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### Groups, Subgroups, Cosets, and Order

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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?

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.

Thanks.
Lara
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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,
then

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
me.

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
http://mathforum.org/dr.math/
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Associated Topics:
College Modern Algebra

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