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Subgroups of the Rational Numbers Under Addition

Date: 02/01/2003 at 23:42:09
From: Cathie
Subject: Subgroups of the rational numbers

I need to describe all the subgroups of the rational numbers under 
addition.

I know the definition of a subgroup:

  1) It has to be closed under the operation (in this case, addition)
  2) It has to contain the identity of the group
  3) Every element has to have an inverse

I've tried obvious choices like the positive or negative rationals 
plus zero, but there was no additive inverse.

I also tried {x, -x : x is in Q} plus zero, but that wasn't 
necessarily closed under addition.

I'm pretty stuck, but I'm going to keep at it. I'd love some input.


Date: 02/02/2003 at 10:02:15
From: Doctor Tom
Subject: Re: Subgroups of the rational numbers

A couple of examples occurred to me right away:

  1)  { ..., -4/3, -3/3, -2/3, -1/3, 0, 1/3, 2/3, ...}
  
  2)  {x : x = k/2^n for all integers k and positive integers n}

Can we generalize these?  For example, consider the set of all numbers 
of the form:

  k/(2^m3^n)

where k is a positive or negative integer, and m and n are 
non-negative integers. This is clearly an additive subgroup of the 
rationals.

In fact, you can pick any subset of the prime numbers to use as 
denominators, for example, all the numbers of the form:

  k/(2^m 5^n 17^p)

And, in fact, not just finite subsets - you can choose ANY subset of 
the prime numbers, finite or infinite, and the set of all numbers that 
look like:

  k/(p1^n1 p2^n2 p3^n3 .... (forever))

such that p1, p2, ... are distinct prime numbers and n1, n2, ... are 
non-negative integers such that only a finite number of them are 
non-zero.

Clearly we can form a subgroup from all the integer multiples of any 
fixed rational number.

In addition, when you look at the various subgroups I mentioned above, 
the numerators could also be any multiple of a number relatively prime 
to all the primes in the denominator.

Also, the trivial subgroup {0} is closed under addition. 

It also seems that you can restrict the exponents on the primes in the 
denominators to be no greater than some fixed number if you want.

In other words,  we could allow 3 and 3^2 down there together with any 
power of 2, so the general form would be:

  k/(2^n3^m), 
  
where k is a positive or negative integer, m is zero, 1, or 2, and n 
is any non-negative integer.

Anyway, after talking this over with a couple of friends, we all agree 
that the following is a complete answer to your question: 

The only finite subgroup is {0}.

All the infinite groups can be characterized as follows:

Let p0, p1, p2, ... be all the prime numbers, so p0 = 2, p1 = 3,
p2 = 5, p3 = 7, and so on.

Let k0, k1, k2, ... be either non-negative integers or a special 
number that I will call "infinity."  In addition, let n be any number 
that is relatively prime to pi^ki for all i. Each different list of 
values of the ki together with the value of n generates a unique 
subgroup of the rationals whose members are defined as follows:

Let m0 <= k0, m1 <= k1, m2 <= k2, ... mi <= ki, where at most a finite 
number of the mi are non-zero. Let q be an integer, positive, negative 
or zero. If the ki is the special number "infinity," then the mi can 
be any non-negative finite integer. Then the number:

  qn/((p0^m0)(p1^m1)(p2^m2)...) 

represents all the elements in that subgroup. (Notice that although 
the thing in the denominator appears to be an infinite product, it is 
not, since at most a finite number of the mi are non-zero.)

That's it!  Much more complicated and interesting than I thought at 
first.

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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