What is a Torsion Subgroup?

Date: 03/05/2003 at 12:42:00
From: Kris
Subject: Torsion subgroup

Let G be an abelian group. Show that the elements of finite order in G 
form a subgroup. This subgroup is called the torsion subgroup of G. 
Now find the torsion subgroup of the multiplicative group R* of 
nonzero real numbers.

Date: 04/04/2003 at 19:30:42
From: Doctor Nitrogen
Subject: Re: Torsion subgroup

Hi, Kris:

A "torsion subgroup" of an Abelian group G is a subgroup of elements 
from G all having finite order for some positive integer m. It is 
known that if G is Abelian and T is a torsion subgroup of G, then the 
factor group

   G/T

is torsion free.

Torsion subgroups are studied somewhat in R-module and L-module 
theory, analogues of a kind of a vector space over a field, where 
instead of having a vector space over a field, you have "a group over 
a Ring." They also have some connection to elliptic curves in number 
theory.

Concerning your question, I'll do part of it for you and you can 
reason out the rest. 

Let T be a subset of elements of finite order m in G. Let "e" denote 
the identity. Then

(1) e is an element of T, so T has an identity.

(2) Since G is an Abelian group which is associative, T is also 
associative (and Abelian) since it is a subset of G.

(3)For "a" any element of T,
                        
             a^m 

             = a(a^m-1)

             = (a^m-1)(a)

             = (a^-1)(a)

             = e

Hence every element in T has an inverse.

(4)For a an element of T, 

             (ae)

             = (ea)

             = a

Hence T satisfies all the requirements for it to be a subgroup of G.

I will leave the rest of the assignment for you to ponder.

You can find more information on torsion subgroups at this sci.math 
posting:

   Re: structure of arbitrary Abelian groups - Dave Rusin
   http://www.math.niu.edu/~rusin/known-math/99/ab_gps 

I hope this helped answer the questions you had concerning your 
mathematics problem.

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/