The Maximal Subgroups of Sylow Subgroups and the Structure of Finite Groups

In this paper we investigate the influence of some subgroups of Sylow subgroups with semi cover-avoiding property and E-supplementation on the structure of finite groups. Some recent results are generalized and unified.


Introduction
All groups considered in this paper will be finite.A subgroup H of a group G is said to be S-quasinormal in G if H permutes with every Sylow subgroup of G.This concept was introduced by Kegel.In 2007, Skiba (see [22]) introduced the concept of S-supplemented subgroup.A subgroup H of G is said to be S-supplemented in G if there is a subgroup K of G such that G = HK and H ∩ K ≤ H sG , where H sG denotes the subgroup of H generated by all those subgroups of H which are S-quasinormal in G.As another generalization of the S-quasinormality, the concept of S-quasinormally embedded subgroup was given by Ballester-Bolinches and Pedraza-Aguilera (see [2]).A subgroup H is said to be S-quasinormally embedded in G if for each prime p dividing |H|, a Sylow p-subgroup of H is also a Sylow p-subgroup of some S-quasinormal subgroup of G.In 2012, Li (see [4]) proposed the definition of E-supplemented subgroup which covers properly both S-quasinormally embedding property and Skiba's weakly Ssupplementation.A subgroup H is said to be E-supplemented in G if there is a subgroup K of G such that G = HK and H ∩ K ≤ H eG , where H eG denotes the subgroup of H generated by all those subgroups of H which are S-quasinormally embedded in G.
On the other hand, we say that a subgroup H of a group G covers G-chief factor A/B if HA = HB, and

Main results
Theorem 3.1.Let p be a prime dividing the order of a group G with (|G|, p−1) = 1.If G has a Sylow p-subgroup P such that every maximal subgroup of P either has the semi cover-avoiding property or is E-supplemented in G, then G is p-nilpotent.
Proof: Assume that the assertion is false and let G be a minimal counterexample.We will derive a contradiction in several steps. ( Then there exists a maximal subgroup P 1 of P such that M = P 1 O p ′ (G).By the hypothesis of the theorem, P 1 either has the semi cover-avoiding property or is ( If there is a maximal subgroup of P which has the semi cover-avoiding property in G, then G is p-solvable by Lemma 2.
Suppose that M/N is a maximal subgroup of P/N .Then M is a maximal subgroup of P .By the hypothesis of the theorem, M either has the semi coveravoiding property or is E-supplemented in G. Then M/N either has the semi cover-avoiding property or is E-supplemented in G/N by Lemmas 2.1 and 2.2.Therefore G/N satisfies the hypothesis of the theorem.The minimal choice of (5) The final contradiction.By the proof in step ( 4), G has a maximal subgroup Furthermore, P ∩ M < P .Thus, there exists a maximal subgroup V of P such that P ∩ M ≤ V .Hence, P = O p (G)V .By the hypothesis, V either has the semi cover-avoiding property or is E-supplemented in G.
Case I: V has the semi cover-avoiding property in G.
Case II: If the latter holds, then M p ′ G and M p ′ is actually the normal p-complement of G, which is contrary to the choice of G. Hence, we may assume M = N G (M p ′ ).By applying Lemma 2.6 and the Feit-Thompson Theorem, there exists g ∈ G such that normalized by T , so g can be considered as an element of V .Thus, G = V T g = V M and P = V (P ∩ M ) = V , a contradiction.Hence T is not p-nilpotent.If V eG = 1, then |T | p = p.By Lemma 2.5, T is p-nilpotent, a contradiction.Thus we may assume that V eG = 1.Let U 1 , U 2 , ..., U s be all the nontrivial subgroups of V which are S-quasinormally embedded in G.For every i ∈ {1, 2, ..., s}, then there is an S-quasinormal subgroup K i of G such that U i is a Sylow p-subgroup of K i .Suppose that for some i ∈ {1, 2, ..., s}, we have

and so
The Maximal Subgroups of Sylow 117 P = O p (G)V = V .This contradiction shows that for all i ∈ {1, 2, ..., s} we have Proof: The necessity is obvious.We only need to prove the sufficiency.Suppose that the assertion is false and let G be a counterexample of minimal order.
(1) G has a minimal normal subgroup N ≤ H and N is an elementary abelian p-group, where p is the largest prime in π(H).
By the hypothesis of the theorem, every maximal subgroup of any noncyclic Sylow subgroup of H either has the semi cover-avoiding property or is E-supplemented in G. Consequently, by Lemmas 2.1 and 2.2 every one also either has the semi cover-avoiding property or is E-supplemented in H. Applying Corollary 3.4, H is a Sylow tower group of supersolvable type.Let p be the largest prime divisor of |H| and P a Sylow p-subgroup of H. Then P is normal in H. Obviously, P is normal in G. Therefore, G has a minimal normal subgroup N ≤ H and N is an elementary abelian p-group.
(2) G/N ∈ F and N = P is the Sylow p-subgroup of H. First, we want to prove that G/N satisfies the hypothesis of the theorem.In fact, (G/N )/(H/N ) ∼ = G/H ∈ F. Let P 1 /N be a maximal subgroup of the Sylow p-subgroup P/N of H/N .Then P 1 is a maximal subgroup of the Sylow p-subgroup P of H.If P/N is noncyclic, then P is also noncyclic.By the hypothesis of the theorem, P 1 either has the semi cover-avoiding property or is E-supplemented in G.By Lemmas 2.1 and 2.2, P 1 /N either has the semi cover-avoiding property or is E-supplemented in G/N .Let M 1 /N be a maximal subgroup of the noncyclic Sylow q-subgroup QN/N of H/N , where q = p and Q is a noncyclic Sylow q-subgroup of H.It is clear that M 1 = Q 1 N , where Q 1 is a maximal subgroup of Q.By the hypothesis of the theorem, Q 1 either has the semi cover-avoiding property or is E-supplemented in G. Hence M 1 /N either has the semi cover-avoiding property or is E-supplemented in G/N by Lemmas 2.1 and 2.2.We now have proved that G/N satisfies the hypothesis of the theorem.By the minimal choice of G, we have G/N ∈ F. Since F is a saturated formation, N is the unique minimal normal subgroup of G contained in P and N Φ(G).By Lemma 2.7, it follows that P = F (P ) = N .
( Case II: M has the semi cover-avoiding property in G. Then there exists a chief series of G Proof: Suppose that the theorem is false and let G be a counterexample of minimal order. (1) Φ(G) ∩ N = 1.
3. Since O p ′ (G) = 1 by step (1), we have O p (G) = 1, a contradiction.Thus we may assume that all maximal subgroups of P are E-supplemented in G.If p = 2, then G is odd from the assumption that (|G|, p − 1) = 1.By the Feit-Thompson Theorem, G is solvable.It follows that O p (G) = 1 by step (1), a contradiction.If p = 2, then we get also G is solvable by [4, Lemma 3.1], the same contradiction.

(
) |N | > p.This follows from Lemma 2.8.(4) The final contradiction.Let M a maximal subgroup of N .By the hypothesis, M either has the semi cover-avoiding property or is E-supplemented in G.Case I:M is E-supplemented in G. Then there is a subgroup T of G such that G = M T and M ∩ T ≤ M eG .Thus G = N T and N = N ∩ M T = M (N ∩ T ).This implies that N ∩ T = 1.But since N ∩ T is normal in G and N is minimal normal in G, N ∩ T = N .It follows that T = G and so M = M eG .In view of Lemma 2.11, M is s-quasinormal in G.By Lemma 2.10, O p (G) ≤ N G (M ).Thus M G p O p (G) = G.It follows that M = 1 and so |N | = p, which contradicts step (3).
If H covers or Lemma 2.6 ( [20, Main Theorem]).Suppose that G has a Hall π-subgroup where π is a set of odd primes.Then all Hall π-subgroups of G are conjugate.Lemma 2.7 ( [21, Lemma 2.6]).Let H = 1 be a solvable normal subgroup of a group G.If every minimal normal subgroup of G which is contained in H is not contained in Φ(G), then the Fitting subgroup F (H) of H is the direct product of minimal normal subgroups of G which are contained in H. Lemma 2.8 ( [22, Lemma 2.16]).Let F be a saturated formation containing U. Suppose that G is a group with a normal subgroup N such that G/N ∈ F. If N is cyclic, then G ∈ F. Lemma 2.9 ( [27, Lemma 2.3]).Suppose that H is S-quasinormal in G, and let P be a Sylow p-subgroup of H.If H G = 1, then P is S-quasinormal in G. Lemma 2.10 ( [28, Lemma A]).If P is an S-quasinormal p-subgroup of a group G for some prime p, then N G and so P = V , a contradiction.✷Corollary 3.2.Let p be a prime dividing the order of a group G with (|G|, p−1) = 1 and H a normal subgroup of G such that G/H is p-nilpotent.If there exists a Sylow p-subgroup P of H such that every maximal subgroup of P either has the semi cover-avoiding property or is E-supplemented in G, then G is p-nilpotent.Proof: In view of Lemmas 2.1 and 2.2, every maximal subgroup of P has the semi cover-avoiding property or is E-supplemented in H.By Theorem 3.1, H is p-nilpotent.Now, let H p ′ be the normal Hall p ′ -subgroup of H. Obviously, H p ′ G. Case I: H p ′ = 1.We consider G/H p ′ .Applying Lemmas 2.1 and 2.2, it is easy to see that G/H p ′ satisfies the hypotheses for the normal subgroup H/H p ′ .Therefore, G/H p ′ is p-nilpotent by induction.It follows that G is p-nilpotent.be the normal Hall p ′ -subgroup of G/P .By the Schur-Zassenhaus Theorem, there exists a Hall p ′ -subgroup K p ′ of K such that K = P K p ′ .A new application of Theorem 3.1 yields that K is p-nilpotent and so K = P × K p ′ .It is easy to see that K p ′ is a normal p-complement of G. Consequently, G is p-nilpotent.By Theorem 3.1, G is p-nilpotent.Let T be the normal Hall p ′ -subgroup of G.In view of Lemmas 2.1 and 2.2, every maximal subgroup of any Sylow subgroup of T has the semi cover-avoiding property or is E-supplemented in T .Thus T satisfies the hypothesis of the corollary.It follows by induction that T , and hence G is a Sylow tower group of supersolvable type.✷ Corollary 3.5 ( [13, Theorem 3.3]).Let G be a group, p a prime dividing the order of G, and P a Sylow p-subgroup of G.If (|G|, p − 1) = 1 and every maximal subgroup of P has the semi cover-avoiding property in G, then G is p-nilpotent.Let P be a Sylow p-subgroup of a group G, where p is the smallest prime divisor of |G|.If P is cyclic or every maximal subgroup of P has the semi cover-avoiding property in G, then G is p-nilpotent.Theorem 3.18.Let F be a saturated formation containing U, where U is the class of all supersolvable groups.A group G ∈ F if and only if there is a normal subgroup H of G such that G/H ∈ F and every maximal subgroup of any noncyclic Sylow subgroup of H either has the semi cover-avoiding property or is E-supplemented in G.
Case II: H p ′ = 1, i.e., H = P is a p-group.Since G/P is p-nilpotent, we can let K/P ✷ Corollary 3.3.Let P be a Sylow p-subgroup of a group G, where p is the smallest prime divisor of |G|.If every maximal subgroup of P either has the semi coveravoiding property or is E-supplemented in G, then G is p-nilpotent.Corollary 3.4.Suppose that every maximal subgroup of any Sylow subgroup of a group G either has the semi cover-avoiding property or is E-supplemented in G. Then G is a Sylow tower group of supersolvable type.Proof: Let p be the smallest prime dividing |G| and P a Sylow p-subgroup of G.
Corollary 3.22 ( [23, Theorem 4.1]).Let F be a saturated formation containing U. If there is a normal subgroup H of G such that G/H ∈ F and every maximal subgroup of any noncyclic Sylow subgroup of H is c-supplemented in G, then G ∈ F. Corollary 3.23 ( [32, Theorem 3.3]).Let F be a saturated formation containing U. If there is a normal subgroup H of a group G such that G/H ∈ F and every maximal subgroup of any Sylow subgroup of H is S-quasinormally embedded in G, then G ∈ F. Corollary 3.24 ( [14, Theorem 3.3]).Let F be a saturated formation containing U. A group G ∈ F if and only if there is a normal subgroup H of G such that G/H ∈ F and every maximal subgroup of any Sylow subgroup of H is either s-quasinormally embedded or c-normal in G. Theorem 3.25.Let F be a saturated formation containing U. Suppose that G is a group with a solvable normal subgroup N such that G/N ∈ F. If every maximal subgroup of each non-cyclic Sylow subgroup of F (N ) either has the semi coveravoiding property or is E-supplemented in G, then G ∈ F.
M = 1.It follows |N | = p, a contradiction.✷ Corollary 3.19 ( [13, Theorem 3.6]).Let F be a saturated formation containing U. If there is a normal Hall subgroup H of G such that G/H ∈ F and every maximal subgroup of any Sylow subgroup of H has the semi cover-avoiding property in G, then G ∈ F. Corollary 3.20 ( [21, Theorem 3.3]).Let H be a normal subgroup of a group G such that G/H is supersolvable.If every maximal subgroup of any Sylow subgroup of H is c-normal in G, then G is supersolvable.Corollary 3.21 ( [16, Theorem 4.2]).Let F be a saturated formation containing U. If there is a normal subgroup H of G such that G/H ∈ F and every maximal subgroup of any Sylow subgroup of H is c-supplemented in G, then G ∈ F.