Re: FLTMA: A little group theory

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.

I am interested in hearing, at this point, whether group theory can say
anything about this triple coincidence that PRNG theory cannot.

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 ...
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  • Re: FLTMA: A little group theory
    ... Gottfried Helms wrote: ... 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 ...
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