Re: FLTMA: A little group theory


Chip Eastham wrote:
The Dougster wrote:
Does anyone have a comment on the lemma?

Doug

I'm not sure I'd call it a "lemma", but it relates, at first
glance to a generalization of FLT commonly known as
the Beal conjecture, that:

x^a + y^b = z^c

has no solution where xyz > 0 and are coprime unless
at least one of a,b,c is equal to 2. More background
here:

http://www.ams.org/notices/199711/beal.pdf

Thanks! Gotta love those cash-prize math problems. We hardly ever get
one in abstract algebra, or even any other class at the CoCo.

We studied finitely generated abelian groups, direct products of
groups, and orders of elements of such groups today and I wrote in my
notes:

(x,y,z) = 1 ==> Z/xZ* x Z/yZ* x Z/zZ* is cyclic.

Now to find | < (x,y,z) > | in that group. Hm. It's got to be LCM (
|<x>|, |<y>|, |<z>| ), right? No, not relevant...

The triple congruence becomes
( z/y, z/x, x/y ) ^ p = ( 1, 1, -1).
Would that imply
( 1 , 1, -1 ) ^ p = ( z/y, z/x, x/y ) ? I think it does because this is
a cyclic group.

No, that's too easy. ( x, y, z ) = 1 doesn't mean (phi(x), phi(y),
phi(z)) = 1. So it's not necesasrily cyclic, and we have only

( z/y, z/x, x/y ) ^ p = ( 1, 1, -1 ), and not the, uh, reverse.

| < ( 1, 1, -1 ) > | = 2. Hmmmmm.....

If it were cyclic we'd have a contradicition on the reverse. Er, I
think we would. Too bad.

We do have
x^p + y^p = z^p & gcd(x,y,z) = 1 ==> p divides gcd(phi(x), phi(y),
phi(z)), or something similar, I think.

Oh, wait.

< ( 1, 1, -1) > <= < ( z/y, z/x, x/y) > so 2 | p. A contradiction
unless p = 2. I think that might be useful. Hell, at this hour of
night, I think that might be *it*!

Whew. With the tobacco cesstion and the frantic schedule, I wonder if I
will ever learn abstract algebra and be able to look into this problem
sometimes.

Doug

.