Re: FLTMA: A little group theory

We have gone from a^n + b^n = c^n to x^p + y^p = z^p with x,y,z coprime
and a few other restrictions, then on to statements in modular
arithmetic culminating in

(z/y, z/x, x/y)^p = (1, 1, -1) in Z/xZ* x Z/yZ* x Z/zZ*

and so the period of this group element is 2p, and

< (z/y, z/x, x/y) > is a cyclic subgroup of the given group.

p shares a factor of p with 2p and so | < (1,1,-1) > | = 2 again.

If there is a prime q smaller than p it shares no factor with 2p and so
generates the same cyclic subgroup. If there is no prime q smaller than
p, and p = 2, and so by the lemma we haven't discussed much, but which
seems clear to me, FLT follows.

Now, can there be such a subgroup? I don't know yet.

If (r,s,t) generates H, a subgroup of G, then (r,s,t)^(-1) generates
the same subgroup. We can write ((y/z, x/z, y/x)^(-1))^p = (1,1,-1) by
symmetry, I think, but can equal powers of presumably distinct
(inverse) elements give the same value? I think not, unless every
component value is self-inverse, so then z/y = y/z = zy = yz, etc.,
commuting. I am not sure of this last point, but we seem to be making
some progress in exploring FLT with group theory and modular
arithmetic.

There is nothing here preventing a restart. That is, for any x, y, z,
the direct product of their corresponding multiplicative groups exists,
so that's an avenue for direct proof, rather than proof by
contradiction, if I recall my proof styles correctly.

Doug

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