Re: FLTMA: A little group theory

Gottfried,

Below you mention a cyclic subgroup we can call cyclic subgroup H of
group G.

What is the set and binary operation that make up group G?

What is the order of H and of G?

What is (are) the generator(s) of H?

Doug

Gottfried Helms wrote:
Am 02.10.2006 11:49 schrieb The Dougster:
Hello. I have completed the first of three tests in the abstract

If so, then x and y, since they are less than z and nonzero, are
elements of this group, and there are some things that can be said
about them and their powers.

I like to write things in the x^n-y^n = z^n way, since the
following is simpler then.
Since f(n) = x^n - y^n (mod p) defines a cyclic subgroup, when
n is varying, this has consequences of the moduli of
*each* of its primefactors.

Right there.


Each prime q (coprime to both x and y) can be one
of its primefactors and f(n) can be examined (mod q).
For each prime q there is cyclic occurence in f(n)
with cycle-length of m= q-1 or m= a divisior of q-1.

Let m be the divisor of q-1, at which q occurs first
time in f(n) (==> with minimal n) and a be that power,
to which it occurs (mostly it is that a=1; if a>1 this
is a rare special case, considering all p<f(n))
Let this fact be denoted as

{f(m),q} = a <==> f(m) = q^a * x where gcd(p,x)=1

For the pair (x,y) = (2,1) primes q, for which a>1 are
called Wieferich primes, and are extremely rare, upt to
something like n=10^15 there are only two such primes
known, q=1093 and q=3511 (?) , and they both have a=2.

Since the occurence of a prime q is cyclic with increasing n
with a cycle length m and a first occurence of q^a, we find
using Eulers Phi-function, that

{f(m*q^k),q} * a+k


Now if n=p is an odd prime, then all that primefactors qk occur, which
either occur already in f(1), or for which f(n) (mod qk) has order qk-1,
means, p is a divisor of q-1.
Since n=p prime, for this primefactor q also n=p=m , and
its power is a.
Also for being primefactor in z^p the "a" for the prime q
must equal n=p =a - so we have, that f(n)'s order (mod q)
must equal p or 1 *and* q's a-value must also equal p.

With (x,y)=/=(2,1) there are more such "wieferich-" like
primes q, with even higher degree of a>2, and if z is
not just q^p but is assumed to consist of more primefactors,
then we need this condition on all that primefactors
simultaneously. (One can find modular conditions on x and
y, which allow a certain prime q to have the "wieferich" effect,
but I didn't complete my search for such conditions to generate
a "wieferich"-case with a given a) One example is
f(n)= 3^n - 1^n , then q=11 is "wieferich" with m=5 and a=2

From Wiles follows, that m=p and a=p for all primefactors of
z^n simultanously actually does not happen,
but this "wieferich"-effect and its occurence is not yet
fully understood, as I recall the relevant entries in
the math-resources, which are widely available.

So a core-key may be to understand about this "wieferich"
analog effect (may be it is simply a translation of the
fermat problem...)

Gottfried Helms

.

Quantcast