# On a New Criterion for the Solvability of Non-Simple Finite Groups and m-Abelian Solvability

Authors: Hasan Sankari; Mohammad Abobala
Journal: Journal of Mathematics
 (2021)

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## Abstract

This paper is devoted to introduce a sufficient condition for the solvability of finite groups. Also, it presents the concepts ofm-abelian and m-cyclic solvability as new generalizations of solvability and polycyclicity, respectively. These new generalizations show a connection between prime powers of elements in a finite group G and its solvability.


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## Body

## 1. Introduction

An important problem in the theory of groups came to light after Galois’ work [4]. This problem is concerned with determining whether a group G is solvable or not. According to the literature, many conditions and criteria were introduced to deal with this problem. Feit and Thompson had proved that each finite group of odd order is solvable (see [4, 5]).Arad and Ward had proved Hall’s Conjecture about solvability in [6].In [7], Dolfi et al. introduced the following criterion.G is solvable if, for all conjugacy classes C andD of G consisting of elements of prime power order, there exist x∈C and y∈Dsuch that x and y generate a solvable group.Through this paper, we use the concept of (m-power closed group) [3] to introduce a sufficient condition that implies the solvability of a finite group.On the other hand, we define two new solvability systems (m-abelian/m-cyclic solvability) based on m-power closed groups. These notions will provide us an interesting connection between prime divisors of the orderG of a finite group G and its solvability. In fact, m-abelian solvability turns out to be equivalent to the classical solvability whenever m is a prime divisor of G. We recall some basic definitions and theorems.All groups throughout this paper are considered finite.Definition 1 (see [1]).
(a)
Let G be a group andm be a fixed integer. We say that G is of exponent type m if for any x,y∈G, there exists z∈G such that xmym=zm.(b)
Let G be a group and H►G, and we say that H is m-normal in G if for anyx,y∈G, there exists z∈G such that zmxmym∈H. If this is the case, we denote H ►mG.Theorem 1 (see [1]).
Let G be a group andm be a fixed integer; then,(a)
G is of exponent typem if and only if Gm=gm;g∈G is a subgroup of G.(b)
The homomorphic image of a group of exponent typem is also of exponent type m.Theorem 2 (see [1]).
Suppose that G is a finite group of exponent typem, where m is a prime divisor of G. Then, G is solvable if and only if Gm is solvable.Definition 2 (see [1]).
Suppose that G is a finite group of exponent typem. We say G is m-abelian if Gm is abelian.Theorem 3.
Let G be an m-abelian group. Then, the homomorphic image of G is also m-abelian.Proof.
It is clear by Theorem1 and the fact that the homomorphic image of an abelian group is abelian.Definition 3.
Let G be a group. The m-th commutator ofx, y∈ G is defined as(1)x,ym=x−my−mxmym.
The m-derived subgroup of G denoted byGm′ is defined to be the subgroup generated by all m-th commutators of G.Lemma 1.
Let G be a group of exponent typem. Then,(a)
Gm′► G.(b)
G is m-abelian if and only ifGm′=e.(c)
For H ►mG, G/H is m-abelian if and only if Gm′≤H.Proof.
(a)
Letφ be a homomorphism on G. ∀z∈Gm′, we have(i)
z=∏i=1i=nxi−myi−mximyim; xi,yi∈G, so φG=∏i=1i=nφxi−mφyi−mφximφyim, and hence φz∈Gm′ and Gm′ is fully invariant and then normal.(b)
Gm is abelian if and only if Gm′=Gm′=e.(c)
It is easy and clear.Lemma 2.
The direct product of two m-abelian groups is m-abelian again.Proof.
Let G, H be two m-abelian groups; then, G× H is a group of exponent type m and G×Hm=Gm×Hm. Since the direct product of abelian groups is abelian, G×Hm is abelian. Thus, G×H is m-abelian.Theorem 4.
Let G be an m-abelian group withm being a prime divisor of G. Then, G is solvable.Proof.
The result holds directly from Theorem2.


## 2. m-Abelian Solvable Groups

Definition 4.
(a)
Lete=H0≤H1≤⋯≤Hn=G be a subnormal series of a group G.(i)
We say that it is an m-abelian solvable series ifHi/Hi−1 is m-abelian for every 1≤i≤n.(b)
We say that G is m-abelian solvable if it has an m-abelian solvable series.Lemma 3.
Let G be a group, and we have(a)
If G is (m-abelian), then it is (m-abelian solvable).(b)
If G is solvable, then G is (m-abelian solvable) for each integerm.(c)
The homomorphic image of (m-abelian solvable) group is (m-abelian solvable).Proof.
(a)
If G is (m-abelian), then it has an (m-abelian solvable) seriese≤G.(b)
If G is solvable, then it has a subnormal seriese=H0≤H1≤⋯≤Hn=G with abelian factors. Since every abelian group is (m-abelian) for each integer m, G must be (m-abelian solvable).(c)
Letφ:G⟶K be a group homomorphism; first of all we will prove that φGm′=φGm′. ∀h∈φGm′,h=φ∏i=1i=nxi−myi−mximyim=∏i=1i=nφxi−mφyi−mφximφyim∈φGm′, and thus φGm′≤φGm′. The other inclusion can be proved by the same.
Suppose that G is (m-abelian solvable); then, it has an (m-abelian solvable) seriese=H0≤H1≤⋯≤Hn=G. We have Hi−1▶mHi and Him′≤Hi−1; this implies that φHim′=φHim′≤φHi−1. It is easy to show thatφHi−1▶mφHi, and thus we obtain an (m-abelian solvable) series e≤φH0≤φH1≤⋯≤φHn=φG, and hence G is (m-abelian solvable).Theorem 5.
Let G be a group, and we have(a)
If H▶mG and H is solvable with Gm′≤H H, then G is (m-abelian solvable).(b)
If there is a positive integerk such Gk is (m-abelian), then G is (m-abelian solvable).(c)
If H ► G and H, G/H are (m-abelian solvable), then G is (m-abelian solvable).Proof.
(a)
Lete=H0≤H1≤⋯≤Hn=H be the solvable series of H; we have that e=H0≤H1≤⋯≤Hn=H≤G is (m-abelian solvable) series of G, and our proof is complete.(b)
It is easy to see thate≤Gk≤Gk−1≤⋯≤G0=G is an (m-abelian solvable) series of G.(c)
Suppose that H, G/H are (m-abelian solvable), and we have the following two (m-abelian solvable) series:e=H0≤H1≤⋯≤Hn=HandH=K0¯≤K1¯≤⋯≤Km¯=G/H. For each 0≤j≤ m we can find a subgroup Kj≤G such that K0=H,Km=G,Kj▶Kj+1,Kj¯=Kj/H; by isomorphism theorem, we obtain Kj¯/Kj−1¯≅Kj/H/Kj−1/H≅Kj/Kj−1, and thus Kj/Kj−1 is (m-abelian), 1≤j≤m, and this implies that the series e=H0≤H1≤⋯≤Hn=H≤K0≤K1≤⋯≤Km=G is (m-abelian solvable).Theorem 6.
(a)
The direct product of two (m-abelian solvable) groups is (m-abelian solvable).(b)
The direct product of finite number of (m-abelian solvable) groups is (m-abelian solvable).Proof.
(a)
Let G, H be two (m-abelian solvable) groups, and we have the following (m-abelian solvable) series:e1=H0≤H1≤⋯≤Hn=Hande2=K0≤K1≤⋯≤Km=G; without affecting the generality, we can assume that n ≥ m; let the series (∗) be H0×K0≤H1×K1≤⋯≤Hm×Km≤Hm+1×Km≤⋯≤Hn×Km=H×G; since Hi×Kj/Hi−1×Kj−1≅Hi/Hi−1×Kj/Kj−1for1≤i≤n,1≤j≤mandHi×Km/Hi−1×Km≅Hi/Hi−1×Km/Km;m+1≤i≤n, each factor of series (∗) is (m-abelian), and hence H × G is (m-abelian solvable).(b)
It holds directly by an easy induction.Theorem 7.
Let G be a finite (m-abelian solvable) group where primem divides its order; then, G is solvable.Proof.
There is an (m-abelian solvable) seriese=H0≤H1≤⋯≤Hn=G, and we have that H1/H0≅H1 is (m-abelian), so H1 is solvable and H2/H1 is (m-abelian); thus, it is solvable, and H2 is solvable. By the same argument, we find that G is solvable.Example 1.
Consider the finite groupG=D4; we have Z(G) as a normal subgroup of order 2, and hence G/ZG is of order 4.
Z(G), G/ZG are 2-abelian groups since they are abelian; this implies that G is 2-abelian solvable, and then it is solvable.


## 3. (m-Cyclic Solvable) Groups

Definition 5.
(a)
Let G be an (m-group); it is (m-cyclic) ifGm is cyclic.(b)
Let G be a group with a subnormal seriese=H0≤H1≤⋯≤Hn=G, and we say that it is (m-cyclic solvable) series if Hi/Hi−1 is (m-cyclic) for each 1 ≤i≤ n.(c)
We say that G is (m-cyclic solvable) if it has an (m-cyclic solvable) series.Lemma 4.
Let G be a group, and we have(a)
If G is cyclic, then it is (m-cyclic) for each integerm.(b)
If G is an (m-cyclic) group, then the homomorphic image of G is (m-cyclic).(c)
If G is an (m-cyclic) group with a primem/G, then G is solvable.Proof.
(a)
A subgroup of cyclic group is cyclic, so it is clear.(b)
It is known that the homomorphic image of any cyclic group is cyclic and by Theorem1, the proof is complete.(c)
It holds easily, since each m-cyclic group is m-abelian group.Theorem 8.
Let G be a group, and we have(a)
If G is (m-cyclic), then it is (m-cyclic solvable).(b)
If G is polycyclic, then G is (m-cyclic solvable) for each integerm.(c)
The homomorphic image of any (m-cyclic solvable) group is (m-cyclic solvable).Proof.
(a)
If G is (m-cyclic), then it has an (m-cyclic solvable) seriese≤G.(b)
If G is polycyclic, then it has a subnormal seriese=H0≤H1≤⋯≤Hn=G with cyclic factors. Since every cyclic group is (m-cyclic) for each integer m, G must be (m-cyclic solvable).(c)
Assume that G is (m-cyclic solvable); then, it has an (m-cyclic solvable) seriese=H0≤H1≤⋯≤Hn=G; suppose that H is a normal subgroup of G, and let Ki=HiH/H;0≤i≤n; because of Hi−1▶Hi, we get e=K0≤K1≤⋯≤Kn=G/H. By isomorphism theorem, we get Ki/Ki−1≅HiH/H/Hi−1H/H≅HiH/Hi−1H≅HiHi−1H/Hi−1H≅Hi/Hi∩Hi−1H. Now we must prove that Hi∩Hi−1H▶mHi. We have Hi−1▶mHi, so ∀x,y∈Hi∃k∈Hi;kmxmym∈Hi−1≤Hi−1Handkmxmym∈Hisokmxmym∈Hi∩Hi−1H, and this means that Hi∩Hi−1H▶mHi; thus, Ki/Ki−1 is an (m-group).
By using isomorphism theorem, we obtainHi/Hi∩Hi−1H≅Hi/Hi−1/Hi∩Hi−1H/Hi−1, which is a homomorphic image of (m-cyclic) group, and hence Ki/Ki−1 is (m-cyclic) and G/H must be (m-cyclic solvable).Theorem 9.
Let G be a group and H►G. If H and G/H are (m-cyclic solvable), then G is (m-cyclic solvable).Proof.
Suppose that H, G/H are (m-cyclic solvable). We have the following two (m-cyclic solvable) series:e=H0≤H1≤⋯≤Hn=HandH=K0¯≤K1¯≤⋯≤Km¯=G/H; for each 0≤j≤ m, we can find a subgroup Kj≤G such that K0=H,Km=G,K▶Kj+1,Kj¯=Kj/H.
By isomorphism theorem, we obtainKj¯/Kj−1¯≅Kj/H/Kj−1/H≅Kj/Kj−1, so Kj/Kj−1 is (m-cyclic); 1≤j≤m; this implies that the series e=H0≤H1≤⋯≤Hn=H≤K0≤K1≤⋯≤Km=G is (m-cyclic solvable).Theorem 10.
Let G be a finite (m-cyclic solvable) group where a primem divides its order; then, G is solvable.Proof.
Since every (m-cyclic solvable group) is an (m-abelian solvable), the proof holds.


## 4. Sufficient Condition for Solvability

Lemma 5 (see [1]).
Let G be an (m∗-group) with m/G, and let G = mk1p2k2,…,psks; piare distinct primes for each2≤i≤s; then,(a)
p2k2,…,psks/Gm.(b)
G/Gm is a (p-group) with m=p.(c)
G is solvable if and only ifGm is solvable.Proof.
(a)
For each primepi, the (pi-Sylow) subgroup Hi has order piki with gcd(piki, m) = 1.(i)
So,Him=Hi≤Gm; then, piki/Gm for each i, and thus p2k2,…,psks/Gm.(b)
G/Gm = mk, k≤k1, so G/Gm is a (p-group) with m=p.(ii)
We meant by (p-group) a group with orderps; s∈N and p is prime.(c)
Assume thatGm is solvable; then, G/Gm is solvable because it is a (p-group). This means that G is solvable, and the converse is clear.Lemma 6 (see [1]).
Let G be an (m∗-group) with m/G; then,(a)
If G is simple, then it is cyclic of orderm.(b)
If H►G, then H/H∩Gm is a (p-group) with p=m.Proof.
(a)
We have m/G, so that G≠Gm, but Gm▶G, so Gm=e, and G/Gm is a (p-group), and in this case, G/Gm≅G, which means that G is a simple (p-group); thus, G is cyclic with order m.(b)
Suppose that H►G; then,Gm∩H▶H and H/(H∩Gm≅GmH/Gm≤G/Gm, and thus H/H∩Gm is a (p-group).Remark 1.
If we consider that the finite group G is (mk-group) with G = mk1p2k2,…,psks, where m,p2,p3,…,ps are distinct primes and k≤k1, then Lemmas (5) and 6 are still true.Theorem 11.
Let G be a finite group. If every normal subgroup H of G is (m∗-group) where a prime m divides O(H), then G is solvable.Proof.
G is (m∗-group); then, Gm▶G and G/Gm is (p-group), and it is solvable. Gm is n∗−group where prime n divides OGm, so Gm/Gmn is a (p-group) and is solvable. By the same argument, we get a series e≤H1≤⋯≤Hl≤G such Hi=Hi+1m where prime m divides O(Hi+1) and each factor Hi+1/Hi is a p-group and is solvable for each i. This implies that H1 is solvable and then H2 is solvable and so on, and thus G is solvable.
Theorem11 is still true if every normal subgroup is (mk-group) with some prime m dividing its order.
The previous theorem can be described by the following form.
If G is a finite m-power closed group with respect tom=pk;pis a prime, suppose that for every normal subgroup H of G, there exists a fixed positive integer n=qs;qis a prime andm/H, with the following property: for each x,y∈H, there is z∈H such thatxnyn=zn . Then, G is solvable.
The previous theorem can be considered as a new criterion to determine if a finite group G is solvable.Conjecture 1.
Each finite group G with odd order ismk-group with some prime m dividing O(G).
By Theorem11, we can find that if this conjecture is true, then Feit–Thompson theorem holds.
Conjecture1 is very important, since if it is true, we will get an easy proof to a famous basic theorem in algebra.Example 2.
This example is devoted to clarify the validity of our criterion in Theorem5.
ConsiderG=S3, the symmetric group of order 6. G is a 2−group, since G2≅Z3.
The only normal subgroup of G isH≅Z3 which is a 3-group since it is abelian, and thus G is solvable according to Theorem 11.


## 5. Conclusion

In this article, we have introduced the concept of (m-abelian solvability) and (m-cyclic solvability) as two new generalizations of classical solvability and polycyclicity, respectively.We have discussed some elementary properties of these concepts and proved the main result through this paper which ensures that m-abelian solvability is equivalent of solvability in finite groups ifm is a prime number that divides the order of the group.This result shows a kind of connection between primes and solvability in finite groups. An interesting question came to light according to this work. This question can be asked as follows:If G is an infinite m-abelian solvable group for a primem, then is G solvable?Also, we have introduced a new sufficient condition for the solvability of finite non-simple group G based on m-power closed groups concept.As a future research direction, m-abelian solvability can be extended to AH-subgroups defined in [8] and neutrosophic groups in [9].



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