Re: FLTMA: A little group theory


Chip Eastham wrote:
Chip Eastham wrote:
The Dougster wrote:
Hello. I have completed the first of three tests in the abstract
algebra course at the CoCo. We are up to cyclic subgroups in Fraliegh's
seventh edition of A First Course in Abstract Algebra.

From FLT we have, by way of contradiction (I think that is how you say
it)

a^n + b^n = c^n

and so

x^p + y^p = z^p where gcd(x,y,z)=1, one of {x,y,z} is even, p is prime,
and x<y<z<(x+y).

I'd like to look at (x^p + y^p) (mod z). This implies x^p == -y^p (mod
z). x^p is the additive inverse of y^p, in addition modulo z.


Hi, Doug:

I don't see what Fermat's Last Theorem has to do with
any of the rest of your post.

In cyclic group notation in our text we write <x> for all the powers of
x modulo some implied z. <x> is the cyclic subgroup generated by x.

The phi function is written phi(n) = | { x | gcd(x,n)=1, 0<x<n } |.
That is, the measure of the set of numbers coprime to n. We have a
theorem in our text that if element a generates G, that is, <a> = G,
then | <a^s> | = n / (gcd(n,s)). Each of <a^s> is a subgroup of G, and
so contains the inverse of element a^s under the group operation, which
we call *. There are phi(n) generators of an arbitrary group isomorphic
to Zn.

For every z, is there an Abelian group, Zz - 0, of numbers from 1 to
z-1 under multiplication modulo z, with associativity, an identity, 1,
in the group, and a muliplicative inverse for every element in the
group?

So, for arbitrary n, there is a multiplicative group Z/nZ* consisting
of the phi(n) residues coprime to n. This group is cyclic only in
some special cases. Consider for example the multiplicative
group Z/15Z* which has phi(15) = phi(3)*phi(5) = 2*4 = 8 elements.
This is not cyclic, being in fact Z/2Z X Z/4Z.

The cases where the multiplicative group Z/nZ* is cyclic are:

n = 2, n = 4, n = p^m, n = 2*p^m

for odd prime p and integer m > 0.

Note that as an additive (abelian) group Z/nZ does have phi(n)
generators, namely any residue x mod n which is coprime to
n will have order n as an element of Z/nZ.

This should not be confused with the abelian group Z/nZ*,
which is a group of order phi(n) < n. It has no generator
unless it is one of the cyclic cases identified above.

regards, chip

I understand that. Perhaps you will explain the notation used in your
earlier post in a reply to my reply to that post.

Doug

.

Relevant Pages

  • Re: FLTMA: A little group theory
    ... algebra course at the CoCo. ... We are up to cyclic subgroups in Fraliegh's ... So, for arbitrary n, there is a multiplicative group Z/nZ* consisting ...
    (sci.math)
  • Re: FLTMA: A little group theory
    ... algebra course at the CoCo. ... We are up to cyclic subgroups in Fraliegh's ... x modulo some implied z. ... So, for arbitrary n, there is a multiplicative group Z/nZ* consisting ...
    (sci.math)
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