Two generator subgroups of free products with commuting subgroups

نویسندگان
چکیده

برای دانلود باید عضویت طلایی داشته باشید

برای دانلود متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

منابع مشابه

on the free profinite products of profinite groups with commuting subgroups

in this paper we introduce the construction of free profinite products of profinite groups withcommuting subgroups. we study a particular case: the proper free profinite products of profinite groups with commuting subgroups. we prove some conditions for a free profinite product of profinite groups with commuting subgroups to be proper. we derive some consequences. we also compute profinite comp...

متن کامل

Fixed Subgroups of Endomorphisms of Free Products

Let G = ∗ i=1 Gi and let φ be a symmetric endomorphism of G. If φ is a monomorphism or if G is a finitely generated residually finite group, then the fixed subgroup Fix(φ) = {g ∈ G : φ(g) = g} of φ has Kurosh rank at most n.

متن کامل

Frobenius Subgroups of Free Profinite Products

We solve an open problem of Herfort and Ribes: Profinite Frobenius groups of certain type do occur as closed subgroups of free profinite products of two profinite groups. This also solves a question of Pop about prosolvable subgroups of free profinite products.

متن کامل

The Two - Generator Subgroups of One - Relator Groups with Torsion

The main aim of this paper is to show that every two-generator subgroup of any one-relator group with torsion is either a free product of cyclic groups or is a one-relator group with torsion. This result is proved by using techniques for reducing pairs of elements in certain HNN groups. These techniques not only apply to one-relator groups with torsion but also to a large number of other groups...

متن کامل

Two subgroups and semi-direct products

Proof. Clearly, the image of F is HK by definition. To see that F is injective, suppose that F (h1, k1) = F (h2, k2). Then by definition h1k1 = h2k2. Thus h −1 2 h1 = k2k −1 1 . Since H is a subgroup, h−1 2 h1 ∈ H, and since K is a subgroup, k2k −1 1 ∈ K. Thus h−1 2 h1 = k2k −1 1 ∈ H ∩ K = {1}, and so h −1 2 h1 = k2k −1 1 = 1. It follows that h−1 2 h1 = 1, so that h1 = h2, and similarly k2k −1 ...

متن کامل

ذخیره در منابع من


  با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید

ژورنال

عنوان ژورنال: Journal of Pure and Applied Algebra

سال: 2002

ISSN: 0022-4049

DOI: 10.1016/s0022-4049(01)00152-9