All groups considered in this paper are finite. Let $ G $ be a group, by $ \pi(G) $, we denote the set of all prime divisors of $ |G| $. Following [1], by $ \mathcal{M}_{d}(P)=\{P_{1}, \dots, P_{d}\} $, we denote a set of maximal subgroups of a $ p $-group $ P $ such that $ \bigcap\limits_{i=1}^{d}P_{i}=\Phi(P) $, where $ d $ and $ \Phi(P) $ are the minimum number of generators and the Frattini subgroup of $ P $, respectively. Other notations and terminology are mostly standard and can be found in [2, 3].

Let $ G $ be a group. Two subgroups $ H $ and $ K $ of $ G $ are said to permute if $ HK = KH $, that is, $ HK $ is a subgroup of $ G $. According to Kegel [4], a subgroup $ H $ of a group $ G $ is called an $ S $-quasinormal ($ S $-permutable or $ \pi $-quasinormal) in $ G $ if $ H $ permutes with every Sylow subgroup of $ G $. Later, Ballester-Bolinches and Pedraza-Aguilera [5] introduced the concept of $ S $-quasinormally embedded subgroups. Let $ H $ be a subgroup of a group $ G $, $ H $ is said to $ S $-quasinormally embedded in $ G $ if for each prime $ p $ dividing $ |G| $, a Sylow $ p $-subgroup of $ H $ is also a Sylow $ p $-subgroup of $ S $-quasinormal subgroup of $ G $. In 2008, Li et al. [1] extended $ S $-quasinormal subgroup to $ SS $-quasinormal subgroup. A subgroup $ H $ of a group $ G $ is called an $ SS $-quasinormal (Supplement-Sylow-quasinormal) subgroup of $ G $ if there exists $ B\leq G $ such that $ G=HB $ and $ H $ permutes with every Sylow subgroup of $ B $. It is obvious that every $ S $-quasinormal subgroup of $ G $ is an $ SS $-quasinormal subgroup of $ G $, but the converse is not true (see [1, 6]).

Some authors have been studied the structure of finite groups under assumptions that some subgroups are $ SS $-quasinormal. For example, Li et al. in [1] gave a sufficient condition for a finite group $ G $ to be $ p $-nilpotent with $ d(P) $ $ SS $-quasinormal maximal subgroups of $ P $, where $ P\in $ Syl$ _p(G) $ and $ d(P) $ is the minimum number of generators of $ P $. Li et al. in [7] obtained some sufficient and necessary conditions for a finite group to be in a saturated formation with some $ SS $-quasinormal subgroups of the generalized Fitting subgroup.

We remark that the $ SS $-quasinormality mentioned above are global. As we all know, it is significant to characterize the global structure of a group by using some local properties of subgroups. From this perspective, the present paper investigates the influence of local $ SS $-quasinormal maximal subgroups of Sylow subgroups on the structure of finite groups. In detail, we study the relationship between $ d(P) $ local $ SS $-quasinormal maximal subgroups of some Sylow $ p $-subgroup $ P $ and the structure of a given group $ G $. In this work, we derive certain sufficient conditions for $ p $-nilpotency of a group. As applications, we establish some sufficient or necessary conditions for a group to be in a saturated formation containing the class of supersolvable groups.

The following Lemmas are crucial to the proofs of main results in Section 3.

Lemma 2.1  [1, Lemma 2.1] Let $ H $ be a $ SS $-quasinormal subgroup of a group $ G $, $ K\leq G $ and $ N \unlhd G $. Then

$ (1) $ If $ H\leq K\leq G $, then $ H $ is $ SS $-quasinormal in $ K $.

$ (2) $ $ HN/N $ is $ SS $-quasinormal in $ G/N $.

$ (3) $ If $ N\leq K $ and $ K/N $ is $ SS $-quasinormal in $ G/N $, then $ K $ is $ SS $-quasinormal in $ G $.

Lemma 2.2  Let $ G $ be a group, $ H\leq G $, $ L\leq G $ and $ H\leq \Phi(L) $. If $ H $ is $ SS $-quasinormal in $ G $, then $ H $ is $ S $-quasinormal in $ G $.

Proof  By the hypotheses, $ H $ is $ SS $-quasinormal in $ G $, there exists a supplement $ B $ of $ H $ to $ G $ satisfying the condition that $ H $ permutes with each Sylow subgroup of $ B $. Hence $ HX=XH $ for all Sylow subgroups $ X $ of $ B $. Then $ L=H(B\cap L) $. Since $ H\leq \Phi(L) $, we have $ L=B\cap L $ and $ L\leq B $. It follows that $ G=B $ and so $ H $ permutes with every Sylow subgroup of $ G $. Hence $ H $ is $ S $-quasinormal in $ G $.

Lemma 2.3  [8, Ⅰ, Satz 17.4] Let $ N $ be a normal abelian subgroup of a group $ G $ and let $ N\le M\le G $ such that $ (|N|, |G:M|)=1 $. If a complement subgroup of $ N $ in $ M $ exists, then $ N $ possesses a complement subgroup in $ G $.

Lemma 2.4  [9, Lemma 2.2] If $ A $ is a subnormal subgroup of a group $ G $ and $ A $ is a $ \pi $-group, then $ A\leq O_{\pi}(G) $.

Lemma 2.5  [10, Lemma 2.5] Let $ G $ be a group, $ K $ an $ S $-quasinormal subgroup of $ G $ and $ P $ a Sylow $ p $-subgroup of $ K $, where $ p $ is a prime. If either $ P\leq O_{p}(G) $ or $ K_{G}=1 $, then $ P $ is $ S $-quasinormal in $ G $.

Lemma 2.6  [11, Proposition B] If $ H $ is a nilpotent subgroup of a group $ G $, then the following two statements are equivalent:

$ (1) $ $ H $ is $ S $-quasinormal in $ G $.

$ (2) $ The Sylow subgroups of $ H $ are $ S $-quasinormal in $ G $.

Lemma 2.7  [11, Lemma A] Let $ G $ be a group and $ P $ a $ S $-quasinormal $ p $-subgroup of $ G $, where $ p\in\pi(G) $. Then $ O^{p}(G)\leq N_{G}(P) $.

Lemma 2.8  [12, Corollary] Let $ G $ be a group, $ P\in $ Syl$ _p(G) $ and $ N $ is a normal subgroup of $ G $ such that $ P\cap N\leq \Phi(P) $, then $ N $ is $ p $-nilpotent.

Lemma 2.9  [10, Lemma 2.8] Let $ G $ be a group and $ p\in\pi(G) $ with $ (|G|, p-1) = 1 $. Then

$ (1) $ If $ N $ is normal in $ G $ of order $ p $, then $ N $ lies in $ Z(G) $.

$ (2) $ If $ G $ has cyclic Sylow $ p $-subgroups, then $ G $ is $ p $-nilpotent.

$ (3) $ If $ M $ is a subgroup of $ G $ with index $ p $, then $ M $ is normal in $ G $.

By Lemma 2.9 and with similar arguments as in the proof of [1, Theorem 1.1], we obtain Lemma 2.10.

Lemma 2.10  Let $ G $ be a group and $ P\in $ Syl$ _p(G) $, where $ p\in\pi(G) $ with $ (|G|, p-1) = 1 $. If every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ G $, then $ G $ is $ p $-nilpotent.

By Lemma 2.10 and with similar arguments as in the proof of [13, Corollary 3.5], we obtain Lemma 2.11.

Lemma 2.11  Let $ H $ be a normal subgroup of a group $ G $ such that $ G/H $ is $ p $-nilpotent, $ P\in $ Syl$ _p(H) $, where $ p\in\pi(G) $ with $ (|G|, p-1)=1 $. If every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ G $, then $ G $ is $ p $-nilpotent.

Lemma 2.12  [1, Theorem 1.3] Let $ G $ be a $ p $-solvable group for a prime $ p $, $ P\in $ Syl$ _p(G) $. If every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ G $, then $ G $ is $ p $-supersolvable.

Lemma 2.13  [7, Theorem 3.3] Let $ G $ be a group and let $ \mathcal{F} $ be a saturated formation containing the class of supersolvable groups $ \mathcal{U} $. Then $ G \in \mathcal{F} $ if and only if there exists $ H\unlhd G $ such that $ G/H \in \mathcal{F} $, and for every Sylow subgroup $ P $ of $ F^{*}(H) $, all maximal subgroups of $ P $ are $ SS $-quasinormal in $ G $.

Lemma 2.14  [14, Theorem 3.1] Let $ H $ be a solvable normal subgroup of a group $ G $ such that $ G/H \in \mathcal{F} $, where $ \mathcal{F} $ is a saturated formation containing $ \mathcal{U} $. If for every maximal subgroup $ M $ of $ G $, either $ F(H)\leq M $ or $ F(H)\cap M $ is maximal in $ F(H) $, then $ G \in \mathcal{F} $. The converse also holds, in the case where $ \mathcal{F=U} $.

Recall that the generalized Fitting subgroup $ F^*(G) $ is the unique maximal normal quasinilpotent subgroup of $ G $ (see [15]). For a prime $ p\in\pi(G) $, the generalized $ p $-Fitting subgroup $ F^*_p(G) $ is defined to be as the normal subgroup of $ G $ such that $ F^*_p(G)/O_{p'}(G)=F^*(G/O_{p'}(G)) $. In the following, we would like to give some basic properties of $ F^*(G) $ and $ F^*_p(G) $.

Lemma 2.15  [14, Theorem 3.1] Let $ G $ be a group and $ N $ a subgroup of $ G $. Then

$ (1) $ If $ N $ is normal in $ G $, then $ F^*(N)\le F^*(G) $.

$ (2) $ $ F^*(G)\ne 1 $ if $ G\ne 1 $; in fact, $ F^*(G)/F(G)=Soc(F(G)C_G(F(G))/F(G)) $.

$ (3) $ $ F^*(F^*(G))=F^*(G)\ge F(G) $; if $ F^*(G) $ is solvable, then $ F^*(G)=F(G) $.

$ (4) $ $ C_G(F^*(G))\le F(G) $.

$ (5) $ If $ N\leq Z(G) $, then $ F^{*}(G/N)=F^{*}(G)/N $.

$ (6) $ If $ N\leq O_{p}(G) $ for some $ p\in \; \pi(G) $, then $ F^{*}(G/\Phi(N))=F^{*}(G)/\Phi(N) $.

Lemma 2.16  [15, Lemma 2.3] Let $ G $ be a group. Then

$ (1) $ $ Soc(G)\le F^*_p(G) $.

$ (2) $ $ O_{p'}(G)\le F^*_p(G) $. In fact, $ F^*(G/O_{p'}(G))=F^*_p(G/O_{p'}(G))=F^*_p(G)/O_{p'}(G) $.

$ (3) $ If $ F^*_p(G) $ is $ p $-solvable, then $ F^*_p(G)=F_p(G) $.

$ (4) $ If $ C=C_G(F_p(G)/O_{p'}(G)) $, then $ F^*_p(G)/F_p(G)=Soc(CF_p(G)/F_p(G)) $.

Theorem 3.1  Let $ H $ be a normal subgroup of a group $ G $ such that $ G/H $ is $ p $-nilpotent and $ P\in $ Syl$ _p(H) $, where $ p\in\pi(G) $ with $ (|G|, p-1)=1 $. If every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_{H}(P)) $, then $ G $ is $ p $-nilpotent.

Proof  Suppose otherwise and let $ G $ be a minimal counterexample. Write $ \mathcal{M}_{d}(P)=\{P_{1}, \dots, P_{d}\} $, where $ P_{i} $ is maximal in $ P $ with $ \bigcap\limits_{i=1}^{d}P_{i}=\Phi(P) $. Then

(1) $ N_{H}(P)< H $.

If not, then $ P\unlhd G $. We have $ N_{G}(P)=G $. It follows from Lemma 2.11 that $ G $ is $ p $-nilpotent.

(2) $ O_{p'}(G)=1 $.

Write $ N=N_{G}(P) $ and $ T=O_{p^{'}}(G) $. If $ T\neq 1 $, we consider the quotient group $ \overline{G}=G/T $. Clearly, $ \overline{H}=HT/T\unlhd \overline{G} $ and $ \overline{G}/\overline{H}\cong G/HT $, hence $ \overline{G}/\overline{H} $ is $ p $-nilpotent. It is obvious that $ P\cong\overline{P}=PT/T\in $ Syl$ _p(\overline{H}) $, so $ PT/T $ has the same smallest generator number as $ P $, i.e., $ d $, and $ \mathcal{M}_{d}(PT/T)=\{P_{1}T/T, \dots, P_{d}T/T\} $ with $ \bigcap\limits_{i=1}^{d}\overline{P_{i}}=\Phi(\overline{P}) $. Meanwhile,

$ N_{\overline{G}}(\overline{P})=N_{G/T}(PT/T)=N_G(P)T/T=NT/T=\overline{N}. $

By the hypotheses, $ P_{i} $ is $ SS $-quasinormal in $ N $, there exists a supplement $ B_{i} $ of $ P_{i} $ to $ N $ such that $ P_{i} $ permutes with each Sylow subgroup of $ B_{i} $. Hence $ P_{i}X=XP_{i} $ for all Sylow subgroups $ X $ of $ B_{i} $. Moreover,

$ NT/T=(P_{i}B_{i})T/T=(P_{i}T/T)(B_{i}T/T). $

For any $ q\in\pi(G) $, by [8, Ⅵ. 4.7], there exist respectively Sylow $ q $-subgroup $ (B_{i})_{q} $ of $ B_{i} $ and $ T_{q} $ of $ T $ such that $ Y=(B_{i})_{q}T_{q} $ is a Sylow $ q $-subgroup of $ B_{i}T $. We may assume that $ X $ is a Sylow $ q $-subgroup of $ B_i $. Since both $ XT/T $ and $ YT/T $ are Sylow $ q $-subgroups of $ B_{i}T/T $, by Sylow's theorem, we have

$ XT/T=(YT/T)^{bT}=((B_{i})_{q}T/T)^{bT}=(B_{i})_{q}^{b}T/T $

for some $ b\in B_{i} $. Under the assumptions of the theorem, $ P_{i} $ permutes with every Sylow $ q $-subgroup of $ B_{i} $, $ P_{i}(B_{i})_{q}^{b}=(B_{i})_{q}^{b}P_{i}, \; \; b\in B_{i} $. We obtain

$ P_{i}T/T\cdot XT/T=P_{i}(XT)/T=P_{i}((B_{i})_{q}^{b}T)/T =(B_{i})_{q}^{b}P_{i}T/T=XT/T\cdot P_{i}T/T, $

which shows that $ P_{i}T/T $ is $ SS $-quasinormal in $ NT/T $. On the other hand, since $ \Psi $ is $ SS $-quasinormal in $ G $, by Lemma 2.1(2), $ \Psi T/T $ is $ SS $-quasinormal in $ G/T $. Meanwhile,

$ (PT/T)^{'}\leq \Psi T/T\leq \Phi(N_{H}(P))T/T\leq \Phi(N_{H}(P)T/T)\leq \Phi(N_{HT/T}(PT/T)). $

Hence $ G/T $ satisfies the conditions of the theorem. By the minimality of $ G $, $ G/T $ is $ p $-nilpotent, and therefore $ G $ is $ p $-nilpotent, a contradiction.

(3) $ H=G $.

If not, $ H<G $. Since $ H $ satisfies the hypotheses of the theorem, $ H $ is $ p $-nilpotent. Suppose that $ K $ is a normal $ p $-complement of $ H $. Then $ K\unlhd G $, hence $ K=1 $ by (2) and $ H=P\unlhd G $, which is contrary to (1).

(4) $ \Phi(P)_{G}=1 $.

Write $ N=\Phi(P)_{G} $. If $ N\neq 1 $, we consider the quotient group $ \overline{G}=G/N $. Clearly, $ N_{G/N}(P/N)=N_G(P)/N $ and $ \mathcal{M}_{d}(P/N)=\{P_{1}/N, \dots, P_{d}/N\} $ with $ \bigcap\limits_{i=1}^{d}\overline{P_{i}}=\Phi(\overline{P}) $. Since $ P_{i} $ is $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $, in view of Lemma 2.1(2), $ P_{i}/N $ is $ SS $-quasinormal in $ N_{G/N}(P/N) $ and $ \Psi N/N $ is $ SS $-quasinormal in $ G/N $. Meanwhile,

$ (P/N)'=P'N/N\le \Psi N/N\le\Phi(N_{G}(P))N/N\le\Phi(N_{G}(P)/N)=\Phi(N_{G/N}(P/N)). $

By the minimality of $ G $, $ G/N $ is $ p $-nilpotent. Since $ N\le\Phi(P) $, $ N\le\Phi(G) $ by [8, Ⅲ, 3.3], it follows that $ G $ is $ p $-nilpotent, which is impossible.

(5) $ N_{G}(P) $ is $ p $-nilpotent and $ \Psi \cap \Phi(P)\neq 1 $.

By Lemma 2.11, $ N_{G}(P) $ satisfies the conditions of the theorem. By the choice of $ G $, $ N_{G}(P) $ is $ p $-nilpotent. Let $ K $ be the normal $ p $-complement of $ N_{G}(P) $, then $ N_{G}(P)=P\times K $. Write $ N=N_{G}(P) $, we have $ \Phi(N)=\Phi(P)\times \Phi(K) $. From this equation, we have $ \Psi=(\Psi\cap \Phi(P))\times (\Psi\cap \Phi(K)) $. If $ \Psi\cap \Phi(P)=1 $, then $ P^{'}=1 $ as $ P^{'}\leq \Psi\cap \Phi(P) $, which implies that $ P $ is an abelian group so that $ N_{G}(P)=C_{G}(P) $ and $ G $ is $ p $-nilpotent by Burnside's Theorem, a contradiction.

(6) The final contradiction.

By Lemma 2.2, we see that $ \Psi $ is $ S $-quasinormal in $ G $. From the proof of (5), we obtain $ \Psi_{G}=(\Psi_{G}\cap \Phi(P))\times (\Psi_{G}\cap \Phi(K)) $, hence both $ \Psi_{G}\cap \Phi(P) $ and $ \Psi_{G}\cap \Phi(K) $ are normal in $ G $, it follows from (2) and (4) that $ \Psi_{G}=1 $. By Lemma 2.5 and Lemma 2.7, we have $ O^{p}(G)\leq N_{G}(\Psi \cap \Phi(P)) $. However, $ P^{'}\leq \Psi \cap \Phi(P) $, hence $ P\leq N_{G}(\Psi \cap \Phi(P)) $, i.e., $ \Psi \cap \Phi(P) $ is normal in $ G $. But $ \Psi \cap \Phi(P)\ne 1 $ by (5) and this is contrary to (4). The proof is complete.

Theorem 3.2  Let $ H $ be a subnormal subgroup of a group $ G $ containing $ F_{p}^{*}(G) $ and let $ P\in $ Syl$ _p(H) $, where $ p\in\pi(G) $ with $ (|G|, p-1)=1 $. If every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_{H}(P)) $, then $ G $ is $ p $-nilpotent.

Proof  Let $ G $ be a minimal counterexample. Then

(1) $ O_{p'}(G)=1 $.

If not, then $ T=O_{p^{'}}(G)>1 $. We consider the quotient group $ \overline{G}=G/T $. In fact,

$ F^{*}_{p}(\overline{G})=F^{*}(\overline{G})=F^{*}_{p}(G)/T\leq H/T=\overline{H}. $

Clearly, $ \overline{H}\lhd \lhd \overline{G} $ and $ \overline{P}=PT/T\in $ Syl$ _p(\overline{H}) $. Noticing that $ P\cong \overline{P} $, hence $ \overline{P} $ has the same smallest generator number, i.e., $ d $, and $ \mathcal{M}_{d}(\overline{P})=\mathcal{M}_{d}(PT/T)=\{P_{1}T/T, \dots, P_{d}T/T\} $ with $ \bigcap\limits_{i=1}^{d}\overline{P_{i}}=\Phi(\overline{P}) $. Furthermore,

$ N_{\overline{G}}(\overline{P})=N_{G/T}(PT/T)=N_G(P)T/T. $

Since $ P_{i} $ is $ SS $-quasinormal in $ N_{G}(P) $, with similar arguments as in the proof of Theorem 3.1, we know that $ P_{i}T/T $ is $ SS $-quasinormal in $ N_{G}(P)T/T $. Moreover, $ \Psi T/T $ is $ SS $-quasinormal in $ G/T $ by Lemma 2.1(2). At the same time,

$ (PT/T)^{'}\leq \Psi T/T\leq \Phi(N_{H}(P))T/T\leq \Phi(N_{H}(P)T/T)\leq \Phi(N_{HT/T}(PT/T)). $

Thereby $ G/T $ satisfies the conditions of the theorem. By the minimality of $ G $, $ G/T $ is $ p $-nilpotent, hence $ G $ is $ p $-nilpotent, a contradiction.

(2) $ H=P=O_{p}(G)=F^{*}(G) $.

By Theorem 3.1 and Lemma 2.1(1), we have that $ H $ is $ p $-nilpotent. Hence $ H $ has a normal $ p $-complement subgroup $ H_{p^{'}} $. Clearly, $ H_{p^{'}}\lhd\lhd G $. By Lemma 2.4, we obtain $ H_{p^{'}}\leq O_{p^{'}}(G)=1 $. Hence $ H_{p^{'}}=1 $ and $ H=P $. Consequently, $ H=P\leq O_{p}(G) $. Moreover, $ O_{p}(G)\leq F_{p}(G)=F_{p}^{*}(G)=F^{*}(G)\leq H $, so $ H=P=O_{p}(G)=F^{*}(G) $.

(3) The final contradiction.

Let $ Q $ be a Sylow $ q $-subgroup of $ G $ with $ q\neq p $. We consider $ H_{1}=PQ $. By Lemma 2.1(1), every member of $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ H_{1} $ and $ \Psi $ is $ SS $-quasinormal in $ H_{1} $. Hence $ H_{1} $ is $ p $-nilpotent by Theorem 3.1. From Lemma 2.15(4), we have

$ Q\le C_G(P)=C_G(F^*(G))\le F(G)=P. $

This is the final contradiction and the proof is complete.

When $ p $ is not a special prime, we may omit $ d(P) $ $ SS $-quasinormal maximal subgroups of $ P $ but need to employ the $ p $-nilpotency of $ N_G(P) $.

Theorem 3.3  Let $ H $ be a normal group of a group $ G $ such that $ G/H $ is $ p $-nilpotent and $ P\in $ Syl$ _p(H) $, where $ p\in\pi(G) $. Suppose that $ N_{G}(P) $ is $ p $-nilpotent and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_H(P)) $, then $ G $ is $ p $-nilpotent.

Proof  Suppose otherwise and let $ G $ be a minimal counterexample.

(1) $ O_{p^{'}}(G)=1 $.

Denote $ T=O_{p^{'}}(G) $. If $ T\neq 1 $, we consider the quotient group $ \overline{G}=G/T $. Clearly, $ \overline{P}=PT/T\in $ Syl$ _p(\overline{H}) $, $ \overline{H}=HT/T\unlhd \overline{G} $ and $ \overline{G}/\overline{H}\cong G/HT $, hence $ \overline{G}/\overline{H} $ is $ p $-nilpotent. Noticing that $ N_{\overline{G}}(\overline{P})=N_G(P)T/T $, hence $ N_{\overline{G}}(\overline{P}) $ is $ p $-nilpotent. On the other hand, by the hypotheses and Lemma 2.1(2), $ \Psi T/T $ is $ SS $-quasinormal in $ G/T $. Meanwhile,

$ (PT/T)^{'}\leq \Psi T/T\leq \Phi(N_{H}(P))T/T\leq \Phi(N_{H}(P)T/T)\leq \Phi(N_{HT/T}(PT/T)). $

By the minimality of $ G $, $ G/T $ is $ p $-nilpotent, hence $ G $ is $ p $-nilpotent, a contradiction.

(2) $ \Phi(P)_{G}=1 $.

Write $ N=\Phi(P)_{G} $. If $ N\neq 1 $, we consider the quotient group $ \overline{G}=G/N $. Clearly, $ N_{H/N}(P/N)=N_H(P)/N $. By the hypotheses and Lemma 2.1(2), $ \Psi N/N $ is $ SS $-quasinormal in $ G/N $. Moreover,

$ (P/N)'=P'N/N\le \Psi N/N\le\Phi(N_{H}(P))N/N\le\Phi(N_{H}(P)/N)=\Phi(N_{H/N}(P/N)). $

By the choice of $ G $, $ G/N $ is $ p $-nilpotent. Since $ N\le\Phi(P) $, $ N\le\Phi(G) $ by [8, Ⅲ, 3.3], it follows that $ G $ is $ p $-nilpotent, which is impossible.

(3) $ H=G $.

If not, $ H<G $. Since $ H $ satisfies the conditions of the theorem, $ H $ is $ p $-nilpotent. Suppose that $ K $ is a normal $ p $-complement of $ H $. Then $ K\unlhd G $, hence $ K=1 $ by (1) and $ H=P\unlhd G $ so that $ G=N_{G}(P) $ is $ p $-nilpotent, a contradiction.

(4) The final contradiction.

By the hypotheses, $ N_H(P) $ is $ p $-nilpotent. Let $ K $ be the normal $ p $-complement of $ N_{H}(P) $, then $ N_{H}(P)=P\times K $. Furthermore, $ \Phi(N_{H}(P))=\Phi(P)\times \Phi(K) $. From this equation, we obtain $ \Psi=(\Psi\cap \Phi(P))\times (\Psi\cap \Phi(K)) $. Since $ \Psi\le\Phi(N_H(P)) $ and $ \Psi $ is $ SS $-quasinormal in $ G $, by Lemma 2.2, we see that $ \Psi $ is $ S $-quasinormal in $ G $. Clearly, $ \Psi_{G}=(\Psi_{G}\cap \Phi(P))\times (\Psi_{G}\cap \Phi(K)) $ and $ \Psi_{G}\cap \Phi(P)\unlhd G, \Psi_{G}\cap \Phi(K)\unlhd G $, hence $ \Psi_{G}=1 $ by (1) and (2). By Lemma 2.5 and Lemma 2.7, $ \Psi\cap \Phi(P) $ is $ S $-quasinormal in $ G $, thereby $ O^{p}(G)\leq N_{G}(\Psi \cap \Phi(P)) $. However, $ P^{'}\leq \Psi \cap \Phi(P) $, hence $ P\leq N_{G}(\Psi \cap \Phi(P)) $, i.e., $ \Psi \cap \Phi(P)\unlhd G $. If $ \Psi \cap \Phi(P)=1 $, then $ P^{'}=1 $ and $ P $ is an abelian group, we have $ N_{G}(P)=C_{G}(P) $. By Burnside's Theorem, we know that $ G $ is $ p $-nilpotent, a contradition. So $ \Psi \cap \Phi(P)\neq1 $. This is contrary to (2) and the proof is complete.

In the following, we give some applications of Theorem 3.1 and Theorem 3.2.

Theorem 3.4  Let $ G $ be a group. If for every $ p\in\pi (G) $, there exists a subnormal subgroup $ H $ of $ G $ containing $ F_{p}^{*}(G) $ and some $ P\in $ Syl$ _p(H) $ such that every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_{H}(P)\cap P^{H}) $, then $ G $ is supersolvable.

Proof  Suppose otherwise, let $ G $ be a minimal counterexample. Then

(1) $ G $ has a Sylow tower of supersolvable type.

Let $ q=min\ \pi(G) $. By Theorem 3.2, $ G $ has a normal $ q $-complement subgroup $ K $. Obviously, $ G $ is a solvable group by Odd Order Theorem. Now let $ p=min\ \pi(K) $. By Lemma 2.16, $ F_{p}^{*}(K)=F_p(K) $. Clearly, $ F_p(K)\le F_p(G) $. By the hypotheses, there exists $ H\lhd\lhd G $ such that $ F_p(G)=F_{p}^{*}(G)\le H $ and $ P\in $ Syl$ _p(H) $ such that every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{G}(P) $. Furthermore,

$ F_p(K)\le F_p(G)\cap K\le H\cap K\lhd\lhd K. $

Clearly, $ P\leq K $. Since $ P\leq H\cap K\unlhd H $, we have $ P^{H}\leq H\cap K $. Therefore,

$ \Phi(N_{H}(P)\cap P^{H})=\Phi(N_{H}(P)\cap P^{H}\cap H\cap K) =\Phi(N_{H\cap K}(P)\cap P^{H})\leq \Phi(N_{H\cap K}(P)). $

By Lemma 2.1(1), every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{K}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ K $ for some $ P'\le\Psi\le\Phi(N_{H\cap K}(P)) $. Thus $ K $ is $ p $-nilpotent by Theorem 3.2. In this way, we see that $ G $ has a Sylow tower of supersolvable type.

(2) Let $ p=max \; \pi(G) $ and $ P\in $ $ Syl_{p}(G) $. Then $ G/P $ is supersolvable.

By (1), $ P\unlhd G $. Clearly, $ P\le F_p(G)=F_{p}^{*}(G) $. For every $ p\ne q\in\pi(G) $, we have

$ F_{q}^{*}(G/P)=F_q(G/P)=F_q(G)/P=F_q^*(G)/P. $

Consider the quotient group $ G/P $. By the hypotheses, there exist $ H\lhd\lhd G $ such that $ F_{q}^{*}(G)\le H $ and $ Q\in $ Syl$ _q(H) $ such that every member of some fixed $ \mathcal{M}_{d}(Q) $ is $ SS $-quasinormal in $ N_{G}(Q) $. Let $ \mathcal{M}_{d}(Q)=\{Q_{1}, \dots, Q_{d}\} $, where $ Q_{i} $ is maximal in $ Q $ with $ \bigcap\limits_{i=1}^{d}Q_{i}=\Phi(Q) $, $ d=d(Q) $. Clearly, $ QP/P\cong Q $, so $ QP/P $ has the same smallest generator number as $ Q $, i.e., $ d $, and $ \mathcal{M}_{d}(QP/P)=\{Q_{1}P/P, \dots, Q_{d}P/P\} $ with $ \bigcap\limits_{i=1}^{d}\overline{Q_{i}}=\Phi(\overline{Q}) $. Also, $ N_{G/P}(QP/P)=N_G(Q)P/P $. Since every member of $ \mathcal{M}_{d}(Q) $ is $ SS $-quasinormal in $ N_{G}(Q) $, with similar arguments as in the proof of Theorem 3.1, we obtain every member of $ \mathcal{M}_{d}(QP/P) $ is $ SS $-quasinormal in $ N_{G/P}(QP/P) $. By Lemma 2.1(2), $ \Psi P/P $ is $ SS $-quasinormal in $ G/P $. Moreover,

$ (QP/P)^{'}\leq \Psi P/P\leq \Phi(N_{H}(Q)\cap Q^{H})P/P\leq \Phi(N_{H/P}(QP/P)\cap (QP/P)^{H/P}). $

Hence $ G/P $ satisfies the hypotheses of the theorem and $ G/P $ is supersolvable.

(3) $ P\in $ $ Syl_{p}(H) $.

Since $ P\unlhd G $, we have $ P\leq F_{p}(G)=F^{*}_{p}(G)\leq H $. Hence $ P\in $ $ Syl_{p}(H) $.

(4) $ \Phi(P)=1 $.

Write $ T=\Phi(P) $. If $ T\neq 1 $, we consider the quotient group $ \overline{G}=G/T $. Let $ q\in\pi(G) $. Since

$ F_{q}^{*}(G/T)=F_q(G/T)=F_q(G)/T=F_q^*(G)/T $

and

$ (P/T)^{'}\leq \Psi T/T\leq \Phi(N_{H}(P)\cap P^{H})/T\leq \Phi(N_{H/T}(P/T)\cap(P/T)^{H/T}), $

we see easily that $ G/\Phi(P) $ satisfies the conditions of the theorem. By the minimality of $ G $, $ G/\Phi(P) $ is supersolvable. Since $ \Phi(P)\le\Phi(G) $ by [8, Ⅲ, 3.3], it follows that $ G $ is supersolvable, a contradiction.

(5) Conclusion of the proof.

First of all, we will show that every minimal subgroup $ N $ of $ G $ is of order $ p $, where $ p=max \; \pi(G) $. In fact, since $ G $ is solvable by (1), $ N $ is an elementary abelian $ q $-group for some $ q\in\pi(G) $. Suppose $ q\ne p $. Clearly, $ (G/N)/(PN/N)\cong G/PN $ is supersolvable by (2). Since every member of some fixed $ \mathcal{M}_{d}(P) $ is $ SS $-quasinormal in $ N_{G}(P)=G $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(P) $, every member of $ \mathcal{M}_{d}(PN/N) $ is $ SS $-quasinormal in $ G/N $ and $ \Psi N/N $ is $ SS $-quasinormal in $ G/N $, where $ (PN/N)'=P'N/N\le\Psi N/N\le\Phi(P)N/N\le \Phi(PN/N) $. Hence $ G/N $ is supersolvable by the minimality of $ G $. It follows from (2) that $ G $ is supersolvable, a contradiction. Hence $ q=p $ and $ N\le P $. Since $ P_{i}\leq P\unlhd G $, we have $ P_{i}\leq O_{p}(G) $, hence $ P_{1} $ is $ S $-quasinormal in $ G $ by [7, Lemma 2.2]. By Lemma 2.7, we have $ O^{p}(G)\leq N_{G}(P_{i}) $. However, $ P_{i} $ is normalized by $ P $. We have $ G=PO^{p}(G)\leq N_{G}(P_{i}) $, that is, $ P_{i}\unlhd G $. Hence $ N\cap P_{i}\unlhd G $. The minimality of $ N $ implies that $ N\cap P_{i}=1 $ or $ N $. If $ N\cap P_{i}=N $ for each $ i $, then $ N\leq P_{i} $. It follows that $ N\leq \bigcap\limits_{i=1}^{d}P_{i}=\Phi(P)=1 $ and consequently $ N=1 $, a contradiction. Hence $ N\cap P_{i}=1 $ for some $ i $. As $ NP_{i}=P $, we obtain $ |N|=p $. Now, it is easy to see that $ N $ is complemented in $ P $, hence it is complemented in $ G $ by Lemma 2.3. Consequently, $ P\cap\Phi(G)=1 $. By [10, Lemma 2.9], we obtain $ P=N_1\times N_2\times\cdots\times N_s $, where $ N_i $ $ (i=1, \ldots, s) $ is minimal normal in $ G $ of order $ p $. It follows from $ G/P $ is supersolvable that $ G $ is supersolvable. This is the final contradiction and the proof is complete.

In the following, we give some sufficient and necessary conditions for a finite group to be in a saturated formation.

Theorem 3.5  Let $ G $ be a group and let $ \mathcal{F} $ be a saturated formation containing the class of supersolvable groups $ \mathcal{U} $. Then the following two statements are equivalent:

$ (1) $ $ G \in \mathcal{F} $.

$ (2) $ There exists $ H\unlhd G $ such that $ G/H \in \mathcal{F} $, and for every Sylow subgroup $ P $ of $ F^{*} $, all maximal subgroups of $ P $ are $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_{F^{*}}(P)) $, where $ F^{*}=F^{*}(H) $.

Proof  Clearly, $ (1) $ implies $ (2) $. We only need to prove that $ (2) $ implies $ (1) $. Assume that the theorem is false and let $ G $ be a minimal counterexample. Then

(1) $ F^{*}=F(H). $ Let $ q = min \; \pi(F^{*}) $. It is clear that $ F^{*} $ is $ q $-nilpotent by Theorem 3.1, hence $ F^{*} $ is solvable and $ F^{*}=F(H) $ by Lemma 2.15(3).

(2) $ F(H) $ is elementary abelian.

Let $ P\in $ Syl$ _p(F(H)) $, where $ p\in \pi(F(H)) $. Clearly, $ \Phi(P)\unlhd G $ and $ (P/\Phi(P))'=1 $. By Lemma 2.15(6), $ F^{*}(H/\Phi(P))=F^{*}/\Phi(P) $, so all maximal subgroups of Sylow subgroups of $ F^{*}(H/\Phi(P)) $ are $ SS $-quasinormal in $ G/\Phi(P) $ by Lemma 2.1(2). If $ \Phi(P)\neq 1 $, then $ G/\Phi(P) \in \mathcal{F} $ by the choice of $ G $. However, $ \Phi(P)\leq \Phi(G) $, thereby $ G\in \mathcal{F} $, a contradiction. So $ \Phi(P)=1 $ and $ P $ is elementary abelian. This proves (2).

(3) $ H $ is solvable.

By Lemma 2.1(1) and Lemma 2.13, $ H $ is supersolvable, of course, $ H $ is solvable.

(4) Finish of the proof.

Let $ M $ be a maximal subgroup of $ G $ such that $ F(H)\nsubseteq M $. Then there exists $ P\in $ Syl$ _p(F(H)) $ such that $ G=PM $. Set $ M_{p}\in $ Syl$ _p(M) $. Clearly, $ G_{p}=PM_{p}\in $ Syl$ _p(G) $. Take a maximal subgroup $ G_{1} $ of $ G_{p} $ containing $ M_{p} $ and let $ P_{1}=P\cap G_{1} $. Then $ G_{1}=P_{1}M_{p} $ and

$ P_{1}\cap M=(P\cap M)\cap G_{1}=P\cap M\unlhd G, $

hence $ P\cap M\leq (P_{1})_{G} $ and

$ |P:P_{1}|=|PM_{p}:P_{1}M_{p}|=|G_{p}:G_{1}|=p, $

that is, $ P_{1} $ is maximal in $ P $. Futhermore, $ (P_{1})_{G}M<G $ implies that $ (P_{1})_{G}\leq M $, thus $ P\cap M=(P_{1})_{G} $. Clearly, $ P\le O_p(G) $, hence $ P_{1}\leq O_{p}(G) $. By the hypotheses of the theorem and [7, Lemma 2.2], $ P_{1} $ is $ S $-quasinormal in $ G $. Noticing that $ P_{1}=P\cap G_{1} $, thereby $ P_{1} $ is normalized by $ G_{p} $. It follows from $ O^{p}(G)\leq N_{G}(P_{1}) $ that $ P_{1}\unlhd G $. Consequently, $ P_{1}=(P_{1})_{G}\leq M $ and therefore $ |F(H):F(H)\cap M|=|G:M|=p. $ By Lemma 2.14, $ G\in $ $ \mathcal{F} $, a final contradiction. This completes our proof.

Theorem 3.6  Let $ G $ be a group and let $ \mathcal{F} $ be a saturated formation containing the class of supersolvable groups $ \mathcal{U} $. Then the following two statements are equivalent:

$ (1) $ $ G \in \mathcal{F} $.

$ (2) $ There exists $ H\unlhd G $ such that $ G/H \in \mathcal{F} $, and for every Sylow subgroup $ P $ of $ H $, all maximal subgroups of $ P $ are $ SS $-quasinormal in $ N_{G}(P) $ and $ \Psi $ is $ SS $-quasinormal in $ G $ for some $ P'\le\Psi\le\Phi(N_{H}(P)) $.

Proof  Clearly, $ (1) $ implies $ (2) $. We only need to prove that $ (2) $ implies $ (1) $. We distinguish two cases:

Case 1  $ H=G $.

In this case, we claim that $ G $ is supersolvable. Assume $ G $ is not supersolvable with minimal order. It is clear that $ G $ is $ q $-nilpotent by Theorem 3.1, where $ q = min \; \pi(G) $. Of course, $ G $ is solvable. Let $ N $ be a minimal normal subgroup of $ G $. Clealy, $ N $ is an elementary $ p $-group for some prime $ p $. With similar arguments as in the proof of Theorem 3.1, we observe that $ G/N $ satisfies the conditions of the theorem. By the minimality of $ G $, $ G/N $ is supersolvable. Now we may assume that $ N $ is the unique minimal normal subgroup of $ G $ and $ \Phi(G)=1 $. Moreover, $ N=F(G) $ and $ C_{G}(N)=N $. Let $ R\in $ Syl$ _r(G) $, where $ r = max \; \pi(G) $. Since $ G/N $ is supersolvable, $ RN/N \unlhd G/N $. If $ p = r $, then we obtain $ N \leq R $ and $ R \unlhd G $. Hence $ G/R $ is supersolvable. Noticing that $ \Phi(R)\unlhd G $ and $ R^{'}\leq \Phi(R)\leq \Phi(G) $, by Theorem 3.5, $ G $ is supersolvable, a contradiction. Hence we may assume $ p\neq r $. Let $ M/N $ be a minimal normal subgroup of $ G/N $ contained in $ RN/N $. Then $ M/N $ is of order $ r $, hence $ M $ has the form of $ R_{0}N $, where $ R_{0}\leq R $ and $ |R_{0}|=r $. Thus $ MP=R_{0}P\leq G $, where $ P\in $ Syl$ _p(G) $.

Suppose $ P^{'} = 1 $. Then we see that $ R_{0}P $ satisfies the hypotheses of the theorem. If $ R_{0}P<G $, then $ R_{0}P $ is supersovable and consequently, $ R_{0}\unlhd R_{0}P $. This yields that $ R_{0}\leq C_{G}(N)=N $, which is impossible. Hence $ R_{0}P=G $ and $ p=q $. It follows from $ G $ is $ q $-nilpotent that $ R_{0}\unlhd G $. We get a contradiction.

Now suppose $ P^{'}\neq 1 $. By the hypotheses, there exists $ P'\le\Psi\le\Phi(N_{G}(P)) $ such that $ \Psi $ is $ SS $-quasinormal in $ G $. Obviously, $ \Psi\cap P\in $ Syl$ _p(\Psi) $. By Lemma 2.2 and Lemma 2.6, $ \Psi \cap P $ is S-quasinormal in $ G $. It follows from Lemma 2.7 that $ O^{p}(G)\leq N_{G}(\Psi \cap P) $. However, $ P^{'}\leq \Psi \cap P $, hence $ P\leq N_{G}(\Psi \cap P) $, i.e., $ \Psi \cap P\unlhd G $. We have $ N\leq \Psi \cap P $. Furthermore, $ N \leq \Psi \cap P \leq \Phi(N_{G}(P)) $ and thereby $ N \leq \Phi(G) $ [8, Ⅲ, 3.3], contradicting $ \Phi(G) = 1 $. This shows that $ G $ must be supersolvable.

Case 2  $ H\neq G $.

By Case 1, $ H $ is supersolvable. Let $ P\in $ Syl$ _p(H) $, where $ p= max \; \pi(H) $. Then $ P\unlhd H $ so that $ P \unlhd G $. Since $ G/P $ satisfies the hypotheses of the theorem, $ G/P \in \mathcal{F} $ by the choice of $ G $. Thus $ G \in \mathcal{F} $ by Theorem 3.5. This is the final contradiction and the proof is complete.

Remark  (1) The condition that $ \Psi $ is $ SS $-quasinormal in $ G $ in theorems cannot be removed. For example, let $ G=PSL(2, 17) $ be the projective special group of degree 17 over a field of order 2 and $ P\in $ Syl$ _2(G) $. Then every maximal subgroup of $ P $ is normal in $ N_{G}(P)=P $, however, $ G $ is a nonabelian simple group.

(2) The condition that all maximal subgroups of $ P $ in Theorem 3.6 cannot be replaced by $ d $ maximal subgroups in some fixed $ \mathcal{M}_{d}(P) $ in general. The example can be seen in [1, Proof of Example 1.6].