Re: FLTAA: Field structure p = 2, p>2


Chip Eastham wrote:
The Dougster wrote:
From
a^n + b^n = c^n in Z we have, BWOC,
x^p + y^p = z^p in Z+ and
(z/y, z/x, x/y)^p = ( 1, 1, -1) in the multiplicative group Mx X My X
Mz.

(z/y, z/x, x/y) ^ 2p = unity in this group.

I recall a theorem from abstract algebra proposing that
Zmn is cyclic and isomorphic to
Zm X Zn if and only if gcd(m.n) = 1.

OK. Zmn is always cyclic. It is ~= Zm x Zn iff (m,n)=1.


<(z/y, z/x, x/y)> are cyclic and isomorphic to Z2 X Zp if p>2.

If p = 2, then are these powers cyclic? Surely for x = 3, y= 4, z = 5,
p = 2 these powers are cyclic. What gives? That is, is there a
contradiction?

I think you are confounding, as before, the roles of the
multiplicative groups, eg. Mx in your notation or in mine
Z/xZ*, and the additive groups, eg. Zx or more verbosely
Z/xZ.

In fact the multiplicative group "Mx" is an abelian group
of order phi(x) and we previously showed how to express
it in accordance with FT of Finite Abelian Groups as a
direct sum of additive groups. We also characterized
the special cases in which Mx = Z/xZ* is itself cyclic,
which is necessary for any direct sum involving Mx to
be cyclic.

But if x,y,z > 1 are coprime, it is impossible that phi(x),
phi(y), phi(z) should all be coprime. In fact any odd
prime factor of x induces phi(x) even.

So even if all three factors Mx,My,Mz were cyclic,
their product is cyclic only in trivial cases.

Yes. And hey, thanks for dropping by on this lonely, lost little
thread, Chip. I appreciate your patience above all else.

I do not assert here, although I may have earlier, that
Mx x My x Mz or Mx x Mx x My x My x Mz x Mz is cyclic. I do understand
that such direct products are of order phi(xyz) and phi( (xyz)^2 )
respectively, and that excepting trivially small x, phi(x) is even. I
get that, or, as one mathematician says it, I accept it without
actually understanding it.

What I am trying to get folks here to focus on is that

in the multiplicative group
Mx x My x Mz = Mxyz. the powers of
(z/y, z/x, x/y) are a cyclic subgroup of Mxyz iff
(z/y, z/x, x/y) ^ p = ( 1, 1, -1 ).

I'd love to read here in this group that there is some reason *why*

(z/y, z/x, x/y) ^ p =/= (1, 1, -1).

although I have failed to write an FLTMA Explorer showing this is not
true in all cases with x,y,z < 65,536 since it takes so long to run. I
don't code all that well. I have made some progress, though, and
through this new year I will continue to try.


I'd love to read here in this group an argument of this form:

========== FLT ==========

If, by way of contradiction,
1) a^n + b^n = c^n in Z, then
2) x^p + y^p = z^p in Z+ with (x,y,z) = 1 and p prime, and
3) (z/y, z/x, x/y) ^ p = (1,1,-1) in Mx x My x Mz = Mxyz

but if 3) is true, then

4) { < (z/y, z/x, x/y) > } is a cyclic subgroup M2p <= Mxyz,

but that group has no such cyclic subgroup, making 3), 2), and 1)
false, and FLT true.

========== QED ==========

I would love to read that, or something like it, even if I was not the
one who wrote it! Then I could sleep through a night for the first time
in over three years, maybe. Or maybe I just have insomnia and it isn't
this problem that keeps me awake. :)


Note that if x and y are coprime, then phi(x) * phi(y) = phi(xy). I am
not as sure that (x,y,z) = 1 implies Mx x My x Mz ~= Mxyz, but I do
think that is true. I am not saying Mxyz is cyclic. Rather I am saying
that M2p is not cyclic, although we'd think it would be. To say that
(x,y) = 1 ==> phi(x) * phi(y) = phi(xy) is not to say ( phi(x), phi(y)
) = 1. Not at all. Is that clear?

How can I show M2p </= Mxyz ?

(how to show that the set of such powers of the given element
is *not* a cyclic subgroup of Mxyz?)

I could try to show it doesn't contain unity, but it does. I could try
to show it isn't associative. I could try to show that there exists
some element within the subgroup such that the element has no inverse.
Are there any other possiblities for proof that I have missed?

I mean, can you imaging replacing the 200 page proof of Wiles et al
with:

M2p </= Mxyz, ergo FLT

or something similar. Wow. That's like 20 characters, not 200 pages.

What fun that would be!

I guess we have to show

exists (z/y, z/x, x/y) ^ m such that ... at this point I grow very
tired and need to return to bed. Heck, I am so sleeeeepy that what I
have written here may not even make any sense.

Doug

.