([6], 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) Φ(L) ≤ Φ(G).
(2) If L ≤ Φ(G), then F (G/L) = F (G)/L.
(3) If L ≤ H ∩ Φ(G), then F (H/L) = F (H)/L.
(4) If H is a p-group and L ≤ Φ(H), then F ∗(G/L) = F ∗(G)/L.
(5) If L ≤ Φ(G) with |L| = p, then F ∗(G/L) = F ∗(G)/L.
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ON NEARLYM-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 65
(6) If L ≤ H ∩ Φ(G) with |L| = p, then F ∗(H/L) = F ∗(H)/L.
3. Main results.
Theorem 3.1. Let G be a finite group and 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 and only if every maximal subgroup of
P having no p-nilpotent supplement is nearlyM-supplemented in G.
Proof. As the necessity part is obvious, we only need to prove the sufficiency part. Assume that
the theorem is false and choose G to be a counterexample of minimal order. Moreover, we have
(1) Op′(G) = 1.
In fact, if Op′(G)6=1, then we consider the quotient group G/Op′(G). By Lemmas 2.1 and 2.7,
it is easy to see that G/Op′(G) satisfies the hypotheses of our theorem. The minimal choice of G
implies that G/Op′(G) is p-nilpotent and hence G is p-nilpotent, a contradiction.
(2) Op(G) 6= 1.
Assume that Op(G)=1. If every maximal subgroup P1 of P has a p-nilpotent supplement in G,
then G is p-nilpotent, a contradiction. So there at least exists a maximal subgroup P1 of P such
that P1 is nearly M-supplemented in G. Then there exists a normal subgroup K1 of G such that
P1K1 � G and TK1 < ·P1K1 for every maximal subgroup T of P1. Furthermore, if P1K1 < G,
then (P1K1)p = P1 or (P1K1)p = P. If (P1K1)p = P1, then P1 is M-supplemented in P1K1. By
Lemma 2.3, P1K1 is p-nilpotent, it follows from (P1K1)p′ char P1K1�G and (1) that (P1K1)p′ = 1.
Therefore P1K1 = P1 � G, a contradiction. Hence we have (P1K1)p = P. Obviously, P1K1
satisfies the hypotheses of the theorem and hence P1K1 is p-nilpotent by the choice of G. With
the similar discussion as above, we also get a contradiction. So we assume that P1K1 = G. This
means every maximal subgroup of P having no p-nilpotent supplement in G isM-supplemented in
G. By Lemma 2.8, G is p-nilpotent also a contradiction.
(3) G has a unique minimal normal subgroup N contained in Op(G) such that N = Op(G) =
= F (G).
Let N be a minimal normal subgroup contained in Op(G). Obviously, G/N satisfies the condition
of the theorem and hence G/N is p-nilpotent by the choice of G. Since the class of all p-nilpotent
groups is a saturated formation, we have N is the unique minimal normal subgroup of G contained
in Op(G) and Φ(G) = 1. By Lemma 2.4 N = Op(G) = F (G).
(4) Final contradiction.
Since N is the unique minimal normal subgroup of G contained in Op(G) and N � Φ(G),
there exists a maximal subgroup M of G such that N � M. Then G = NM and N ∩M = 1,
P = NMp where Mp is a Sylow p-subgroup of M. Since G/N ∼= M is p-nilpotent and (3), we have
G = NNG(Mp′) = PNG(Mp′) where Mp′ is a Hall p′-subgroup of M and of course of G. So we
may assume Mp ≤ P ∩NG(Mp′) ≤ L2 < ·L1 < ·P. Otherwise, if P∩NG(Mp′)=P, then Mp′ �G, a
contradiction. If |P : P ∩NG(Mp′)| = |G : NG(Mp′)| = p and hence NG(Mp′)�G by Lemma 2.2,
a contradiction. If L1 has a p-nilpotent supplement K in G such that G = L1NG(Kp′) where Kp′
is a Hall p′-subgroup of K and of course of G. By Lemma 2.6, there exists an element x of L1
such that NG(Mp′) = (NG(Kp′))
x. Therefore G = L1NG(Kp′) = (L1NG(Kp′))
x = L1NG(Mp′).
Moreover, P = P ∩ L1NG(Mp′) = L1(P ∩ NG(Mp′)) = L1, a contradiction. So we may assume
L1 is nearlyM-supplemented in G, there exists a normal subgroup B of G such that L1B �G and
TB < L1B for every maximal subgroup T of L1. We will divide into the following two cases.
(a) L1B < G.
If (L1B)p = L1, then L1 isM-supplemented in L1B and hence L1B is p-nilpotent by Lemma 2.3.
It follows from (L1B)p′ char L1B�G and (1) that L1B = L1�G, a contradiction. If (L1B)p = P,
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 1
66 J. GUO, J. ZHANG, L. MIAO
L1B satisfies the condition of the theorem by Lemmas 2.1 and 2.7, the minimality of G im-
plies that L1B is p-nilpotent. So (L1B)p′ char L1B � G. With the similar discussion as above,
L1B = P = N�G by (3) and L1 is the maximal subgroup of N with L1 is nearlyM-supplemented
in G, hence we have |N | = p by Lemma 2.1(4) and G/N is p-nilpotent. Therefore G is p-nilpotent,
a contradiction.
(b) L1B = G.
That is, L1 isM-supplemented in G. For every maximal subgroup T of L1, |G : TB| = p and
hence TB � G by Lemma 2.2. Set T = L2 and N ≤ L2B or N ∩ L2B = 1. If N ∩ L2B = 1,
then |N | = |G : L2B| = p, a contradiction. If N ≤ L2B, since Mp ≤ P ∩ NG(Mp′) ≤ L2 and
P = NL2, it follows that L1B = PB = NL2B = L2B < G, a contradiction.
Theorem 3.1 is proved.
Theorem 3.2. Let G be a finite group where p is an odd prime divisor of |G|. Then G is
p-nilpotent if and only if NG(P ) is p-nilpotent and every maximal subgroup of P is nearly M-
supplemented in G.
Proof. As the necessity part is obvious, we only need to prove the sufficiency part. Assume that
the assertion is false and choose G to be a counterexample of minimal order. Then
(1) Op′(G) = 1.
Suppose that L = Op′(G) 6=1, we consider the factor group G/L. Clearly, P1L/L is a maximal
subgroup of Sylow p-subgroup of G/L where P1 is a maximal subgroup of Sylow p-subgroup of P.
Since P1 is nearlyM-supplemented in G, we have P1L/L is also nearlyM-supplemented in G/L
by Lemma 2.1(3). On the other hand, NG/L(PL/L) = NG(P )L/L ([6], Lemma 3.6.10) and so it is
p-nilpotent. Therefore G/L satisfies the hypotheses of the theorem. The minimal choice of G implies
that G/L is p-nilpotent, and hence G is p-nilpotent, a contradiction.
(2) If M is a proper subgroup of G with P ≤M < G, then M is p-nilpotent.
Clearly, NM (P ) 6 NG(P ) and hence NM (P ) is p-nilpotent. Applying Lemma 2.1(1), we find
that M satisfies the hypotheses of our theorem. Now, the minimal choice of G implies that M is
p-nilpotent.
(3) G = PQ, where Q is a Sylow q-subgroup of G with q 6= p.
Since G is not p-nilpotent, by Thompson ([11], Corollary), there exists a characteristic subgroup
H of P such that NG(H) is not p-nilpotent. Since NG(P ) is p-nilpotent, we may choose a charac-
teristic subgroup H of P such that NG(H) is not p-nilpotent, but NG(K) is p-nilpotent for every
characteristic subgroup K of P with H < K ≤ P. Since NG(H) ≥ NG(P ) and NG(H) is not
p-nilpotent, we must have NG(P ) < NG(H). Then by our claim (2), we obtain NG(H) = G. This
leads to Op(G) 6= 1 and NG(K) is p-nilpotent for every characteristic subgroup K of P satisfy-
ing Op(G) < K ≤ P. Now, by using the result of Thompson ([11], Corollary) again, we see that
G/Op(G) is p-nilpotent and therefore G is p-soluble. Since G is p-soluble for any q ∈ π(G) with
q 6= p, there exists a Sylow q-subgroup Q of G such that G1 = PQ is a subgroup of G. Invoking
our claim (2) above, G1 is p-nilpotent if G1 < G. This leads to Q ≤ CG(Op(G)) ≤ Op(G), a
contradiction. Thus we have proved that G = PQ.
(4) G has a unique minimal normal subgroup N such that N = Op(G) = CG(N) = F (G).
Let N be a minimal normal subgroup of G. By (1) and (3), N is an elementary abelian p-group.
Obviously G/N satisfies the condition of the theorem, the minimal choice of G implies that G/N
is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, we have N is the
unique minimal normal subgroup and Φ(G) = 1. By Lemma 2.4 N = Op(G) = CG(N) = F (G)
and N � Φ(G).
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ON NEARLYM-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 67
(5) Final contradiction.
Since N is the unique minimal normal subgroup of G and N � Φ(G), there exists a maximal
subgroup M of G such that N � M. Then G = NM and N ∩M = 1. Clearly, P = NMp and
Mp ≤ P2 < P1 where P1 is the maximal subgroup of P and P2 is the maximal subgroup of P1.
If Mp = P1 then |N | = p and hence Aut (N) is a cyclic group of order p − 1. If p < q, by ([9],
10.1.9) then NQ is p-nilpotent. Consequently, Q ≤ CG(N) = Op(G), a contradiction. If q < p, then
M ∼= G/N ∼= NG(N)/CG(N) is isomorphic to a subgroup of Aut (N). It follows that Q is a cyclic
group that G is q-nilpotent by ([9], 10.1.9) and hence P �G. Therefore NG(P ) = G is p-nilpotent,
a contradiction.
By hypotheses, P1 is nearlyM-supplemented in G. There exists a normal subgroup B1 such that
P1B1 �G and TB1 < P1B1 for every maximal subgroup T of P1. Furthermore,
(a) P1B1 < G.
Since Mp ≤ P1 and N ≤ P1B1, we have (P1B1)p = P. By Lemma 2.1(1) and the hypotheses,
P1B1 satisfies the condition of the theorem, the minimal choice of G implies that P1B1 is p-nilpotent.
Since (P1B1)p′ char P1B1 � G and (P1B1)p′ � G. We have (P1B1)p′ = 1 by (1), and hence
(P1B1) = P �G, also is a contradiction.
(b) P1B1 = G.
That is, P1 is M-supplemented in G. For every maximal subgroup T of P1, TB1 < G and
|G : TB1| = p by ([6], Lemma 2.2). Set T = P2, if N ≤ P2B1, then P2B1 = NP2B1 = PB1 =
= P1B1 = G, a contradiction. So we may have that N � P2B1 and hence we have G = P2B1N
and |N | = |G : P2B1| = p, a contradiction.
Theorem 3.3. Let F be a saturated formation containing U , suppose G has a soluble normal
subgroup N with G/N ∈ F . If every maximal subgroup of noncyclic Sylow subgroup of F (N) having
no supersoluble supplement is nearlyM-supplemented in G, then G ∈ F .
Proof. Assume that the theorem is false and let G be a counterexample of minimal order.
Furthermore, we have that
(1) N ∩ Φ(G) = 1.
If N ∩ Φ(G) 6= 1, then there exists a minimal normal subgroup L of G such that L ≤ N ∩
∩ Φ(G). Since N is soluble, we know that L is an elementary abelian p-group, moreover, we have
F (N/L) = F (N)/L. By Lemmas 2.1(2) and 2.7, every maximal subgroup of noncyclic Sylow
subgroup of F (N/L) having no supersoluble supplement is nearly M-supplemented in G/L and
(G/L)/(N/L) ∼= G/N ∈ F . Clearly,G/L satisfies the condition of the theorem and henceG/L ∈ F
by the minimal choice of G. Therefore G ∈ F , a contradiction.
(2) Every minimal normal subgroup of G contained in Op(N) is cyclic of order p where p is a
prime divisor of |N |.
If N = 1, the assertion is ture. So we may assume that N 6= 1, the solubility of N implies that
F (N) 6= 1. By Lemma 2.4, F (N) is the direct product of minimal normal subgroups of G contained
in N. There at least exists a maximal subgroup W of G not containing F (N) and hence there at
least exists a prime p of π(|N |) with Op(N) � W by Lemma 2.5. Applying Lemma 2.5 again, we
have |G : W | is not prime order .
Denote P = Op(N). Then P is the direct product of some minimal normal subgroups of G. We
assume that P = R1 ×R2 × . . .×Rt where Ri is a minimal normal subgroup of G, i = 1, 2, . . . , t.
Since N∩Φ(G) = 1, for every minimal normal subgroup R of G contained P, there exists a maximal
subgroup M of G such that G = RM = PM and R ∩M=1. Let Mp be a Sylow p-subgroup of M
and P = (P ∩M) × R. Then Gp = PMp is a Sylow p-subgroup of G. Now, let P1 be a maximal
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 1
68 J. GUO, J. ZHANG, L. MIAO
subgroup of Gp containing Mp and set P2 = P1∩P. Since |P : P2| = |P : P1∩P | = p, that is, P2 is a
maximal subgroup of P. On the other hand, P2 = P2∩P = P2∩ (P ∩M)R = (P ∩M)× (P2∩R).
Similarly, we know that P2 ∩ R is a maximal subgroup of R. If P2 is nearly M-supplemented
in G, then there exists a normal subgroup K of G such that P2K E G and TK < P2K for
every maximal subgroup T of P2. Set K2 = (P ∩M)K E G, therefore (P2 ∩ R)K2 E G and
T ′K2 < (P2 ∩ R)K2 = P2K for every maximal subgroup T ′ of P2 ∩ R, that is, P2 ∩ R is nearly
M-supplemented in G and hence |R| = p by Lemma 2.1(4). If P2 has supersoluble supplement
in G. Then there exists a subgroup H of G such that G = P2H and H is supersoluble. Let L =
= (P ∩M)H, so we have G = P2H = P2(P ∩M)H = P2L = PL. If P � L, then L < G. Since
P ∩M ≤ P2 ∩ L ≤ P ∩ L = (P ∩M)R ∩ L = (P ∩M)(R ∩ L) = P ∩M, then |P | = |P2|, a
contradiction. So we assume P ≤ L, then L = G, G/(P ∩M) = L/(P ∩M) ∼= H/(P ∩M ∩H)
is supersoluble. Since M/(P ∩M) is a maximal subgroup of G/(P ∩M), and hence |G : M | = p
and |R| = p.
(3) Final contradiction.
For every Ri, i = 1, 2, . . . , t, is of prime order, G = RiM and Ri ∩M = 1. It is clearly that
|G : M | = p and hence G ∈ F by Lemma 2.5, a contradiction.
Theorem 3.3 is proved.
The final contradiction completes our theorem.
Corollary 3.1. Let F be a saturated formation containing U and G be a soluble group. If every
maximal subgroup of noncyclic Sylow subgroup of F (G) having no supersoluble supplement is nearly
M-supplemented in G, then G ∈ F .
Corollary 3.2. Let F be a saturated formation containing U . Suppose G has a soluble normal
subgroup N with G/N ∈ F . If every maximal subgroup of noncyclic Sylow subgroup of N having
no supersoluble supplement is nearlyM-supplemented in G, then G ∈ F .
Proof. By Theorem 3.1, we know that N has supersoluble type Sylow tower. Let P be the
Sylow p-subgroup of N, where p is the largest prime divisor of |N |. Then P char N and hence
P E G. It is easy to know G/P satisfies the hypotheses, therefore G/P ∈ F . Since every maximal
subgroup of noncyclic Sylow subgroup of F (P ) = P having no supersoluble supplement is nearly
M-supplemented in G, then G ∈ F by Theorem 3.1.
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 supersoluble supplement is nearlyM-supplemented in G, then G ∈ F .
Proof. Suppose that the theorem is false and choose G to be a counterexample of the minimal
order, so in particular, H 6= 1. Furthermore, we have
Case I. F = U .
By Corollary 3.2, we easily verify that F ∗(H) is supersoluble and hence F (H) = F ∗(H) 6= 1.
Since the pair (H,H) satisfies the hypotheses of the theorem in place of (G,H), the minimal choice
of G implies that H is supersoluble if H < G; then G ∈ U by Theorem 3.3, a contradiction. Hence
(1) H = G is nonsoluble and F ∗(G) = F (G) 6= 1.
Let N be a proper normal subgroup of G containing F ∗(G). By Lemma 2.9, F ∗(G) =
= F ∗(F ∗(G)) ≤ F ∗(N) ≤ F ∗(G), so F ∗(N) = F ∗(G). Moreover, every maximal subgroup
of every non-cyclic Sylow subgroup of F ∗(N) having no supersoluble supplement is nearly M-
supplemented in N by Lemma 2.1(1). Hence N is supersoluble by the minimal choice of G. So we
have
(2) Every proper normal subgroup of G containing F ∗(G) is supersoluble.
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ON NEARLYM-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 69
Suppose now that Φ(Op(G)) 6= 1 for some p ∈ π(F (G)). By Lemma 2.9 we have
F ∗(G/Φ(Op(G))) = F ∗(G)/Φ(Op(G)). Using Lemma 2.1 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 then get G ∈ U , a contradiction. Thus we
have
(3) If p ∈ π(F (G)), then Φ(Op(G)) = 1 and so Op(G) is elementary abelian; in particular,
F ∗(G) = F (G) is abelian and CG(F (G)) = F (G).
Suppose that L is a minimal normal subgroup of G contained in F (G) and that |L| = p for some
p ∈ π(F (G)); also set C := CG(L). Clearly, F (G) ≤ C E G. If C < G, then C is solvable by
(2). Since G/C is cyclic, we get that G is solvable, a contradiction. So we have C = G and hence
L ≤ Z(G). Then we consider the group G/L. By Lemma 2.9, we have F ∗(G/L) = F ∗(G)/L =
= F (G)/L. In fact, G/L satisfies the hypotheses of the theorem by Lemma 2.1. Therefore the
minimal choice of G implies that G/L ∈ U and hence G is supersoluble, a contradiction. This proves
(4) There is no minimal normal subgroup of prime order in G contained in F (G).
If F (G) = H1× . . .×Hr with cyclic Sylow subgroups H1, . . . ,Hr of of F (G), then G/CG(Hi)
is abelian for any i ∈ {1, . . . , r} and so G/
⋂r
i=1CG(Hi) = G/CG(F (G)) = G/F (G) is abelian.
Therefore G is solvable, a contradiction. This proves
(5) P := Op(G) ∈ Sylp(F (G)) is non-cyclic for some p ∈ π(F (G)).
Let P1 be a maximal subgroup of Op(G). If P1 has a supersoluble supplement in G, then there
exists a supersoluble subgroup K of G such that G = P1K = Op(G)K. Clearly, G/Op(G) ∼=
∼= K/K ∩Op(G) is supersoluble and hence G is soluble, a contradiction. So we obtain that
(6) Every maximal subgroup of every noncyclic Sylow subgroup of F (G) has no supersoluble
supplement in G.
Furthermore, if P ∩ Φ(G) = 1, then P = R1 × . . . × Rt with minimal normal subgroups
R1, . . . , Rt of G by Lemma 2.4. Clearly, P2 = R1
∗R is the maximal subgroup of P where R∗1 is the
maximal subgroup of R1 and R = R2× . . .×Rt. By hypotheses, P2 is nearlyM-supplemented in G.
There exists a normal subgroup of K of G such that P2K E G and TK < P2K for every maximal
subgroup T of P2. Let K1 = RK. Clearly, K1 E G and P2K = R1
∗K1 and T1K1 < R1
∗K1 for
every maximal subgroup T1 of R1
∗. Therefore R1
∗ is also nearlyM-supplemented in G and hence
|R1| = p by Lemma 2.1(4), contrary to (4). So we get that
(7) R := P ∩ Φ(G) 6= 1.
Now suppose that Q ∈ Sylq(F (G)) for some prime q 6= p and let L be a minimal normal
subgroup ofG contained in R. ThenQ 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 by Lemma 2.10. 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) E G. If CG(Q) < G, then CG(Q) is supersoluble by (2); thus E(G/L) =
= E/L is supersoluble and consequently F ∗(G/L) = F (G)/L. Clearly, we see that G/L satisfies
the hypotheses of the theorem. By the minimal choice of G, G/L is supersoluble and so is G, a
contradiction. Henceforth we have CG(Q) = G, i.e. Q ≤ Z(G). Obviously, using the same argument
as in the proof of (7), G/Q ∈ U and hence G is supersoluble, also is a contradiction. Thus we have
(8) F (G) = P, in particular, 1 < R = Φ(G) ≤ P.
On the other hand, let X be a minimal normal subgroup of G contained in P with X 6= L.
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 by
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 1
70 J. GUO, J. ZHANG, L. MIAO
Lemma 2.10. Now set E/L = E(G/L). Since X is normal in G and [F (G)/L,E/L] = 1, [X,E] ≤
≤ X ∩ L = 1, i.e., [X,E] = 1. Therefore F (G)E ≤ CG(X) E G. If CG(X) < G, then CG(X) is
supersoluble by (2); thus E(G/L) = E/L is supersoluble and consequently F ∗(G/L) = F (G)/L.
Using the same argument as in the proof of (3) we see that G/L satisfies the hypotheses of the
theorem. By the minimal choice of G, G/L is supersoluble and so is G, a contradiction. Henceforth
we have CG(X) = G, i.e., X ≤ Z(G). Clearly, this also violates (4). Thus we have
(9) L is the unique minimal normal subgroup of G contained in P.
By (3), there exists a maximal subgroup P1 of P with L � P1. By hypotheses, P1 is nearly
M-supplemented in G. So there exists a normal subgroup K1 in G such that P1K1 E G and
TK1 < P1K1 for every maximal subgroup T of P1. If P1K1 = G, since L ∩ P1 6= 1, we may
choose a maximal subgroup P2 of P1 with L ∩ P1 � P2 and P2K1 < G. On the other hand,
L ≤ Φ(G) and hence P2K1 = LP2K1 = G, a contradiction.
So we may assume P1K1 < G. Since L is the unique minimal normal subgroup, we have
L ∩ P1K1 = 1 or L. If L ∩ P1K1 = 1, then L ∩ P1 = 1 and hence |L| = p, contrary to (4).
Therefore L ≤ P1K1 and P ≤ P1K1. By Lemma 2.9 and (2), P1K1 is supersoluble. Particularly,
K1 is supersoluble and (K1)q E K where q is the largest prime divisor of |K1|. On the other hand,
P1 ∩ K1 ≤ Φ(P1) = 1 and P ∩ K1 = 1, otherwise, |P ∩ K1| = p, contrary to (4). It follows
from (K1)q char K1 and K1 E G imply that (K1)q E G. So we have (K1)q ≤ P, a contradiction.
Therefore K1 = 1 and P1K1 = P1 E G, contrary to the choice of P1.
Case II. F 6= U .
By Case I, H is supersoluble. Particularly, H is soluble and hence F ∗(H) = F (H). Therefore
G ∈ F by Theorem 3.3.
Theorem 3.4 is proved.
Corollary 3.3. Let G be a group with a normal subgroup H such that G/H ∈ U . If every
maximal subgroup of every non-cyclic Sylow subgroup of F ∗(H) is nearly M-supplemented in G,
then G ∈ U .
1. Doerk K., Hawkes T. Finite soluble groups. – Berlin; New York: de Gruyter, 1992.
2. Guo W. The theory of classes of groups. – Beijing etc.: Sci. Press-Kluwer Acad. Publ., 2000.
3. Gross F. Conjugacy of odd order Hall subgroups // Bull. London Math. Soc. – 1987. – 19. – P. 311 – 319.
4. Hall P. A characteristic property of soluble groups // J. London Math. Soc. – 1937. – 12. – P. 188 – 200.
5. Huppert B. Endliche Gruppen, I. – Berlin etc.: Springer-Verlag, 1967.
6. Miao L., Lempken W. OnM-supplemented subgroups of finite groups // J. Group Theory. – 2009. – 12. – P. 271 – 289.
7. Miao L. On p-nilpotency of finite groups // Bull. Braz. Soc. – 2007. – 38, № 4. – P. 585 – 594.
8. Miao L. Finite groups with some maximal subgroups of Sylow subgroupsM-supplemented // Math. Note. – 2009. –
86. – P. 655 – 664.
9. Robinson D. J. S. A course in the theory of groups. – Berlin; New York: Springer-Verlag, 1993.
10. Srinivasan S. Two sufficeient conditions for supersolvability of finite groups // Isr. J. Math. – 1980. – 35, № 3. –
P. 210 – 214.
11. Thompson J. G. Normal p-complement for finite groups // J. Algebra. – 1964. – 1. – P. 43 – 46.
12. Wang Y. Finite groups with some subgroups of Sylow subgroups c-supplemented // J. Algebra. – 2000. – 224. –
P. 467 – 478.
13. Wang Y., Guo W. Nearly s-normality of groups and its properties // Communs Algebra. – 2010. – 38, № 10. –
P. 3821 – 3836.
14. Wang Y., Wei H., Li Y. A generalization of Kramer’s theorem and its applications // Bull. Austral. Math. Soc. – 2002. –
65. – P. 467 – 475.
Received 03.01.12
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 1
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| institution | Ukrains’kyi Matematychnyi Zhurnal |
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| language | English |
| last_indexed | 2026-03-24T02:18:57Z |
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| publisher | Institute of Mathematics, NAS of Ukraine |
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| spelling | umjimathkievua-article-21112019-12-05T10:24:29Z On Nearly ℳ-Supplemented Subgroups of Finite Groups Про майже ℳ-доповнені підгрупи скінченних груп Guo, J. Miao, L. Zhang, Jiajia Го, Й. Мяо, Л. Чжан, Їайія A subgroup H is called nearly ℳ-supplemented in a finite group G if there exists a normal subgroup K of G such that HK ⊴ G and TK < HK for every maximal subgroup T of H. We obtain some new results on supersoluble groups and their formation by using nearly ℳ-supplemented subgroups and study the structure of finite groups. Підгрупу H називаємо майже M-доповненою в скінчєнній групі G, якщо існує нормальна підгрупа K групи G така, що HK ⊴ G i TK < HK для кожної максимальної підгрупи T групи H. Отримано деякі нові результати про суперрозв'язні групи та їх утворення за допомогою майже ℳ-доповнених підгруп та досліджено структуру скінченних груп. Institute of Mathematics, NAS of Ukraine 2014-01-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2111 Ukrains’kyi Matematychnyi Zhurnal; Vol. 66 No. 1 (2014); 63–70 Український математичний журнал; Том 66 № 1 (2014); 63–70 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2111/1224 https://umj.imath.kiev.ua/index.php/umj/article/view/2111/1225 Copyright (c) 2014 Guo J.; Miao L.; Zhang Jiajia |
| spellingShingle | Guo, J. Miao, L. Zhang, Jiajia Го, Й. Мяо, Л. Чжан, Їайія On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title | On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title_alt | Про майже ℳ-доповнені підгрупи скінченних груп |
| title_full | On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title_fullStr | On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title_full_unstemmed | On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title_short | On Nearly ℳ-Supplemented Subgroups of Finite Groups |
| title_sort | on nearly ℳ-supplemented subgroups of finite groups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2111 |
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