Re: Self Study problem help - Group theory

In article <1122339624.677755.300860@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
abe.buckingham@xxxxxxxxx wrote:

You've pretty nearly got it right. See below for a comment or two.

> I am studying the third edition of 'Abstract Algebra' by Dummit and
> Foote on my own and have come across a question which I not certain I
> have solved correctly, or rigorously. As always even though this is
> self study I would ask that you give as little help as possible and
> avoid giving me an answer, only guideance on how to improve my proof,
> or where the reasoning went wrong rather then an explicit solution.
>
> The problem is (paraphased) 'Exhibit a proper subgroup of the rationals
> under addition which is not cyclic'.
>
> I had trouble at first, but I think I have a good answer, I'm just not
> certain how rigorous it is. This text defined a cyclic group to be a
> group that's generated by a single element.
>
> So, consider X which is generated by the infinite collection <1/p_1,
> 1/p_2, ...> where p_n is a prime greater then 2. Now note that adding
> any two two together gives the product (p_i + p_k)/(p_i * p_k). So, all
> the elements generated by this collection will have an integer for a
> numerator and the denominator will decompose into powers of the primes
> p_n listed in the numerators of X.

Where do the *powers* of primes come from? Is 1/9 in X?

> Now, no matter what the numerator
> is, when we reduce the fraction we can only divide out by the primes in
> the bottom, and none of them are 2, so no matter what we can never
> generate 1/2. (I'm not certain if my reasoning is solid here, I keep
> thinking somehow I can create 1/2 by a clever combination of additions
> and subtractions on the rationals I have).

You should be able to prove that if a, b, c, d are integers
with b, d odd then the denominator of (a/b) + (c/d), in lowest
terms, is odd.

> Now this means we have a
> proper subset, since we chose to generate the group we know that it's
> closed.
>
> Now assume that X is cyclic. This would mean some element, well call it
> k = (v / [(p_a)^x*(p_b)^y*...(p_c)^z] would generate X. But, since k^m
> for all integers m would only contain powers of a finite number of
> prime denominators there would exist elements of X that k would not
> construct.

The operation being addition, k^m is not germaine.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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