$S\Phi$-Supplemented subgroups of finite groups
We call $H$ an $S\Phi$-supplemented subgroup of a finite group $G$ if there exists a subnormal subgroup $T$ of $G$ such that $G = HT$ and $H \bigcap T \leq \Phi(H)$, where $\Phi(Н)$ is the Frattini subgroup of $H$. In this paper, we characterize the $p$-nilpotency and supersolubility of a finite g...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508471980457984 |
|---|---|
| author | Li, Xianhua Zhao, Tao Лі, Хіаньхуа Чжао, Тао |
| author_facet | Li, Xianhua Zhao, Tao Лі, Хіаньхуа Чжао, Тао |
| author_sort | Li, Xianhua |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:29:28Z |
| description | We call $H$ an $S\Phi$-supplemented subgroup of a finite group $G$ if there exists a subnormal subgroup $T$ of $G$ such that $G = HT$ and $H \bigcap T \leq \Phi(H)$,
where $\Phi(Н)$ is the Frattini subgroup of $H$. In this paper, we characterize
the $p$-nilpotency and supersolubility of a finite group $G$ under the assumption that every subgroup of a Sylow $p$-subgroup of $G$ with given order is $S\Phi$-supplemented in $G$.
Some results about formations are also obtained. |
| first_indexed | 2026-03-24T02:25:45Z |
| format | Article |
| fulltext |
UDC 512.5
Xianhua Li (School Math. Sci., Soochow Univ., Suzhou, China),
Tao Zhao (School Sci., Shandong Univ. Technology, Zibo, China)
SΦ-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS*
SΦ-ДОПОВНЮВАНI ПIДГРУПИ СКIНЧЕННИХ ГРУП
We call H an SΦ-supplemented subgroup of a finite group G if there exists a subnormal subgroup T of G such that
G = HT and H ∩ T ≤ Φ(H), where Φ(H) is the Frattini subgroup of H. In this paper, we characterize the p-nilpotency
and supersolubility of a finite group G under the assumption that every subgroup of a Sylow p-subgroup of G with given
order is SΦ-supplemented in G. Some results about formations are also obtained.
Пiдгрупу H називають SΦ-доповнюваною пiдгрупою скiнченної групи G, якщо iснує така субнормальна пiдгрупа
T групи G, що G = HT i H∩T ≤ Φ(H), де Φ(H) є пiдгрупою Фраттiнi пiдгрупи H. У цiй статтi охарактеризовано
p-нiльпотентнiсть та надрозв’язнiсть скiнченної групи G за припущення, що кожна пiдгрупа силовської p-пiдгрупи
групи G заданого порядку є SΦ-доповнюваною в G. Отримано також деякi результати щодо формацiй.
1. Introduction. All groups considered in this paper are finite. F denotes a formation, a normal
subgroup N of a group G is said to be F-hypercentral in G provided N has a chain of subgroups
1 = N0�N1� . . .�Nr = N such that each Ni+1/Ni is an F-central chief factor of G, the product
of all F-hypercentral subgroups of G is again an F-hypercentral subgroup of G. It is denoted by
ZF (G) and called the F-hypercenter of G. U and N denote the classes of all supersoluble groups and
nilpotent groups respectively. The other terminology and notations are standard, as in [7] and [13].
We know that for every normal subgroup N of G, the minimal supplement H of N in G satisfies
H ∩N ≤ Φ(H). Then naturally, we consider the converse case, i.e., if for some subgroup H of G,
there exists a subnormal subgroup N of G such that HN = G and H ∩ N ≤ Φ(H), what can we
say about G? To study this question, we introduce the concept of SΦ-supplemented subgroups of a
finite group.
Definition 1.1. A subgroup H of a group G is said to be SΦ-supplemented in G if there exists
a subnormal subgroup T of G such that G = HT and H ∩ T ≤ Φ(H), where Φ(H) is the Frattini
subgroup of H.
From the Definition 1.1, we can easily deduce that the minimal supplement of any minimal normal
subgroup and every Sylow subgroup of a nilpotent group G are SΦ-supplemented in G. We can also
deduce that every non-trivial subgroup of G contained in Φ(G) can not be SΦ-supplemented in G.
Meanwhile, a group with a non-trivial SΦ-supplemented subgroup cannot be a non-abelian simple
group.
Inspired by [1] and [11], for each prime p dividing the order of G, let P be a Sylow p-subgroup
of G and D a subgroup of P such that 1 < |D| < |P |, we study the structure of G under the
assumption that each subgroup H of P with |H| = |D| is SΦ-supplemented in G. We get some
characterizations about formation.
2. Preliminaries. In this section, we list some basic results which will be used below.
Lemma 2.1. Let H be an SΦ-supplemented subgroup and N a normal subgroup of G.
*This work was supported by the National Natural Science Foundation of China (Grant No. 11171243, 10871032), the
Natural Science Foundation of Jiangsu Province (No. BK2008156).
c© XIANHUA LI, TAO ZHAO, 2012
92 ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
SΦ-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 93
(1) If H ≤ K ≤ G, then H is SΦ-supplemented in K.
(2) If N ≤ H, then H/N is SΦ-supplemented in G/N.
(3) Let π be a set of primes, H a π-subgroup and N a π′-subgroup. Then HN/N is SΦ-
supplemented in G/N.
Proof. By the hypothesis, there exists a subnormal subgroup T of G such that G = HT and
H ∩ T ≤ Φ(H). Then
(1) K = K ∩ G = H(K ∩ T ) and H ∩ (K ∩ T ) = (H ∩ T ) ∩ K ≤ Φ(H) ∩ K ≤ Φ(H).
Obviously, K ∩ T is a subnormal subgroup of K. Hence H is SΦ-supplemented in K.
(2) G/N = (H/N)(TN/N) and
H/N ∩ TN/N = (H ∩ TN)/N = (H ∩ T )N/N ≤ Φ(H)N/N ≤ Φ(H/N),
TN/N is subnormal in G/N. Hence H/N is SΦ-supplemented in G/N.
(3) Since T contains a Hall π′-subgroup of G and T is subnormal in G, it is easy to see that
N ≤ T and G/N = (HN/N)(T/N). Since HN/N ∩ T/N = (H ∩ T )N/N ≤ Φ(H)N/N ≤
≤ Φ(HN/N) and T/N is subnormal in G/N, HN/N is SΦ-supplemented in G/N.
Lemma 2.1 is proved.
Lemma 2.2. Let P be a Sylow p-subgroup of a group G, where p is a prime dividing |G|. If
every subgroup of P with order p is SΦ-supplemented in G, then G is p-nilpotent.
Proof. We use induction on |G|. Let H be a subgroup of P of order p. By the hypothesis, there
exists a subnormal subgroup T of G such that G = HT and H ∩ T ≤ Φ(H) = 1. Then T is a
maximal subgroup of G and so T �G. Hence if p - |T |, then G is p-nilpotent, the result holds. Thus
we may suppose that p||T | and P = H(P ∩ T ). Clearly P ∩ T is a Sylow p-subgroup of T. Then
every subgroup of P ∩T with order p is SΦ-supplemented in G and so in T by Lemma 2.1. Hence T
is p-nilpotent by induction. Since T is p-nilpotent and T �G, we have G is p-nilpotent, as required.
Lemma 2.2 is proved.
From [2] (Theorem A) or [3] (Theorem A or B), we can easily deduce that:
Lemma 2.3. Let P be a normal p-subgroup of a group G, where p is a prime dividing |G|. If
every subgroup of P with order p is SΦ-supplemented in G, then P ≤ ZU (G).
Lemma 2.4. Let P be a normal p-subgroup of a group G, where p is a prime dividing |G|. If
every maximal subgroup of P is SΦ-supplemented in G, then P ≤ ZU (G).
Proof. Assume that the result is false and let (G,P ) be a counterexample for which |G||P | is
minimal. We treat with the following two cases:
Case 1. Φ(P ) 6= 1.
By Lemma 2.1, every maximal subgroup of P/Φ(P ) is SΦ-supplemented in G/Φ(P ). Then
P/Φ(P ) ≤ ZU (G/Φ(P )) by the choice of G. Hence P ≤ ZU (G) by [12] (I, Theorem 7.19), a
contradiction.
Case 2. Φ(P ) = 1.
At this time, P is an elementary abelian group. If |P | = p, then P ≤ ZU (G), a contradiction.
Now we may assume that |P | = pn, n ≥ 2. Let P1 be a maximal subgroup of P. By the hypothesis,
there exists a subnormal subgroup K of G such that G = P1K and P1 ∩K ≤ Φ(P1) = 1. Clearly,
P = P1(P ∩K) and P ∩K is a normal subgroup of G of order p. By Lemma 2.1, every maximal
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
94 XIANHUA LI, TAO ZHAO
subgroup of P/(P ∩K) is SΦ-supplemented in G/(P ∩K). Then P/(P ∩K) ≤ ZU (G/(P ∩K))
by induction. Since |P ∩K| = p, we have P ≤ ZU (G), as required.
Lemma 2.4 is proved.
Now we can prove the following lemma.
Lemma 2.5. Let P be a normal p-subgroup of a group G, D a subgroup of P such that
1 < |D| < |P |. Suppose that every subgroup H of P with |H| = |D| is SΦ-supplemented in G, then
P ≤ ZU (G).
Proof. Assume that the result is false and let (G,P ) be a counterexample for which |G||P | is
minimal. Then:
(1) |P : D| > p.
By Lemma 2.4, it is true.
(2) Φ(P ) = 1.
Suppose that Φ(P ) 6= 1. If |Φ(P )| < |D|, then every subgroup H of P/Φ(P ) with |H| = |D|
|Φ(P )|
is SΦ-supplemented in G = G/Φ(P ) by Lemma 2.1. Then P/Φ(P ) ≤ ZU (G/Φ(P )) by induction.
Hence P ≤ ZU (G) by [12] (I, Theorem 7.19), a contradiction. Thus |Φ(P )| ≥ |D|. Let H be a
subgroup of Φ(P ) with |H| = |D|. By the hypothesis, H is SΦ-supplemented in G and so in P, a
contradiction.
(3) The final contradiction.
Let H be a subgroup of P with |H| = |D|. By the hypothesis, H is SΦ-supplemented in G,
then there exists a subnormal subgroup K of G such that G = HK and H ∩K ≤ Φ(H) = 1. Since
|G : K| is a power of p, there exists a normal subgroup M of G containing K such that |G : M | = p.
Let P1 = P ∩M, then P1 is a maximal subgroup of P and it is normal in G. By (1), |P1| > |D|.
Then every subgroup of P1 with order |D| is SΦ-supplemented in G. So P1 ≤ ZU (G) by induction.
Since |P/P1| = p, we have P ≤ ZU (G), the final contradiction.
Lemma 2.5 is proved.
Lemma 2.6 ([9], Lemma 2.8). Let G be a group and p a prime dividing |G| with (|G|, p− 1) =
= 1. Then:
(1) If N is normal in G and of order p, then N lies in Z(G).
(2) If G has cyclic Sylow p-subgroups, then G is p-nilpotent.
(3) If M ≤ G and |G : M | = p, then M �G.
Lemma 2.7 ([5], X. 13). Let G be a group, then:
(1) F ∗(G) 6= 1 if G 6= 1; in fact, F ∗(G)/F (G) = Soc(F (G)CG(F (G))/F (G)).
(2) If F ∗(G) is soluble, then F ∗(G) = F (G).
(3) CG(F ∗(G)) ≤ F (G).
Lemma 2.8 ([8], Lemma 2.6). Let N be a nontrivial soluble normal subgroup of a group G. If
every minimal normal subgroup of G which is contained in N is not contained in Φ(G), then the
Fitting subgroup F (N) of N is the direct product of minimal normal subgroups of G which are
contained in N.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
SΦ-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 95
3. Main results.
Theorem 3.1. Let P be a Sylow p-subgroup of a group G, where p is a prime dividing |G|
such that (|G|, p − 1) = 1. If every maximal subgroup of P is SΦ-supplemented in G, then G is
p-nilpotent.
Proof. Assume that the result is false and let G be a counterexample of minimal order. Then we
have:
(1) G has the unique minimal normal subgroup N such that G/N is p-nilpotent and Φ(G) = 1.
Clearly G is not a non-abelian simple group. Let N be a minimal normal subgroup of G, consider
G/N. Let M/N be a maximal subgroup of PN/N, by Lemma 2.6 we may suppose that |PN/N | ≥
≥ p2. Clearly M = P1N for some maximal subgroup P1 of P and P ∩ N = P1 ∩ N is a Sylow
p-subgroup of N. By the hypothesis, P1 is SΦ-supplemented in G, then there exists a subnormal
subgroup K of G such that G = P1K and P1 ∩K ≤ Φ(P1). Thus G/N = (P1N/N)(KN/N) =
= (M/N)(KN/N). It is easy to see that K ∩ N contains a Hall p′-subgroup of N and then
(|N : (P1 ∩N)|, |N : (K ∩N)|) = 1, so (P1 ∩N)(K ∩N) = N = N ∩G = N ∩P1K. By [10] (A,
Lemma 1.2), we have P1N ∩KN = (P1∩K)N. Thus (P1N)/N ∩ (KN)/N = (P1N ∩KN)/N =
= (P1 ∩K)N/N ≤ Φ(P1)N/N ≤ Φ(P1N/N), i.e., M/N is SΦ-supplemented in G/N. Therefore,
G/N satisfies the hypothesis of the theorem. The minimal choice of G implies that G/N is p-
nilpotent. The uniqueness of N and Φ(G) = 1 are obvious.
(2) Op′(G) = 1.
If Op′(G) 6= 1, then N ≤ Op′(G) and G/N is p-nilpotent by (1). Hence G is p-nilpotent, a
contradiction.
(3) Op(G) = 1 and so N is not p-nilpotent.
If Op(G) 6= 1, then N ≤ Op(G). Since Φ(G) = 1, there exists a maximal subgroup M of G
such that G = MN and M ∩N = 1. Since Φ(Op(G)) ≤ Φ(G) = 1, Op(G) is an elementary abelian
group. It is easy to see that Op(G) ∩M is normalized by N and M, hence Op(G) ∩M � G. If
Op(G) ∩M 6= 1, by the uniqueness of N, we have N ≤ Op(G) ∩M, hence G = MN = M, a
contradiction. This contradiction shows that Op(G) ∩M = 1. By N ≤ Op(G) and G = MN, we
have N = Op(G). Let Mp be a Sylow p-subgroup of M. If Mp = 1, then N is a Sylow p-subgroup of
G. Let P1 be a maximal subgroup of N, then Φ(P1) = 1. By the hypothesis, there exists a subnormal
subgroup K of G such that G = P1K and P1 ∩K ≤ Φ(P1) = 1. Since any Sylow r-subgroup of
G with r 6= p is a Sylow r-subgroup of K, we have Op(G) ≤ K. By the uniqueness of N, we
obtain that N ≤ Op(G). So G = P1K = NK = K, then P1 = 1 and |G|p = p, so G is p-nilpotent
by Lemma 2.6, a contradiction. Thus Mp 6= 1. Let P0 be a maximal subgroup of P containing
Mp, then by the hypothesis, there exists a subnormal subgroup T of G such that G = P0T and
P0 ∩ T ≤ Φ(P0). By the previous argument, N ≤ Op(G) ≤ T. Then P0 = P0 ∩ P = P0 ∩NMp =
= Mp(P0 ∩ N) ≤ Mp(P0 ∩ T ) ≤ MpΦ(P0) ≤ P0. Thus we have Mp = P0 and so |N | = p, then
N ≤ Z(G) by Lemma 2.6. Since G/N is p-nilpotent, G is p-nilpotent, a contradiction.
If N is p-nilpotent, then Np′char N �G, so Np′ ≤ Op′(G) = 1 by (2). Thus N is a p-group and
then N ≤ Op(G) = 1, a contradiction. Thus (3) holds.
(4) The final contradiction.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
96 XIANHUA LI, TAO ZHAO
By Lemma 2.1, we know that every maximal subgroup of P is SΦ-supplemented in PN. Thus
if PN < G, then PN is p-nilpotent and N is p-nilpotent, contradicts to (3), so we have PN = G.
Since G/N is a p-group, N = Op(G). It is easy to see that G is not a non-abelian simple group, so we
have G 6= N. Hence there exists a maximal normal subgroup M/N of G/N such that |G : M | = p.
Since P ∩M is a maximal subgroup of P, by the hypothesis, there exists a subnormal subgroup T of
G such that G = (P∩M)T and P∩M∩T ≤ Φ(P∩M). In this case, we still have T ≥ Op(G) = N.
P ∩M �P implies that Φ(P ∩M) ≤ Φ(P ), so P ∩N ≤ P ∩M ∩T ≤ Φ(P ∩M) ≤ Φ(P ). Thus N
is p-nilpotent by Tate’s theorem [4] (IV, Theorem 4.7), contrary to (3). This contradiction completes
the proof.
Theorem 3.1 is proved.
Remark. The hypothesis that (|G|, p−1) = 1 in Theorem 3.1 cannot be removed. For example,
S3, the symmetry group of degree 3 is a counter-example.
Theorem 3.2. Let P a Sylow p-subgroup of a group G, where p is a prime dividing |G| such
that (|G|, p− 1) = 1. Let D be a subgroup of P such that 1 < |D| < |P |. If every subgroup H of P
with |H| = |D| is SΦ-supplemented in G, then G is p-nilpotent.
Proof. Suppose that the result is false and let G be a counterexample of minimal order. Then we
have:
(1) Op′(G) = 1.
If Op′(G) 6= 1, Lemma 2.1 shows that the hypothesis still holds for G/Op′(G). Then G/Op′(G)
is p-nilpotent by our minimal choice of G and so is G, a contradiction.
(2) |P : D| > p.
If |P : D| = p, then by Theorem 3.1, G is p-nilpotent.
(3) The final contradiction.
Let H be a subgroup of P such that |H| = |D|, then by the hypothesis H is SΦ-supplemented
in G. So there exists a subnormal subgroup K of G such that G = HK and H ∩ K ≤ Φ(H).
Since |G : K| is a power of p, there exists a normal subgroup M of G containing K such that
|G : M | = p. Let P1 = P ∩M be a Sylow p-subgroup of M, then P1 is a maximal subgroup of P.
By (2), |P1| > |D|. Lemma 2.1 shows that every subgroup of P1 with order |D| is SΦ-supplemented
in M. Then M is p-nilpotent by our minimal choice of G and so is G, the final contradiction. This
contradiction completes the proof.
Theorem 3.2 is proved.
Obviously, Theorem 3.2 is true when p is the smallest prime divisor of |G|. Then we have the
following corollary.
Corollary 3.1. Let G be a finite group. If for every prime p dividing |G|, there exists a Sylow
p-subgroup P of G such that P has a subgroup D satisfying 1 < |D| < |P | and every subgroup H
of P with |H| = |D| is SΦ-supplemented in G, then G has a Sylow tower of supersoluble type.
If we drop the assumption that (|G|, p− 1) = 1 and add NG(P ) is p-nilpotent, we still have the
similar results.
Theorem 3.3. Let p be a prime dividing |G|, P a Sylow p-subgroup of G. If NG(P ) is p-
nilpotent and there exists a subgroup D of P such that 1 < |D| < |P | and every subgroup H of P
with |H| = |D| is SΦ-supplemented in G, then G is p-nilpotent.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
SΦ-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 97
Proof. Assume that the result is false and let G be a counterexample of minimal order. By
Theorem 3.2, we may suppose that p is not the smallest prime divisor of |G| and so p is an odd
prime. Moreover, we have:
(1) Op′(G) = 1.
If Op′(G) 6= 1, then by Lemma 2.1 it is easy to see that G/Op′(G) satisfies the hypothesis of
the theorem. Thus the minimal choice of G implies that G/Op′(G) is p-nilpotent and hence G is
p-nilpotent, a contradiction.
(2) If L is a proper subgroup of G containing P, then L is p-nilpotent.
It is easy to see that P ∈ Sylp(L) and NL(P ) ≤ NG(P ) is p-nilpotent. Furthermore, by
Lemma 2.1 we know every subgroup of P with order |D| is SΦ-supplemented in G and thus
SΦ-supplemented in L. So L is p-nilpotent by the minimal choice of G.
(3) Op(G) 6= 1.
Let J(P ) be the Thompson subgroup of P, then NG(P ) ≤ NG(Z(J(P ))) and by Lemma 2.1,
we know every subgroup of P with order |D| is SΦ-supplemented in NG(Z(J(P ))). Thus if
NG(Z(J(P ))) < G, then NG(Z(J(P ))) is p-nilpotent by (2). It follows from [6] (VIII, Theo-
rem 3.1) that G is p-nilpotent, a contradiction. Thus we may suppose that NG(Z(J(P ))) = G and
hence Op(G) 6= 1.
Next, we let N be a minimal normal subgroup of G contained in Op(G). Then we have:
(4) |N | < |D| and G/N is p-nilpotent.
If |N | > |D|, pick a subgroup H of N with order |D|. By the hypothesis, there exists a subnormal
subgroup K of G such that G = HK and H∩K ≤ Φ(H) ≤ Φ(N) = 1. Clearly, we have G = NK,
N ∩K 6= 1 and N ∩K �G. Thus N ∩K = N by the minimality of N and so K = G and H = 1,
a contradiction. If |N | = |D|, then N is SΦ-supplemented in G by the hypothesis, so there exists
a subnormal subgroup T of G such that G = NT and N ∩ T ≤ Φ(N) = 1. Since T is subnormal
in G and |G : T | is a power of p, there exists a normal subgroup M of G containing T such that
|G : M | = p. Clearly, G = NM and N ∩M�G; then the minimality of N implies that N ∩M = 1.
So |N | = p and in this case, every minimal subgroup of P is SΦ-supplemented in G. Then G is
p-nilpotent by Lemma 2.2, a contradiction. Thus we have |N | < |D|. It is easy to see that G/N
satisfies the hypothesis of the theorem, therefore G/N is p-nilpotent by the minimal choice of G.
(5) G = PQ, where Q is a Sylow q-subgroup of G with q 6= p. Moreover, N = Op(G) = F (G).
Since G/N is p-nilpotent and N is a p-group, G is p-soluble. By [6] (VI, Theorem 3.5), there
exists a Sylow q-subgroup Q of G such that PQ is a subgroup of G for any q ∈ π(G) with q 6= p.
If PQ < G, then PQ is p-nilpotent by (2). Thus Op(G)Q = Op(G) × Q and Q ≤ CG(Op(G)) ≤
≤ Op(G) by [6] (VI, Theorem 3.2), a contradiction. Hence we may assume that G = PQ. Since the
class of all p-nilpotent subgroups formed a saturated formation, we may assume that N is the unique
minimal normal subgroup of G contained in Op(G). By (1) and the fact that G is p-soluble, we can
conclude that N is the unique minimal normal subgroup of G and Φ(G) = 1. By Lemma 2.8, we
have N = Op(G) = F (G).
(6) |P : D| > p.
Now we assume that |P : D| = p. Since N � Φ(G), there exists a maximal subgroup M of
G such that G = MN and M ∩ N = 1. Obviously, P ∩ M is a Sylow p-subgroup of M and
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
98 XIANHUA LI, TAO ZHAO
P = (P ∩ M)N. Pick a maximal subgroup P1 of P containing P ∩ M, then by hypothesis P1
is SΦ-supplemented in G, so there exists a subnormal subgroup T of G such that G = P1T and
P1 ∩ T ≤ Φ(P1). Since |G : T | = |P1 : (P1 ∩ T )| is a power of p and T is subnormal in G,
Op(G) ≤ T and thus N ≤ T. Then P1 = P1 ∩ P = P1 ∩ (P ∩M)N = (P ∩M)(P1 ∩ N) ≤
≤ (P ∩M)(P1 ∩ T ) ≤ (P ∩M)Φ(P1), therefore P ∩M = P1 and |N | = p. Since G is soluble
by (5), CG(F (G)) ≤ F (G) = N, so CG(N) = N. Then we have M ∼= G/N = NG(N)/CG(N) .
. Aut(N). Since Aut(N) is a cyclic group of order p − 1, M and in particularly Q is cyclic and
hence G is q-nilpotent by Burnside’s Theorem [4] (IV, Theorem 2.8). It follows that G = NG(P ) is
p-nilpotent by the hypothesis, a contradiction.
(7) The final contradiction.
Since G is soluble, there is a normal maximal subgroup M of G such that |G : M | is a prime. If
|G : M | = q, then M is p-nilpotent by (2) and therefore P = M � G by (1), a contradiction. Thus
we may assume that |G : M | = p, then it follows that P ∩M ∈ Sylp(M) is a maximal subgroup
of P. If NG(P ∩M) < G, then NG(P ∩M) ≥ P is p-nilpotent by (2) and so is NM (P ∩M).
Since |P : D| > p by (6), every subgroup of P ∩M with order |D| is SΦ-supplemented in M by
Lemma 2.1. Consequently, M satisfies the hypothesis of the theorem and therefore M is p-nilpotent
by the minimal choice of G. The normal p-complement of M is also the normal p-complement of
G, a contradiction. Hence we may suppose that P ∩M � G and then N = Op(G) = P ∩M is a
maximal subgroup of P. This leads to |D| < |N |, contradicts to (4), the final contradiction.
Theorem 3.3 is proved.
Theorem 3.4. Let F be a saturated formation containing U and E a normal subgroup of a
group G such that G/E ∈ F . If for every prime p dividing |E|, there exists a Sylow p-subgroup
P of E such that P has a subgroup D satisfying 1 < |D| < |P | and every subgroup H of P with
|H| = |D| is SΦ-supplemented in G, then G ∈ F .
Proof. By Lemma 2.1, we know that for every prime p dividing |E|, there exists a Sylow p-
subgroup P of E such that P has a subgroup D satisfying 1 < |D| < |P | and every subgroup H of
P with |H| = |D| is SΦ-supplemented in E, then E is a Sylow tower group of supersoluble type
by Corollary 3.1. Let p be the largest prime dividing |E| and P a Sylow p-subgroup of E, then P
is normal in G. Since (G/P )/(E/P ) ∼= G/E ∈ F and the hypothesis still holds for (G/P,E/P )
by Lemma 2.1, we have G/P ∈ F by induction on |G|. Since P ≤ ZU (G) by Lemma 2.5 and
ZU (G) ≤ ZF (G) by [10] (IV, Proposition 3.11), we have P ≤ ZF (G) and so G ∈ F , as required.
Theorem 3.4 is proved.
Theorem 3.5. Let F be a saturated formation containing U and E a normal subgroup of a
group G such that G/E ∈ F . If for every prime p dividing |F ∗(E)|, there exists a Sylow p-subgroup
P of F ∗(E) such that P has a subgroup D satisfying 1 < |D| < |P | and every subgroup H of P
with |H| = |D| is SΦ-supplemented in G, then G ∈ F .
Proof. We use induction on |G|. By Lemma 2.1, we know that for every prime p dividing
|F ∗(E)|, there exists a Sylow p-subgroup P of F ∗(E) such that P has a subgroup D satisfying
1 < |D| < |P | and every subgroup H of P with |H| = |D| is SΦ-supplemented in F ∗(E).
By Corollary 3.1, F ∗(E) possesses an ordered Sylow tower of supersoluble type. In particular,
F ∗(E) is soluble and so F ∗(E) = F (E) by Lemma 2.7. Lemma 2.5 shows that F (E) ≤ ZU (G).
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
SΦ-SUPPLEMENTED SUBGROUPS OF FINITE GROUPS 99
Since ZU (G) ≤ ZF (G) by [10] (IV, Proposition 3.11), we have F (E) ≤ ZF (G). By [10] (IV,
Theorem 6.10), G/CG(ZF (G)) ∈ F and since F (E) ≤ ZF (G), we have G/CG(F (E)) ∈ F .
By the hypothesis G/E ∈ F , so G/CE(F (E)) ∈ F . But CE(F (E)) = CE(F ∗(E)) ≤ F (E) by
Lemma 2.7, then we have G/F (E) ∈ F . Hence G ∈ F by Theorem 3.4, as required.
Theorem 3.5 is proved.
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Received 06.04.11,
after revision — 25.12.11
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 1
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| id | umjimathkievua-article-2558 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:25:45Z |
| publishDate | 2012 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/4c/b5b2a9a005d9bc52f3aac4c754eb3b4c.pdf |
| spelling | umjimathkievua-article-25582020-03-18T19:29:28Z $S\Phi$-Supplemented subgroups of finite groups $S\Phi$-доповнюванi пiдгрупи скiнченних груп Li, Xianhua Zhao, Tao Лі, Хіаньхуа Чжао, Тао We call $H$ an $S\Phi$-supplemented subgroup of a finite group $G$ if there exists a subnormal subgroup $T$ of $G$ such that $G = HT$ and $H \bigcap T \leq \Phi(H)$, where $\Phi(Н)$ is the Frattini subgroup of $H$. In this paper, we characterize the $p$-nilpotency and supersolubility of a finite group $G$ under the assumption that every subgroup of a Sylow $p$-subgroup of $G$ with given order is $S\Phi$-supplemented in $G$. Some results about formations are also obtained. Пiдгрупу $H$ називають $S\Phi$-доповнюваною пiдгрупою скiнченної групи $G$, якщо iснує така субнормальна пiдгрупа $T$ групи $G$, що $G = HT$ and $H \bigcap T \leq \Phi(H)$, де $\Phi(Н)$ є пiдгрупою Фраттiнi пiдгрупи $H$. У цiй статтi охарактеризовано $p$-нiльпотентнiсть та надрозв’язнiсть скiнченної групи $G$ за припущення, що кожна пiдгрупа силовської $p$-пiдгрупи групи $G$ заданого порядку є $S\Phi$-доповнюваною в $G$. Отримано також деякi результати щодо формацiй. Institute of Mathematics, NAS of Ukraine 2012-01-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2558 Ukrains’kyi Matematychnyi Zhurnal; Vol. 64 No. 1 (2012); 92-99 Український математичний журнал; Том 64 № 1 (2012); 92-99 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2558/1875 https://umj.imath.kiev.ua/index.php/umj/article/view/2558/1876 Copyright (c) 2012 Li Xianhua; Zhao Tao |
| spellingShingle | Li, Xianhua Zhao, Tao Лі, Хіаньхуа Чжао, Тао $S\Phi$-Supplemented subgroups of finite groups |
| title | $S\Phi$-Supplemented subgroups of finite groups |
| title_alt | $S\Phi$-доповнюванi пiдгрупи скiнченних груп |
| title_full | $S\Phi$-Supplemented subgroups of finite groups |
| title_fullStr | $S\Phi$-Supplemented subgroups of finite groups |
| title_full_unstemmed | $S\Phi$-Supplemented subgroups of finite groups |
| title_short | $S\Phi$-Supplemented subgroups of finite groups |
| title_sort | $s\phi$-supplemented subgroups of finite groups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2558 |
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