Answered: Let G be a (not necessarily finite)…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Let G be a (not necessarily finite) group with two subgroups H and K such that K ! H !
G. The purpose of this question is to establish the index formula
|G : K| = |G : H|·|H : K|.
Let T be a transversal to K in H and U be a transversal to H in G.
(a) By considering the coset Hg or otherwise, show that if g is an element of G, then
Kg = Ktu for some t ∈ T and some u ∈ U.
(b) If t, t! ∈ T and u, u! ∈ U with Ktu = Kt!u!, first show that Hu = Hu! and deduce
u = u!, and then show that t = t!.
(c) Deduce that TU = {tu | t ∈ T, u ∈ U } is a transversal to K in G and that
|G : K| = |G : H|·|H : K|.
(d) Show that this formula follows immediately from Lagrange’s Theorem if G is a finite group.

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