Re: FLTMA: A little group theory


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?

For example, from 3^2 + 4^2 = 5^2, with * multiplication modulo 5:

* 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1

Isn't that a 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.

I don't follow this. What is the notation Z/nZ*? The integers divided
by the nonzero multiples of n? I can understand that phi(5) is 4, and
that the table above has 4 elements.

What is the X? In our text, x is the Cartesian product...

Yes, there is some (simple?) notation I am missing to follow this.

I guess I should work out:

* is multiplication modulo 6:

* 1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4

That's not even a group, much less cyclic. Hm. I have answered my own
question. No, there is not such a group for *every* n.

Now, you write there is a group of the phi(n) residues coprime to n. 2
is not coprime to 6. Neither are 3 or 4. 5 and 1 are coprime to 6.

* = * mod 6
* 1 5
1 1 5
5 5 1

Hm. In FLT, for every possible x^n + y^n = z^n there *is* a group G
under multiplication modulo z with elements the phi(z) residues of z
coprime to z, containing all possible values of x and y as elements.
This group is not always cyclic, but is always Abelian. Right?

I think I am getting the content but the notation is new to me.

This group is the powers of x and y mod z only if it's cyclic. Right?

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.

regards, chip

If z is 2, 4, p^m, or 2*p^m, then there is a cyclic group I can call G
containing all possible values of x and y for x^p + y^p = z^p and all
powers of x and y mod z.

Now there are also z^p - x^p = y^p and z^p - y^p = x^p to consider.

Is it true that for a counterexample to FLT, *at least one* of {x,y,z}
must be 2, 4, p^m, or 2*p^m?

I do notice that for all x^2 + y^2 = z^2, at least one of {x,y,z} seems
to always be p^1, but I don't see that that has to be true for the
equation to power p.

This was not the week to try to quit smoking tobacco! Grrrr....

Thanks. Doug

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