Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Groups, Subgroups, Cosets, and Order

Date: 04/20/2000 at 18:44:57
From: Lara 
Subject: Groups, Subgroups, Order, and Beyond

Hello Dr. Math:

I have 3 questions for you:

1) Show that every element of the quotient group G = Q/Z has finite 
   order. Does G have finite order?

NOTE: Q means rational numbers and Z is the integers

So do I need to show that Q has finite order AND Z has finite order? 
How can I even assume anything to show this? Are the integers 
represented within the rationals, so that I don't need to consider 
them? If it happens that Q has finite order and Z has finite order, is 
it necessarily true that G has finite order?

I know that order really means the number of elements in that group.

So do I let Q be a group and let Z be a group, and assume that Q has 
order n and Z has order m? I'm thinking that the order of Z is finite 
because eventually you just get multiples of previous integers. Thus, 
you only require so many integers before everything is represented 
(and for some reason, I'm beginning to think I've proven this before, 
but I don't remember how!) I'm guessing that Z is definitely of finite 
order. But what about Q? Can we say Q = a + bi? Now what?

2) Prove that a group of order n has a proper subgroup if and only if 
   n is composite.

Let G be a group. Suppose G has order n, meaning there are n elements 
in G. We need to show that n is not {e} and n is not ALL of G. But why 
is the compositeness of this important? What can I say? What do I do?

3) Suppose H, K contained in  G are subgroups of orders 5 and 8, 
   respectively. Prove that H intersect K = {e}

Since H has order 5, it has 5 elements. However, H can have subgroups 
of 1 element, namely {e}, or 5 elements, namely {e, a, a^2, a^3, a^4}.

Since K has order 8, K can have subgroups of 1 element, 2 elements, 4 
elements, or 8 elements.

We need to show that H intersect K = {e}.

The only subgroups that are common to both H and K is the subgroup of 
1 element, namely {e}. Thus, H intersect K = {e}

Is this correct?

Please help me! I realize that this is difficult material, but my 
background in this material is not very strong and I would like to 
better understand it, so that I can continue to learn and understand 
more difficult mathematics.


Date: 04/20/2000 at 19:29:02
From: Doctor Wilkinson
Subject: Re: Groups, Subgroups, Order, and Beyond

Hi, Lara. This stuff can be pretty bewildering at first.

Let me see if I can help clear up some of the confusion.

Starting with the third question, your proof is pretty much correct. 
The important fact that you should state more explicitly is that the 
order of a subgroup of a finite group is a divisor of the order of the 
group. Since the intersection of H and K is a subgroup of H and a 
subgroup of K, its order must divide both 5 and 8, so it must be 1. 
This is pretty much what you said, but a little clearer.

Now similarly on the second question, if the order is NOT composite, 
then it must be 1 or a prime, and the order of a subgroup must be 1 or 
n, so it's not a PROPER subgroup. That's the easy part. If the order 
is composite, then you need to look at an element a different from 1, 
and look at the subgroup {1, a, a^2, ...}. If this is not the whole 
group, then it's a proper subgroup, so you can assume it's the whole 
group. Now see how you can find a proper subgroup if n is composite. 
(Hint:  if n = rs, look at a^s.)

The first problem is a bit harder, since it involves the concept of 
quotient group, which is a bit hard to grasp at first. Also it is this 
problem that you seem most confused about.

First of all, the group operation here is addition. Q is the set of 
rational numbers, Z the set of integers. Neither one of these sets is 
finite. There are an infinite number of integers and an infinite 
number of rational numbers (these do include the integers).

So it's not going to do any good to suppose for a minute that either 
of these groups has finite order.

Now what is this mysterious quotient group Q/Z? Quotient groups are 
made up of COSETS. The cosets that make up Q/Z have the form Z + q, 
where q belongs to Q. For example, there is a coset Z + 1/2, which is 
the set of all numbers of the form {n + 1/2}, where n is an integer. 
And there is a coset Z + 2/5, which consists of all numbers of the 
form n + 2/5, where n is an integer. The cosets form a group if you 
define the sum of A and B to be the set of all sums of an element in A 
and an element in B, and this group is the quotient group. The 
identity of Q/Z is just Z. 

Now if you take a rational number r/s, where r and s are integers, 

     s (r/s) = r

which is an integer. Now anything in the coset Z + r/s is an integer 
plus r/s, so if you multiply anything in that coset by s, you get an 
integer. So if you multiply the coset by s (i.e. add it to itself s 
times) you get a coset consisting of all integers, but that's just Z 
itself. That is, the coset is of finite order s (or a divisor of s). 
But hey, that's the answer to the first question! Every element of the 
quotient group Q/Z DOES have finite order.

But Q/Z is not of finite order. For Q/Z to be of finite order, there 
would have to be just a finite number of cosets. But that's certainly 
not true, since for example all the cosets Z + 1/n where n = 2, 3, 4, 
5, ... are distinct.

Study this carefully, and if you're still confused, please get back to 

You need to be very careful with the definitions, especially at the 
beginning, when you haven't developed much of a feeling for what's 
going on.

- Doctor Wilkinson, The Math Forum
Associated Topics:
College Modern Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum