Abstract

This paper is devoted to introduce a sufficient condition for the solvability of finite groups. Also, it presents the concepts of m-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.

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]).

In [7], Dolfi et al. introduced the following criterion.

G is solvable if, for all conjugacy classes C and D of G consisting of elements of prime power order, there exist and such 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 order 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 . 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 and be a fixed integer. We say that G is of exponent type m if for any there exists such that (b)Let G be a group and H►G, and we say that H is m-normal in G if for any there exists such that If this is the case, we denote H ►mG.

Theorem 1 (see [1]). Let G be a group and m be a fixed integer; then,(a)G is of exponent type m if and only if is a subgroup of G.(b)The homomorphic image of a group of exponent type m is also of exponent type m.

Theorem 2 (see [1]). Suppose that G is a finite group of exponent type m, where m is a prime divisor of . Then, G is solvable if and only if is solvable.

Definition 2 (see [1]). Suppose that G is a finite group of exponent type m. We say G is m-abelian if 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 Theorem 1 and the fact that the homomorphic image of an abelian group is abelian.

Definition 3. Let G be a group. The m-th commutator of x, y G is defined asThe m-derived subgroup of G denoted by is defined to be the subgroup generated by all m-th commutators of G.

Lemma 1. Let G be a group of exponent type m. Then,(a)► G.(b)G is m-abelian if and only if .(c)For H ►mG, G/H is m-abelian if and only if .

Proof. (a)Let be a homomorphism on G. z , we have(i); , so , and hence and is fully invariant and then normal.(b) is abelian if and only if .(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 . Since the direct product of abelian groups is abelian, is abelian. Thus, GH is m-abelian.

Theorem 4. Let G be an m-abelian group with m being a prime divisor of . Then, G is solvable.

Proof. The result holds directly from Theorem 2.

2. m-Abelian Solvable Groups

Definition 4. (a)Let be a subnormal series of a group G.(i)We say that it is an m-abelian solvable series if is m-abelian for every 1n.(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 integer m.(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) series .(b)If G is solvable, then it has a subnormal series with abelian factors. Since every abelian group is (m-abelian) for each integer m, G must be (m-abelian solvable).(c)Let be a group homomorphism; first of all we will prove that . , and thus . The other inclusion can be proved by the same.Suppose that G is (m-abelian solvable); then, it has an (m-abelian solvable) series . We have and ; this implies that . It is easy to show that, and thus we obtain an (m-abelian solvable) series , and hence G is (m-abelian solvable).

Theorem 5. Let G be a group, and we have(a)If HG and H is solvable with H, then G is (m-abelian solvable).(b)If there is a positive integer k such 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)Let be the solvable series of H; we have that is (m-abelian solvable) series of G, and our proof is complete.(b)It is easy to see that 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: . For each 0 m we can find a subgroup such that ; by isomorphism theorem, we obtain , and thus is (m-abelian), 1, and this implies that the series 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: ; without affecting the generality, we can assume that n  m; let the series () be ; since , 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 prime m divides its order; then, G is solvable.

Proof. There is an (m-abelian solvable) series , and we have that is (m-abelian), so is solvable and is (m-abelian); thus, it is solvable, and is solvable. By the same argument, we find that G is solvable.

Example 1. Consider the finite group ; we have Z(G) as a normal subgroup of order 2, and hence is of order 4.
Z(G), 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) if is cyclic.(b)Let G be a group with a subnormal series , and we say that it is (m-cyclic solvable) series if is (m-cyclic) for each 1  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 integer m.(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 prime m/, 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 Theorem 1, 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 integer m.(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) series .(b)If G is polycyclic, then it has a subnormal series 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) series ; suppose that H is a normal subgroup of G, and let ; because of , we get . By isomorphism theorem, we get . Now we must prove that . We have , so and this means that ; thus, is an (m-group).By using isomorphism theorem, we obtain , which is a homomorphic image of (m-cyclic) group, and hence 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: ; for each 0 m, we can find a subgroup such that .
By isomorphism theorem, we obtain , so is (m-cyclic); 1; this implies that the series is (m-cyclic solvable).

Theorem 10. Let G be a finite (m-cyclic solvable) group where a prime m 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 (-group) with m/, and let  = ; ; then,(a)/.(b) is a (p-group) with .(c)G is solvable if and only if is solvable.

Proof. (a)For each prime , the (-Sylow) subgroup has order with gcd(, m) = 1.(i)So, ; then, for each , and thus /.(b) = , k, so G/ is a (p-group) with .(ii)We meant by (p-group) a group with order ; s and p is prime.(c)Assume that is solvable; then, G/ 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 (-group) with m/; then,(a)If G is simple, then it is cyclic of order m.(b)If H►G, then H/ is a (p-group) with .

Proof. (a)We have m/, so that G, but , so , and G/ is a (p-group), and in this case, G/G, which means that G is a simple (p-group); thus, G is cyclic with order m.(b)Suppose that H►G; then, and H/(, and thus H/ is a (p-group).

Remark 1. If we consider that the finite group G is (-group) with  = , where m, are distinct primes and k, then Lemmas (5) and 6 are still true.

Theorem 11. Let G be a finite group. If every normal subgroup H of G is (-group) where a prime m divides O(H), then G is solvable.

Proof. G is (-group); then, and G/ is (p-group), and it is solvable. is where prime n divides O, so is a (p-group) and is solvable. By the same argument, we get a series such where prime m divides O() and each factor is a p-group and is solvable for each . This implies that is solvable and then is solvable and so on, and thus G is solvable.
Theorem 11 is still true if every normal subgroup is (-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 to , suppose that for every normal subgroup H of G, there exists a fixed positive integer , with the following property: for each , there is such that . 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 is -group with some prime m dividing O(G).
By Theorem 11, we can find that if this conjecture is true, then Feit–Thompson theorem holds.
Conjecture 1 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 Theorem 5.
Consider , the symmetric group of order 6. G is a , since .
The only normal subgroup of G is 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 if m 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 prime m, 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].

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.