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


The Dougster schrieb:

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

This is valid at most if (x,y)=(x,z)=(y,z)=1, but, ok, in fact that is
the only case you are interested in.

(z/y, z/x, x/y) are a cyclic subgroup of Mxyz iff

The powers of a single element (of a finite group) are always a cyclic
subgroup!

(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

You have (and need) more than (x,y,z)=1, namely (x,y)=(x,z)=(y,z)=1.
Note that (6,10,15)=1.

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,

That "but" is very bold.
Let n=xyz. Then Mn=(Z/nZ)* is a finite group of order phi(n).
If p is a prime then phi(n) is divisible by p as soon as n is divisible
by p^2.
This is because for we have
phi(p^k * <stuff prime to p>) = p^(k-1) * (p-1) * phi(<stuff prime to
p>).
But phi(n) is also divisible by p as soon as n is divisible by a prime
q with the property
that p divides q-1 (seen from phi(q^k * <stuff prime to q>) = q^(k-1) *
(q-1) * phi(<stuff prime to q>) ).
If phi(n) is divisible by p, then Mn has a subgroup isomorphic to Z/pZ.
This is valid also for p=2, i.e. if phi(n) is even ten M has a subgroup
isomorphic to Z/2Z.
Together this implies (if p is an odd prime):
If phi(n) is even and divisible by p, then Mn has a subgroup isomorphic
to Z/2Z, a subgroup isomorphic to Z/pZ and hence (as Mn is abelian) a
subgroup isomorphic to Z/(2p)Z.

As seen above, phi(n) is even as soon as n is divisible by an odd prime
or by 4, i.e. for all n>2.
And we have divisibility by p as soon as either n is divisible by p^2
(i.e. one of x,y,z is divisible by p^2) or n (i.e. one of x,y,z) is
divisible by a prime of the form k*p+1.

Thus even for p=3 you have nothing in your hands for cases where one of
x,y,z is divisible by any of the numbers {7, 9, 13, 19, 31, 37, 43, 61,
67, 73, 79, ...}


and FLT true.

Well that last line can be justified, and I have heard of a wonderful
proof of it.
Alas, this post is too small to have that proof included.

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

First of all, you need (x,y)=(x,z)=(y,z)=1.

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.

But M2p /is/ cyclic:
If p=2, then (Z/4Z) has order 3 and is trivially cyclic.
Thus assume for the rest of the argument that p is odd.
Then (Z/(2p)Z)* is of order phi(2p)=phi(2)*phi(p)=p-1.
(Z/(pZ))* is cyclic because it is a finite subgroup of the
multiplicative group of a field.
Thus, let c be an integer such that (Z/(pZ))* is generated by c+pZ.
We may assume that c is odd (otherwise replace c by c+p).
Just to make sure, it follows that c+2pZ is invertible in Z/(2p)Z
because c*c^(p-2) = c^(p-1) is odd and is 1 modulo p, i.e. it is of the
form k*p+1 and odd, hence k must be even,
k=2m, c^(p-1) = k*p+1 = m*(2p)+1.
Moreover, c,c^2, ..., c^(p-2) are p-1 numbers that are pairwise
incongruent mod p.
Hence they are also incongruent mod 2p and thus cover all p-1 elements
of (Z/(2p)Z)*.

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 ?

Not at all. See above.


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

Not at all. See above.


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?

You should know that one need not check all of the group axioms for
subgroups of finite groups.

Theorem:
Let (G,°) be a finite group and S a subset of G.
Then (S,°) is a subgroup of (G,°) if and only if
S is not empty and S is closed under the group operation.

(In your case: there exists a power of the generating element and the
product of two powers is a power).

It looks like I have to spell that out in detail, thus here is a

Proof:
The "only if" is trivial (note that a subgroup must contain the neutral
element e).
Now for the "if" part:
Assume S is not empty and is closed under °.
Let a,b,c be elements of S.
Then a°b, b°c, (a°b)°c, a°(b°c) are elements of S by assumption
and
(a°b)°c = a°(b°c) holds in S because it holds in G.
Since S is not empty, let a be an element of S.
Then all members of the infinite sequence a, a°a, a°a°a, ... (or
written simpler
a^1, a^2, a^3, ...)
are in S by assumption, but they cannot all be distinct because S (like
G) is finite.
Thus there are n,m>0 such that a^n = a^(n+m).
By associativity, we see that a^n = a^(n+m) = a^n ° a^m.
Since this equality is also valid in G, it follows that a^m=e, hence e
in S
and s°e=e°s=s holds for all s in S because s is also in G.
If a=e, then a is its own inverse.
Otherwise, the m above must be >1, hence we know that b=a^(m-1) is in
S.
But b°a = a^m = e, hence S contains the inverse of a.


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.

Yeah, that would be really amazing, esp. as you would have found the
proof
without ever reading an introductory book on the subject of group
theory.

.

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