Products of two one-parameter subgroups

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 ...

One-parameter subgroups of the holomorphic overshear group on C

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...

Stability and vibration analyses of tapered columns resting on one or two-parameter elastic foundations

This paper presents a generalized numerical method to evaluate element stiffness matrices needed for the free vibration and stability analyses of non-prismatic columns resting on one- or two-parameter elastic foundations and subjected to variable axial load. For this purpose, power series approximation is used to solve the fourth–order differential equation of non-prismatic columns with v...

Maximal subgroups of direct products

A group G is simple if and only if the diagonal subgroup of G ×G is a maximal subgroup. This striking property is very easy to prove and raises the question of determining all the maximal subgroups of G , where G denotes the direct product of n copies of G . The first purpose of this note is to answer completely this question. We show in particular that if G is perfect, then any maximal subgrou...