mathematics Article On the s-Length of Maximal Subgroups of Finite s-Soluble Groups Abd El-Rahman Heliel 1,2,* , Mohammed Al-Shomrani 1 and Adolfo Ballester-Bolinches 3 1 Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia; [email protected] 2 Department of Mathematics and Computer Science, Faculty of Science, Beni-Suef University, Beni-Suef 62511, Egypt 3 Departament de Matemàtiques, Universitat de València, Dr. Moliner 50, Burjassot, 46100 València, Spain; [email protected] * Correspondence: [email protected] Received: 22 September 2020; Accepted: 2 December 2020; Published: 4 December 2020 Abstract: Let s = fsi : i 2 Ig be a partition of the set P of all prime numbers and let G be a finite group. We say that G is s-primary if all the prime factors of jGj belong to the same member of s. G is said to be s-soluble if every chief factor of G is s-primary, and G is s-nilpotent if it is a direct product of s-primary groups. It is known that G has a largest normal s-nilpotent subgroup which is denoted by Fs(G). Let n be a non-negative integer. The n-term of the s-Fitting series of G is defined inductively by F0(G) = 1, and Fn+1(G)/Fn(G) = Fs(G/Fn(G)). If G is s-soluble, there exists a smallest n such that Fn(G) = G. This number n is called the s-nilpotent length of G and it is denoted by ls(G). If F is a subgroup-closed saturated formation, we define the s-F-length ns(G, F) of G as the s-nilpotent length of the F-residual GF of G. The main result of the paper shows that if A is a maximal subgroup of G and G is a s-soluble, then ns(A, F) = ns(G, F) − i for some i 2 f0, 1, 2g. Keywords: finite group; s-solubility; s-nilpotency; s-nilpotent length 1. Introduction All groups considered in this paper are finite. Skiba [1] (see also [2]) generalised the concepts of solubility and nilpotency by introducing S s-solubility and s-nilpotency, in which s is a partition of P, the set of all primes. Hence P = i2I si, with si \ sj = Æ for all i 6= j. In the sequel, s will be a partition of the set of all primes P. A group G is called s-primary if all the prime factors of jGj belong to the same member of s. Definition 1. A group G is said to be s-soluble if every chief factor of G is s-primary. G is said to be s-nilpotent if it is a direct product of s-primary groups. We note in the special case that s is the partition of P containing exactly one prime each, the class of s-soluble groups is just the class of all soluble groups and the class of s-nilpotent groups is just the class of all nilpotent groups. Many normal and arithmetical properties of soluble groups and nilpotent groups still hold for s-soluble and s-nilpotent groups (see [2]) and, in fact, the class Ns of all s-nilpotent groups behaves in s-soluble groups as nilpotent groups in soluble groups. In addition, every s-soluble group has a 0 conjugacy class of Hall si-subgroups and a conjugacy class of Hall si -subgroups, for every si 2 s. Recall that a class of groups F is said to be a formation if F is closed under taking epimorphic images and every group G has a smallest normal subgroup with quotient in F. This subgroup is called Mathematics 2020, 8, 2165; doi:10.3390/math8122165 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 2165 2 of 4 the F-residual of G and it is denoted by GF. A formation F is called subgroup-closed if XF is contained in GF for all subgroups X of every group G; F is saturated if it is closed under taking Frattini extensions. A class of groups F is said to be a Fitting class if F is closed under taking normal subgroups and every group G has a largest normal subgroup in F. This subgroup is called the F-radical of G. The following theorem which was proved in [1] (Corollary 2.4 and Lemma 2.5) turns out to be crucial in our study. Theorem 1. Ns is a subgroup-closed saturated Fitting formation. The Ns-radical of a group G is called the s-Fitting subgroup of G and it is denoted by Fs(G). Clearly, Fs(G) is the product of all normal s-nilpotent subgroups of G. If s is the partition of P containing exactly one prime each, then Fs(G) is just the Fitting subgroup of G. If G is s-soluble, then every minimal normal subgroup N of G is s-primary so that N is s-nilpotent and it is contained in Fs(G). In particular, Fs(G) 6= 1 if G 6= 1. Let n be a non-negative integer. The n-term of the s-Fitting series of G is defined inductively by F0(G) = 1, and Fn+1(G)/Fn(G) = Fs(G/Fn(G)). If G is s-soluble, there exists a smallest n such that Fn(G) = G. This number n is called the s-nilpotent length of G and it is denoted by ls(G) (see [3,4]). The nilpotent length l(G) of a group G is just the s-nilpotent length of G for s the partition of P containing exactly one prime each. The s-nilpotent length is quite useful in the structural study of s-soluble groups (see [3,4]), and allows us to extend some known results. The central concept of this paper is the following: Definition 2. Let F be a saturated formation. The s-F-length ns(G, F) of a group G is defined as the s-nilpotent length of the F-residual GF of G. Applying [5] (Chapter IV, Theorem (3.13) and Proposition (3.14)) (see also [3] (Lemma 4.1)), we have the following useful result. Proposition 1. The class of all s-soluble groups of s-length at most l is a subgroup-closed saturated formation. It is clear that the F-length nF(G) of a group G studied in [6] is just the s-F-length of G for s the partition of P containing exactly one prime each, and the s-nilpotent length of G is just the s-F-length of G for F = f1g. Ballester-Bolinches and Pérez-Ramos [6] (Theorem 1), extending a result by Doerk [7] (Satz 1), proved the following theorem: Theorem 2. Let F be a subgroup-closed saturated formation and M be a maximal subgroup of a soluble group G. Then nF(M) = nF(G) − i for some i 2 f0, 1, 2g. Our main result shows that Ballester-Bolinches and Pérez-Ramos’ theorem still holds for the s-F-length of maximal subgroups of s-soluble groups. Theorem A. Let F be a saturated formation. If A is a maximal subgroup of a s-soluble group G, then ns(A, F) = ns(G, F) − i for some i 2 f0, 1, 2g. 2. Proof of TheoremA Proof. Suppose that the result is false. Let G be a counterexample of the smallest possible order. Then G has a maximal subgroup A such that ns(A, F) 6= ns(G, F)) − i for every i 2 f0, 1, 2g. Since AF is contained in GF because F is subgroup-closed, we have that GF 6= 1. Moreover, Mathematics 2020, 8, 2165 3 of 4 ns(A, F) ≤ ns(G, F) = n and n ≥ 1. We proceed in several steps, the first of which depends heavily on the fact that the F-residual is epimorphism-invariant. Step 1. If N is a normal s-nilpotent subgroup of G, then N is contained in A, ns(A, F) = ns(A/N, F) and ns(G/N, F) = n − 1. Let N be a normal s-nilpotent subgroup of G. Applying [7] (Chapter II, Lemma (2.4)), we have F F that G N/N = (G/N) . Consequently, either ns(G/N, F) = n or ns(G/N, F) = n − 1. Assume that N is not contained in A. Then G = AN and so G/N =∼ A/A \ N. Observe that either ns(A/A \ N, F) = ns(G/N, F) = n or ns(A/A \ N, F) = ns(G/N, F) = n − 1. Therefore n − 1 ≤ ns(A, F) ≤ n. Consequently, either ns(A, F) = n or ns(A, F) = n − 1, contrary to assumption. Therefore, N is contained in A. The minimal choice of G implies that ns(A/N, F) = ns(G/N, F) − i for some i 2 f0, 1, 2g, and so either ns(A/N, F) = n − i or ns(A/N, F) = n − i − 1. Suppose that ns(A, F) 6= ns(A/N, F). Then ns(A, F) = ns(A/N, F) + 1. Hence either ns(A, F) = n − i + 1 or ns(A, F) = n − i. In the first case, i > 0 because n ≥ ns(A, F). Hence ns(A, F) = n − j for some j 2 f0, 1, 2g, which contradicts our supposition. Consequently, ns(A, F) = ns(A/N, F). Suppose that ns(G/N, F) = n. The minimality of G yields ns(A/N, F) = n − i for some i 2 f0, 1, 2g.