--

(Department of Mathematics, Henan University of Technology, Zhengzhou 450001, China )

Abstract: Let G be a group and A and B be subgroups of G. If G=AB, then G is said to be factorized by A and B. Let p be a prime number. The factorization numbers of a 2-generators abelian p-group and a modular p-group have been determined. Further,suppose that G is a finite p-group as follows G=〈a,b|apn=bpm=1, ab=apn-1+1〉, where n≥2, m≥1. In this paper, the factorization number of G is computed completely, which is a generalization of the result of Saeedi and Farrokhi.

Keywords: Finite p-group; Factorization number; Subgroup commutativity degree

§1. Introduction

In this paper,palways is a prime number and the terminologies and notations used are standard, see [1].

LetGbe a group andAandBbe subgroups ofG. IfG=AB,thenGis said to be factorized byAandB. Letf2(G) be the number of factorizations ofG. In [3], T?arn?auceanu defined the subgroup commutativity degreescd(G) ofG. The relation of the factorization number and subgroup commutativity degree is given as follows

whereL(G) is the lattice of all subgroups ofG. Hence to compute the subgroup commutativity degree of a finite group it is enough to know the factorization number of its subgroups.

§2. The proof of Theorem 1.1

Assume thatpcan divides, and (i,p)=(j,p)=(r,p)=1. Similarly, we can obtain the same results.

which contradicts〈a〉∩B/=1.

For convenience, by Lemma 2.1, we consider further the structure ofB.Lemma 2.2.(i) If x1=n, B is a subgroup of 〈b〉.

(ii) If x1/=n and0≤y1≤m-n and y1≤x1, then there exists an integer x(1≤x＜pn)such that B is a subgroup of 〈axb〉 or 〈a2b2〉.

(iii) If x1/=n and0≤x1＜y1≤m-n, then there exist integers x (where x and p are coprime) and y(1≤y≤m-n)such that B is a subgroup of 〈axbpy〉.

(iv) If x1/=n and y1-x1≤m-n＜y1, then there exists an integer x(1≤x＜pn-1)such that B is a subgroup of 〈apxbpm-n+1〉.

Proof. Since (y2,p)=1, there exist the integersy3andy4such thaty2y3+pmy4=1.

(i) The result is clear.

(ii) Ify1=0, then there exits an integerx(1≤x＜pn) such thatBy3=〈(ax2by2)y3〉=〈axb〉.HenceB=〈(axb)y2〉. From now on, lety1/=0.

If|G|=2m+2, that is,p=2=n, then we havex1=1=y1and

This lemma is proved.

By Lemma 2.2, we consider these five types of groups:

(1)〈b〉; (2)〈a2b2〉; (3)〈axb〉, where 1≤x＜pn; (4)〈axbpy〉, where (x,p)=1 and 1≤y≤m-n;(5)〈apxbpm-n+1〉, where 1≤x＜pn-1.

Note that the intersections of the above five types of groups with〈a〉are 1. Furthermore,the fifth type may be incorporated into the others. To observe this, we suppose thatx=pk1k2in (5), where 0≤k1＜n-2 and (k2,p)=1. Ifm-n≤k1, then