Re: Self Study problem help - Group theory
- From: quasi <quasi@xxxxxxxx>
- Date: Mon, 25 Jul 2005 22:57:39 -0700
On 25 Jul 2005 19:38:26 -0700, abe.buckingham@xxxxxxxxx wrote:
>I see what you mean about 1/9. Perhaps it would be better to simple
>define X to be all rationals in lowest terms with an odd denominator? I
>suspect that it would still work this way.
Yes, much simpler, a fine improvement.
> I have convinced myself that if a, b, c, and d, integers with b and d
>odd then (a/b) + (c/d) in lowest terms is odd so I am content that I
>had that portion correct.
You meant to say that (a/b)+(c/d) has odd denominator.
>Unfortunately I do not understand what you mean when you say that 'k^m
>is not germaine' and cannot seem to find a definition for germaine.
Germaine is an older form of German, hardly used any more.
(just kidding)
But if k is an element of Q, what is the cyclic group <k> generated by
k. We are not talking about the multiplicative group Q*, so don't be
switching groups in the middle of a problem.
For example, if k=1/3 does <k> even contain k^2? No, right? What are
the elements of <k>?
>With the revision to the definition of X to include all odd
>denominators could I simply take an odd prime p not in the denominator
>of k and say that 1/p is not constructed therefore k cannot generate
>the group, and therefore X is not cyclic?
Yes, fine.
Although, as a variation on the same proof line, try to build an X
where the denominators have even less distinct prime factors. How much
less can you get? (still requiring X to be non-cyclic)
>Thanks again for all the help, I'm glad to see I was on the right track
>but also glad to see that I was correct about being a bit off the mark.
>Sometimes it's hard to tell if I'm completely off my rocker when I'm
>studying alone.
You're doing well.
quasi
.
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