On the supersolvability of a finite group by the sum of subgroup orders
aa r X i v : . [ m a t h . G R ] F e b On the supersolvability of a finite group by thesum of subgroup orders
Marius T˘arn˘auceanuFebruary 14, 2021
Abstract
Let G be a finite group and σ ( G ) = | G | P H ≤ G | H | . In this paper,we prove that if σ ( G ) < | G | , then G is supersolvable. In particular,some new characterizations of the well-known groups Z × Z and A are obtained. We also show that σ ( G ) < c does not imply thesupersolvability of G for no constant c ∈ (2 , ∞ ). MSC2000 :
Primary 20D60; Secondary 20D10, 20F16, 20F17.
Key words : subgroup orders, supersolvable groups.
Given a finite group G , we consider the function σ ( G ) = 1 | G | X H ≤ G | H | studied in our previous papers [7, 10, 11]. Recall some basic properties of σ :- if G is cyclic of order n and σ ( n ) denotes the sum of all divisors of n ,then σ ( G ) = σ ( n ) n ;- σ is multiplicative, i.e. if G i , i = 1 , , . . . , m , are finite groups ofcoprime orders, then σ ( Q mi =1 G i ) = Q mi =1 σ ( G i );- σ ( G ) ≥ σ ( G/H ) + G : H ) ( σ ( H ) − ≥ σ ( G/H ), for all H E G .1he function σ was used to provide some criteria for a group to belongto one of the well-known classes of groups. For instance, if σ ( G ) ≤ G is cyclic of deficient or perfect order by [7], while if σ ( G ) ≤ then G issolvable by [4, 11]. Also, we note that there is no constant c ∈ (2 , ∞ ) suchthat σ ( G ) < c implies the nilpotency of G , but this can be obtained fromthe condition σ ( G ) < | G | (see [10]).In the current paper, we will first determine the structure of finite groups G satisfying σ ( G ) <
3. This will allow us to prove that if σ ( G ) < | G | then G is supersolvable. Moreover, A is the unique finite non-nilpotentgroup G such that σ ( G ) = 2 + | G | . Then we will show that σ ( G ) < c doesnot imply the supersolvability of G for no constant c ∈ (2 , ∞ ).For the proof of our results, we need the following two theorems (seeTheorem 1 of [6] and Theorem 1 of [8]). Theorem 1.1.
Suppose every non-normal maximal subgroup of a finite group G has the same order. Then G is solvable and the nilpotent length of G is atmost two. Theorem 1.2.
A finite group G contains a cyclic maximal subgroup if andonly if it is of one of the following types: a) G = P × K , where K is an arbitrary finite cyclic group and P is aSylow p -subgroup of G containing a cyclic maximal subgroup. b) G = ( K ⋊ P ) × K , where K is an arbitrary finite cyclic group whichis a Hall subgroup of G , P is a Sylow p -subgroup of G containing acyclic maximal subgroup, K ⋊ P is non-nilpotent, and the centralizer C P ( K ) is a maximal cyclic subgroup of P . c) G = ( P ⋊ K ) × K , where K is an arbitrary finite cyclic group whichis a Hall subgroup of G and G = P ⋊ K is a non-nilpotent groupsatisfying the following conditions: P is a SyIow p -subgroup of G , C P ( K ) ⊇ Φ( P ) , and C P ( K ) is a cyclic invariant subgroup of G suchthat K C P ( K ) /C P ( K ) is a maximal subgroup of G /C P ( K ) . We also need two auxiliary results, taken from [11] and [7], respectively.
Lemma 1.3.
Let G be a finite group and [ M ] be a conjugacy class of non-normal maximal subgroups of G . Then X H ∈ [ M ] | H | = | G | . emma 1.4. Let G be a finite group. Then X H ≤ G, H =cyclic | H | ≥ | G | . Note that similar problems for some other functions related to the struc-ture of a finite group G , for example for the function ψ ( G ) = P x ∈ G o ( x )(where o ( x ) denotes the order of the element x ), have been recently investi-gated by many authors (see e.g. [1, 2, 3]).Most of our notation is standard and will not be repeated here. Basicdefinitions and results on groups can be found in [5]. For subgroup latticeconcepts we refer the reader to [9]. We start with the following important lemma.
Lemma 2.1.
Let G be a finite group with σ ( G ) < . Then either G isnilpotent or a group of type c) in Theorem 1.2.Proof. Assume that G is not nilpotent. If G has at least two conjugacyclasses of non-normal maximal subgroups [ M ] and [ M ], then Lemma 1.3leads to σ ( G ) > | G | | G | + X i =1 X H ∈ [ M i ] | H | = 1 | G | ( | G | + 2 | G | ) = 3 , a contradiction. Thus G possesses a unique conjugacy class of non-normalmaximal subgroups, say [ M ]. Moreover, M is cyclic. Indeed, if this is nottrue, then Lemmas 1.3 and 1.4 imply σ ( G ) ≥ | G | | G | + X H ∈ [ M ] | H | + X H ≤ G, H =cyclic | H | ≥ | G | ( | G | + | G | + | G | ) = 3 , contradicting again our hypothesis.From Theorem 1.1 it follows that G is solvable of nilpotent length two.More precisely, by taking a prime p dividing [ G : M ], we infer that a Sylow p -subgroup P of G must be normal and a p -complement of G must be cyclic.This shows that G is a group of type c) in Theorem 1.2, completing theproof. 3e are now able to prove our first main result. Theorem 2.2.
Let G be a finite group. If σ ( G ) < | G | , then G issupersolvable.Proof. Since all groups of order ≤
11 are supersolvable, we may assume that | G | ≥
12. Then σ ( G ) < G isnilpotent or G = ( P ⋊ K ) × K with P , K and K as in Theorem 1.2, c).In the first case we are done, while in the second case we may assume that G = G = P ⋊ K because σ ( G ) = σ ( G/K ) ≤ σ ( G ) < | G | ≤ | G | . (1)Denote by [ M ] the unique conjugacy class of non-normal maximal subgroupsof G . Let | P | = p r and suppose that P is not cyclic. Then r ≥ P contains at least p + 1 subgroups of order p r − . It follows that σ ( G ) ≥ | G | | G | + X H ∈ [ M ] | H | + X H ≤ P | H | ≥ p + · · · + ( p + 1) p r − + p r | G |≥ p + 1) p + p | G | = 2 + 1 + p + 2 p | G | ≥ | G | , contradicting (1). Thus P is cyclic and consequently all Sylow subgroups of G are cyclic, implying that G is supersolvable. This completes the proof.Next we will determine finite groups G such that σ ( G ) = 2 + | G | . Firstof all, we will focus on p -groups. We observe that for all primes p and allpositive integers n we have σ ( Z p n ) = σ ( p n ) p n = 1 + p + · · · + p n p n = 1 + 1 p − − p n ( p − < . The following lemma deals with the case of non-cyclic p -groups. Lemma 2.3.
Let G be a non-cyclic p -group of order p n . Then: a) σ ( G ) < | G | if and only if either n = 3 , p = 2 and G = Q , or n = 2 , p ∈ { , , , } and G = Z p × Z p ; b) σ ( G ) = 2 + | G | if and only if n = 3 , p = 2 and G = Z × Z . roof. Since G contains at least p + 1 subgroups of order p n − , one obtains2 + 11 p n ≥ σ ( G ) ≥ p + · · · + ( p + 1) p n − + p n p n = 2 + 1 + p + · · · + p n − p n , that is 1 + p + · · · + p n − ≤
11. Thus either n = 3 and p = 2, or n = 2and p ∈ { , , , } . The conclusion follows by checking the correspondinggroups.Our second main result is stated as follows. Theorem 2.4.
Let G be a finite non-cyclic group. If σ ( G ) = 2 + | G | , theneither G = Z × Z or G = A .Proof. We can easily see that G = Z × Z is the unique group G of order ≤
11 satisfying σ ( G ) = 2 + | G | . Assume that | G | ≥
12. Then σ ( G ) < G is nilpotent or G = ( P ⋊ K ) × K with P , K and K as in Theorem 1.2, c).In the first case, let G = G × · · · × G k be the decomposition of G as adirect product of Sylow p -subgroups. For k = 1, we have no solution G of σ ( G ) = 2 + | G | by Lemma 2.3, b). For k ≥
2, we get σ ( G i ) < | G i | , ∀ i = 1 , ..., k . Then Lemma 2.3, a), shows that every G i is of one of thefollowing types: Q , Z p i × Z p i or Z p nii for some prime p i and some positiveinteger n i . Since σ ( Q ) = 238 > , σ ( Z p i × Z p i ) = 2 + p i + 1 p i > , ∀ i = 1 , ..., k and σ ( G ) = σ ( G ) · · · σ ( G k ) , we infer that at most one of G i ’s is non-cyclic. Suppose that G = Q and G i = Z p nii , for all i = 2 , ..., k . Then our condition becomes238 σ ( p n ) p n · · · σ ( p n k k ) p n k k = 2 + 118 p n · · · p n k k , i.e. 23 σ ( p n ) · · · σ ( p n k k ) = 16 p n · · · p n k k + 11 , a contradiction (left side > right side). Similarly, if we suppose G = Z p × Z p and G i = Z p nii , for all i = 2 , ..., k , then we obtain5 p + p + 1 p σ ( p n ) p n · · · σ ( p n k k ) p n k k = 2 + 11 p p n · · · p n k k , i.e. (2 p + p + 1) σ ( p n ) · · · σ ( p n k k ) = 2 p p n · · · p n k k + 11 , a contradiction (left side > right side). Thus every G i is cyclic, and so is G ,contradicting our hypothesis.In the second case, a similar reasoning shows that we have no solution G = ( P ⋊ K ) × K of σ ( G ) = 2 + | G | with K = 1. Consequently, wemay assume G = P ⋊ K . Let | P | = p r and | K | = n . Note that p ∤ n and n | | Aut G | . As in the proof of Theorem 2.2, we get2 + 11 | G | = σ ( G ) ≥ X H ≤ P | H || G | and thus X H ≤ P | H | ≤ . It follows that either r = 2, p = 2 and P ∈ { Z , Z × Z } , or r = 1, p ∈ { , , , } and P = Z p .If P = Z , then n |
2, a contradiction. If P = Z × Z , then n | n = 3; thus G = A , which is a solution of σ ( G ) = 2 + | G | . If P = Z p with p ∈ { , , , } , then n divides 1, 2, 4 or 6, respectively. We infer that theunique possibilities are( p, n ) ∈ { (3 , , (5 , , (5 , , (7 , , (7 , , (7 , } . By checking all these semidirect products G = Z p ⋊ Z n we find no othersolution of σ ( G ) = 2 + | G | . Hence in this case the unique solution is G = A ,as desired.Obviously, Theorem 2.4 leads to the following characterizations of Z × Z and A . Corollary 2.5. Z × Z is the unique finite non-cyclic nilpotent group G satisfying σ ( G ) = 2 + | G | , while A is the unique finite non-nilpotent group G satisfying σ ( G ) = 2 + | G | . σ ( G ) < = σ ( A ) does not imply the super-solvability of G , a counterexample being SmallGroup (56 ,
11) (also knownas the Frobenius group F ). In fact, we are able to prove that there is noconstant c ∈ (2 , ∞ ) such that σ ( G ) < c implies the supersolvability of G . Theorem 2.6.
There are sequences of finite non-supersolvable groups ( G n ) n ∈ N such that σ ( G n ) approaches monotonically from above as n tends to infi-nity.Proof. Let ( q n ) n ∈ N be the sequence of odd prime numbers. By Dirichlet’stheorem we infer that for every n ∈ N there is a prime p n such that q n | p n − G n be the unique non-abelian group of order p n q n . Clearly, G n is notsupersolvable. It contains one subgroup of order 1, p n + 1 subgroups of order p n , one subgroup of order p n , p n subgroups of order q n , and one subgroup oforder p n q n . Then σ ( G n ) = 1 + ( p n + 1) p n + p n + p n q n + p n q n p n q n = 2 + 2 + 1 p n + 1 p n q n , which approaches 2 monotonically from above as n tends to infinity, as de-sired.We end this paper by formulating a natural problem concerning the abovestudy. Open problem.
Find all finite cyclic groups G satisfying σ ( G ) = 2 + | G | .Note that this is equivalent to finding all positive integers n = p n · · · p n k k such that(1 + p + · · · + p n ) · · · (1 + p k + · · · + p n k k ) = 2 p n · · · p n k k + 11 . References [1] S.M. Jafarian Amiri,
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II, Comm. Algebra (2017), 4865-4868.[11] M. T˘arn˘auceanu, On the solvability of a finite group by the sum of sub-group orders , Bull. Korean Math. Soc. (2020), 1475-1479. Marius T˘arn˘auceanuFaculty of Mathematics“Al.I. Cuza” UniversityIa¸si, Romaniae-mail: [email protected]@uaic.ro