Re: FLTMA: A little group theory


Chip Eastham wrote:
The Dougster wrote:

Chip Eastham wrote:
The converse is not true, though. Consider the residue
in Z/15Z. The order of residue 4 is 2, but clearly half that
order does not give a power of 4 equal to -1 in Z/15Z.
In fact you can run through all possible powers of 4 in
Z/15Z without getting to -1.

That follows from 2 prime, I think, if the above was valid.

More directly, if the order of an element w mod z is 2,
then the only possible solution is the trivial one:

w^1 = w = -1 mod z.

since the sequence of positive powers of w repeats:

w, 1, w, 1,...

Other prime orders are oddly enough, odd, and we
know if some power of w is -1, then the order of w
must be divisible by the order of -1, which is 2 (in
that z > 2).

However w may fail to have a power equal to -1
in spite of the order of w mod z being even and
composite. For example in Z/15Z*, the powers
of 2 are:

2, 4, 8, 1, (repeat)

so the order of 2 is 4, but no power of 2 is -1.

regards, chip

Ah. The order of -1 is 2. | < -1 > | = 2. < -1 > = { -1, 1 }.

How do we know that if w^n == -1 mod z that
| < -1 > | divides | < w > | ?

I am fairly sure <x/y> and <y/x> are cyclic subgroups of the set of
numbers relatively prime to z, and <z/x> of the set ... y, and <z/y> of
the set... x.

Groups of prime order are always cyclic. Are these cyclic subgroups of
prime order? It can go either way. There are, of course groups of
composite order that are cyclic.

From

x^n + y^n == 0 mod z

we have both
(x/y)^n == -1 mod z and
(y/x)^n == -1 mod z.

The n in both expressions is the same, but we do not know x/y = y/x. We
do know |<x/y>| = |<y/x>|, though.

To prove that

(1) gcd(x,y,z) = 1, (2) x < y < z < x+y, (3) exactly one of {x, y, z}
is even,

(4) (x/y) ^ n = -1 mod z,
(5) (z/x) ^ n == 1 mod y, and
(6) (z/y) ^ n == 1 mod x

implies
(7) n is 2 or composite

might be done 4 ways, right?

Directly: (1) & (2) & (3) & (4) & (5) & (6) ==> (7)
Indirectly: ~(7) ==> !(1) \/ !(2) ....
By contradiction: ?
By induction: Er, well, not by induction, I certainly don't think so.

Doug

.

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