Research Article | Open Access

# Splitting Groups with Basis Property

**Academic Editor:**E. Bannai

#### Abstract

A finite group is called splitting or splittable if it is a union of some collections of its proper subgroups intersecting pairwise at the identity. A special kind of splitting is known to be normal splitting. Also, a group is said to have the basis property if, for each subgroup , has a basis (minimal generating set), and any two bases have the same cardinality. In this work, I discuss a relation between classes of finite groups that possess both normal splitting and the basis property. This paper shows mainly that any non--group with basis property is normal splitting. However, the converse is not true in general. A counterexample is given. It is well known that any -group has basis property. I demonstrate some types of -groups which are splitting as well.

#### 1. Introduction

Miller [1] was the first one who studied the concept of splitting groups; then it was discussed by many authors such as Young [2] who gave a precise definition of this concept. Hugher and Thompson [3] introduced the definition of HT-groups as follows: a finite nilpotent group (not -group) is called HT-group if there exists a prime number and a subgroup which is generated by all elements of order not equal to with and .

In Baer [4, 5], Kegel [6], and Kontorovich [7], it was given a classification of solvable non--groups that are splitting. Also, Isaacs [8] studied the so-called equally splitting. Zassenhaus [9] had a great role in studying splitting groups, and Suzuki [10] had a crucial contribution in this regard by finding a new class of simple splitting groups, and finishing the topic of classifying unsolvable splitting finite groups. Sozutov and Shlepkin [11] generalized Kegel’s theorem [6] on finite HT-groups and Baer’s theorem [5]. I use the same notation and terminology that appeared in the references, especially [5, 12]. In [13], it is shown that any finite group with basis property is either elementary -group, nonelementary -group, or a group that can be written as , where is of order and is a cyclic -group of order where , are distinct primes and . This work shows that all finite groups with basis property of the first or third kind are splitting, while groups of the second kind may or may not be splitting.

Let be a nontrivial finite group. A collection of nontrivial subgroups of is said to be splitting of if every element of is contained in one and only one subgroup of , and is called nontrivial if for all .

Let be a group with splitting ; then the subgroup is called a component of this splitting. So, one has Let be a subgroup of ; then the collection of subgroups of defined by is a splitting of . This is called the induced splitting. The induced splitting is nontrivial if and only if is contained in no component of the splitting .

Lemma 1 (Young [2]). *If a torsion abelian group has a splitting, then there exists a prime number so that each nonidentity element is of order . Moreover, if (prime number), and the nonidentity elements of have order , then has a nontrivial splitting.*

Lemma 2 (Miller [1]). *An abelian group of finite order has a nontrivial splitting if and only if it is of order (), and it must be an elementary abelian -group.*

#### 2. Some Results on Splitting -Groups

In 1960-1961, Baer [4, 5], Kegel [6], and Suzuki [10] obtained the classification of finite splitting groups that are not -groups, such groups must be either simple or solvable.

Theorem 3 (the classification theorem). *Let be a finite splitting group which is not -group. Then is isomorphic to one of the following groups:*(i)* ;*(ii)

*(iii)*

*where**,**is odd;**(iv)*

*where**;**(v)*

*Suzuki simple group,**,**;**(vi)*

*Frobenius group;*

*HT-group (the Hugher and Thompson group).*Proposition 4. *Consider the HT-group, and keep the notation in its definition. If , then is abelian.*

*Proof. *Right from the definition of HT-group, is generated by all elements of order greater than . Let , so . For an arbitrary element , one has , since, otherwise, would be in , and this contradicts the assumption. Thus is of order or . Now, for , we get . Thus is abelian.

Proposition 5. *Let be a nonelementary -group. Then is splitting if and only if where is abelian -group, containing an element of order , and , .*

*Proof. *Assume that ; then is a Frobenius group with abelian kernel, so by Theorem 3, is splitting.

Conversely, assume that is splitting; then byKontorovich [7], all elements of whose order is greater than are contained in one component of the splitting of . If an element , then for an arbitrary element , the element . Hence . Since , then . Thus is abelian as we have seen in Proposition 4. Since is nonelementary -group, must have an element of order , . Thus the element is of order 4.

Finally, we will prove that . Suppose that . Then, as mentioned before, , . Hence, , so, commutes with every element in . Since contains elements with fourth order, then by Kontorovich [14], . Hence, and .

A splitting is called abelian if it consists of two components, one of them ( say) is abelian, containing all elements of nonprime order, and can be written as , where and for , .

Proposition 6. *Let be a -group with minimal abelian splitting which contains elements of order . Let be an elementary abelian group such that . Then, is a direct product of dihedral group of order 8 and elementary 2-group. In particular .*

*Proof. *Assume that is not 2-group. Since has abelian splitting and it contains a normal subgroup of index , then where is an abelian component containing all elements of composite order. And all cyclic subgroups that are not contained in form abelian splitting, see [15]. Hence, by [7], the center of is contained in the abelian components of which is generated by all elements of composite order. If is of order , then we prove that . If , then is -group with abelian splitting containing element of order and . This leads to a contradiction. So, . On the other hand, because, otherwise, would be an abelian splitting of . Thus . In this case, by [15], corresponds to the Wreath product of two cyclic subgroups of with prime orders. Hence, is not splitting if . But this forces to be equal to . This means that where and , , and is abelian 2-group, containing an element of order . So, by Proposition 5, is 2-group and where is dihedral group of order 8 and is an elementary 2-group.

#### 3. Normal Splitting

Let be a splitting of and another splitting of ; then the collection of intersections is a splitting of . A splitting is said to be a refinement of another splitting if every component is contained in a component of . Thus, a splitting is a trivial refinement of , if every of the splitting is either trivial group; that is, is the identity or for some . Hence if the splitting admits no nontrivial refinement, then it is called a minimal splitting.

*Definition 7. *A normal splitting is a splitting in which every conjugate subgroup of a component is again a component of .

*Remark 8. *The sum of subgroups, which are conjugate of , forms the following normal subgroup:
Frobenius theorem gives the sufficient conditions to decompose Frobenius group (Frobenius group is a group which has a subgroup , such that and for every ) into a direct sum. Then by Frobenius theorem, is decomposable into a direct sum of subgroups; that is, , such that is normal subgroup of and

*Example 9. *The symmetric group is normal splitting group, since it is equal to and . Take , and verify that for all .

Any splitting can be refined. A splitting is called minimal if it cannot be refined anymore. This has been shown in the following.

Lemma 10 (see [10]). *Every splitting has a refinement which is normal.*

Proposition 11. *Any minimal splitting is normal.*

*Proof. *It is straightforward that if is splitting, then is again splitting for any . Now consider the refinement splitting and notice that is minimal. Then, is either 1 or . In both cases . Similarly one has , so and thus . This shows that is normal.

*Definition 12. *A finite group is said to have the basis property if any subgroup has a base (minimal generating set) and any two bases of have the same cardinality.

*Definition 13. *A subgroup of is said to be isolated if for every cyclic subgroup one has implies that .

In [12, 13], the sufficient and necessary conditions for a finite group to have the basis property were stated and proved. A special attention was given to groups that have abelian fitting subgroups. The following two theorems summarize such results.

Theorem 14. *If a finite group possesses the basis property, then it is a -group or where is a -group and is a cyclic group of order , are primes and .*

I remind the reader with some basic definitions. The isotopic representation is the representation which can be written as a direct sum of equivalent irreducible representation.

The Frattini subgroup of a group is the intersection of all maximal proper subgroups of . Frattini of higher order is defined inductively as . The fitting subgroup of a finite group is the unique maximal nilpotent normal subgroup. So we have the following theorem.

Theorem 15. *Let a finite group be a semidirect product of -group by a cyclic -group of order , are primes and . Then, the group has the basis property if and only if for any element and for any normal subgroup , the induced automorphism must define an isotopic representation on every quotient Frattini subgroup.*

As in [13], finite groups with basis property are either -groups which are elementary, that is, the order of any element is less than or equal to , or -groups which are not elementary, or of mixed order where are distinct primes and .

Theorem 16. *Let be a finite non--group with basis property such that , where is the maximal nilpotent normal subgroup of of order , and is a cyclic -group of order where , are primes, , and . Then has a normal splitting.*

*Proof. *Since , then is a -Sylow subgroup of , which contains all -elements of . Therefore, the order of every element in is either power of or . So, we know that by [13], -elements do not commute with -elements. Thus for every element , , we have . is the centralizer of in . Hence, is isolated subgroup in , so by [14], is splitting group. That is, such that where .

Now, we will show that for any . If for some and , then . So, we may assume that . If where , one can find , such that or

Since , there is a number with . If we apply the homomorphism on (4), we get . Thus, or , which leads to . This later result, given that possesses the basis property, contradicts, according to [13], the fact that -elements do not commute with -elements. So, must be the identity. This means that has a normal splitting.

For the sake of the converse of Theorem 16, one notices that in Theorem 3, does not have the basis property and cases (ii), (iii), and (iv) are simple, so they do not have basis property. The class of HT-group is nilpotent but the only nilpotent groups with the basis property are -groups. The only possible true converse is Frobenius groups. In fact, this is not true in general. Here is a counterexample.

*Example 17. *The group , where is the quaternion with 8 elements, is normal splitting group that does not have the basis property. We demonstrate this situation.

The quaternion , and the semidirect product is defined by an automorphism , which gives rise to a corresponding linear matrix representation for some base on the field . Such a representation can be given in terms of
by considering as a vector space on . If are operators represented by , for some base, respectively, then we get a linear representation for whose kernel is 1. And we get that
So, we get all, linear representations for the elements of which induce nonidentical minimal polynomials. So, we get a nonequivalent irreducible representation, and thus we do not have isotopic representation, but this violates one of the necessary conditions for finite groups with basis property as in the proof of Theorem 15 (see [13]). Thus does not have the basis property. From the other side, we show that this is a Frobenius group and so, it is a normal splitting group. Consider the subgroup , and let . If , , then we have
where and . So, , . That is to say
Applying the homomorphism on (7) and (8) leads to for all , , and hence . This implies that or commutes with . That is or which contradicts the assumption. Henceforth, and is a Frobenius group with abelian kernel.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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#### Copyright

Copyright © 2014 Abdullah Aljouiee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.