Re: FLTMA: A little group theory


Gottfried Helms wrote:
Am 05.10.2006 21:27 schrieb The Dougster:
The Dougster (I) wrote:
Hello, all. Hello, Chip.
We are talking here about all cyclic multiplicative groups of integers
with order phi(n), n in Z+, aren't we?

With x, y, and z pairwise coprime, is
phi(z) / gcd( phi(z), x ) ever equal to
phi(z) / gcd( phi(z), y )? ( I can start some checks )

Hm.

For every triple (x,y,z) with gcd(x,y,z)=1 there are the three groups
of integers excluding 0 with multiplcation modulo x,y, or z. Each of
these groups has phi(m) members, where m is one of {x,y,z}. Each member
is relatively prime to m.

How do we know how many generators each of these groups has? Well, if
there is any generator a in a cyclic group of order m, then we have a
theorem and proof that says that |<a^s>| = m / gcd(m,s).

Knowing what we know can we now say there are phi(phi(m)) numbers
coprime to phi(m) where n is one of {x,y,z}? Well, yes, we can say that
much. These are the "primitive roots" of m, I think, the numbers {a}
which when raised to powers n, produce all possible values of a^n mod
m..

Since x and y are coprime, their GCDs with phi(z) are distinct, and the
quotient of phi(z) and their GCDs are also distinct. So the cyclic
subgroups have different periods.

Did I get it right, that you are saying, with varying n,

length of cyclic group x^n = a (mod fixed p)
length of cyclic group y^n = b (mod same p)
--------------------------------------------------------------
length of cyclic group x^n + y^n = c = lcm(a,b) (mod same p)

where a and b are divisors of phi(?)

?
Gottfried Helms

Yes, something similar to this was established in a post to sci.math. I
need to look it up to see who contributed. Would you like to know?

Not only that but the location of one known zero of (x^n - y^n) mod z,
gcd(x,y,z)=1, is exactly at the lcm of the periods of x^n mod z and y^n
mod z. However, since both these periods divide phi(z) we do not know
they are distinct and so do not know that this zero is at composite n.
With group theory it may be possible to show that gcd(x,y) = 1 ==> n
composite, and thus, FLT. That is the whole point of this thread. Neat,
huh?

But does it work out? I don't know yet.

Doug

.

Relevant Pages

  • Re: FLTMA: A little group theory
    ... We are talking here about all cyclic multiplicative groups of integers ... With x, y, and z pairwise coprime, is ... Since x and y are coprime, their GCDs with phiare distinct, and the ... I am interested in hearing, at this point, whether group theory can say ...
    (sci.math)
  • Re: FLTMA: A little group theory
    ... We are talking here about all cyclic multiplicative groups of integers ... With x, y, and z pairwise coprime, is ... Since x and y are coprime, their GCDs with phiare distinct, and the ... I am interested in hearing, at this point, whether group theory can say ...
    (sci.math)