[QUOTE=blob100;217692]Given three units of order 2, a1, a2, a3, it is [B]sometimes[/B]
true that a1*a2 *a3 = 1 or -1 mod m, but it is not always true.
Given any two such units a1, a2, we can find a unit a3, such
that a1 * a2 * a3 = 1 or -1, but it will not be true for (say)
a1 * a2 * a4 where a4 is a unit of order 2 different from a3.

It is true for [B]some[/B] triples of units, but not for all

OK, This is the same as I said:
"We see that every product of two units ai of order 2 is antoher units of order 2.
The product of these three units (the two and the third we got by the product) is +/-1."[/QUOTE]


No. What you said and what I said are NOT the same thing.

You said that the product of two units of order 2 is another unit of
order 2. This is correct. You then say "The product of these three units...", but you do not SPECIFY the 3rd unit. You do say "the third we
got by the product", but this phrase is meaningless gibberish.

In any event, you can not prove the result by the mathod you are attempting. Consider: What if the total number of units of order 2 is not
divisible by 3? Then one cannot match them up in the way you suggest.

Perhaps this is just a language difficulty. Would someone else
care to help me explain it? Apparently, my message is not getting through.

[quote=R.D. Silverman;217693]

In any event, you can not prove the result by the mathod you are attempting. Consider: What if the total number of units of order 2 is not
divisible by 3? Then one cannot match them up in the way you suggest.

[/quote]

I'm sorry for not expressing myself legally.
I knew there are naturals m with a number of units of order 2 not divisible by 3.
You need to agree there are sometimes "triangles" with a same unit.
For example:
a1a2=a3(mod m)
a5a4=a1(mod m)

Let's say A(a1,a2,a3) and B(a5,a4,a1) which both contain a1.
The point is that every unit of order 2 can be located in a tripe of that kind.
My propisition is that:
For g1=a1a2a3=+/-1(mod m)
The product of every such g we have:
P=(-1)^k(mod m)
Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m).

Let me attempt to "translate". If I fail to restate you argument in a proper manner, then you should attempt to clarify the language.

[QUOTE=blob100;217695]I'm sorry for not expressing myself legally.[/QUOTE]
Rather than "legally", "succinctly" would be a better choice. In mathematics, the only "law" that comes to mind is "The Law of Large Numbers".
[QUOTE]I knew there are naturals m with a number of units of order 2 not divisible by 3.
You need to agree there are sometimes "triangles" with a same unit.[/QUOTE]
A "triangle" is a two-dimensional geometric shape.
I think that you are trying to discuss some kind of 3-tupes.
[QUOTE]
For example:
a1a2=a3(mod m)
a5a4=a1(mod m)

Let's say A(a1,a2,a3) and B(a5,a4,a1) which both contain a1.[/QUOTE]
For an appropriate integer m, within {a}, the set of integers mod m,
consider the set of 3-tuples of the form {a[SUB]i[/SUB], a[SUB]j[/SUB], a[SUB]k[/SUB]} where
a[SUB]i[/SUB] * a[SUB]j[/SUB] = a[SUB]k[/SUB] (mod m).
[QUOTE]The point is that every unit of order 2 can be located in a trip[B]l[/B]e of that kind.[/QUOTE]

Let A = {a1,a2,a3} and B={a5,a4,a1} be such -tuples.

(Here I am "lost" because there is nothing "special" about the triples, as defined. [B]Every[/B] "a" appears in many such triples (as the a[SUB]i[/SUB], a[SUB]j[/SUB], and/or a[SUB]k[/SUB]).

[QUOTE]
My prop[B]o[/B]sition is that:
For g1=a1a2a3=+/-1(mod m)
The product of every such g we have:
P=(-1)^k(mod m)
Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m).[/QUOTE]

[quote=Wacky;217702]Let me attempt to "translate". If I fail to restate you argument in a proper manner, then you should attempt to clarify the language.


Rather than "legally", "succinctly" would be a better choice. In mathematics, the only "law" that comes to mind is "The Law of Large Numbers".

A "triangle" is a two-dimensional geometric shape.
I think that you are trying to discuss some kind of 3-tupes.

For an appropriate integer m, within {a}, the set of integers mod m,
consider the set of 3-tuples of the form {a[sub]i[/sub], a[sub]j[/sub], a[sub]k[/sub]} where
a[sub]i[/sub] * a[sub]j[/sub] = a[sub]k[/sub] (mod m).


Let A = {a1,a2,a3} and B={a5,a4,a1} be such -tuples.

(Here I am "lost" because there is nothing "special" about the triples, as defined. [B]Every[/B] "a" appears in many such triples (as the a[sub]i[/sub], a[sub]j[/sub], and/or a[sub]k[/sub]).[/quote]

Thank you verry much for teaching me how to build a mathematical argument.:bow:

"My propisition is that:
For g1=a1a2a3=+/-1(mod m)
The product of every such g we have:
P=(-1)^k(mod m)
Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m). "

I'll try to write it better:
We will denote the product of such tripe as gi.
The product of the whole products gi of tripes will be denoted as P.
P=g1g2g3...gk where we have k combinations for triples.
P=(-1)^k(mod m), Where we assume k as the number of tripes g such that g=-1(mod m).

We see a connection between P and g(m) is the product of the units of order 2.P is the product of the units of order 2 powers>/=1 and g(m) is the product of the units of order 2.

[QUOTE=blob100;217707]"My propisition is that:
For g1=a1a2a3=+/-1(mod m)
[/QUOTE]
Whoa! Back up. What are you trying to say? How does this relate to the 3-tuples; or the -tuples "A" or "B"? And, just to make it clear, repeat, from the top, all of the clauses that you are using.

I don't see "a4" or "a5". Why did you introduce them if you are not going to use them?

Are you saying:

For some/any/all a[SUB]1[/SUB], a[SUB]2[/SUB],
let
a[SUB]3[/SUB] = a[SUB]1[/SUB] * a[SUB]2[/SUB]
then
(g[SUB]1[/SUB] = a[SUB]1[/SUB] * a[SUB]2[/SUB] * a[SUB]3[/SUB]) == +/-1 (mod m)

??

Perhaps you meant to restrict the "a" in some way? Perhaps to units of order 2?

If the relationship is meant to represent a number of such "g", then rather than using integer subscripts, using letters for the "independent" selections might make your intention more clear-- as in:

For some/any/all a[SUB]i[/SUB], a[SUB]j[/SUB],
let
a[SUB]ij[/SUB] = a[SUB]i[/SUB] * a[SUB]j[/SUB]
then
(g[SUB]ij[/SUB] = a[SUB]i[/SUB] * a[SUB]j[/SUB] * a[SUB]ij[/SUB]) == +/-1 (mod m)


[QUOTE]The product of every such g we have:
P=(-1)^k(mod m)
Where P is the product of the whole gs and k is the number of such gs mod m -1 (-1 as a residue of g mod m). "

I'll try to write it better:
We will denote the product of such tripe as gi.
The product of the whole products gi of tripes will be denoted as P.
P=g1g2g3...gk where we have k combinations for triples.
P=(-1)^k(mod m), Where we assume k as the number of tripes g such that g=-1(mod m).

We see a connection between P and g(m) is the product of the units of order 2.P is the product of the units of order 2 powers>/=1 and g(m) is the product of the units of order 2.[/QUOTE]

I think that I see where you are going. This argument relies on the "definition" of which -tuples you are considering.
Therefore, you need to state, precisely, which g get included in the product P. In particular, how does that selection process relate to the a ?

[quote=Wacky;217711]Whoa! Back up. What are you trying to say? How does this relate to the 3-tuples; or the -tuples "A" or "B"? And, just to make it clear, repeat, from the top, all of the clauses that you are using.

I don't see "a4" or "a5". Why did you introduce them if you are not going to use them?

Are you saying:

For some/any/all a[sub]1[/sub], a[sub]2[/sub],
let
a[sub]3[/sub] = a[sub]1[/sub] * a[sub]2[/sub]
then
(g[sub]1[/sub] = a[sub]1[/sub] * a[sub]2[/sub] * a[sub]3[/sub]) == +/-1 (mod m)

??

Perhaps you meant to restrict the "a" in some way? Perhaps to units of order 2?

If the relationship is meant to represent a number of such "g", then rather than using integer subscripts, using letters for the "independent" selections might make your intention more clear-- as in:

For some/any/all a[sub]i[/sub], a[sub]j[/sub],
let
a[sub]ij[/sub] = a[sub]i[/sub] * a[sub]j[/sub]
then
(g[sub]ij[/sub] = a[sub]i[/sub] * a[sub]j[/sub] * a[sub]ij[/sub]) == +/-1 (mod m)




I think that I see where you are going. This argument relies on the "definition" of which -tuples you are considering.
Therefore, you need to state, precisely, which g get included in the product P. In particular, how does that selection process relate to the a ?[/quote]

OK, I understand everything was deffinietly not understandable.
Again:
We denote g as the product of ai, aj and av.
Where ai*aj=av(mod m) for ai,j,v units of order 2.
This is the specified definition: g=ai*aj*av.

The product P contains (includes) every g as defined before.
And the proposition is:
P=(-1)^k(mod m), where k is the number of numbers g such that
g=-1(mod m).

This proposition is directed to the actual problem becuase of P product of the units of order 2 powers and we are on the actual problem we deal with the product of the units of order 2.

[QUOTE=blob100;217718]OK, I understand everything was deffinietly not understandable.
Again:
We denote g as the product of ai, aj and av.
Where ai*aj=av(mod m) for ai,j,v units of order 2.
This is the specified definition: g=ai*aj*av.

The product P contains (includes) every g as defined before.
[/QUOTE]


No, it does NOT. What if the number of order-2 units is not a multiple of 3???

What if ai*aj*av = 1 (say), but when you multiply ax * ay, you get
ai? You have already used ai in a different triplet. It is quite possible
that aj * av = ax * ay.

You can not prove the result by combining them as 3-tuples. I already told
you this.

[QUOTE=R.D. Silverman;217375]Try proving the following:

(working mod m)
If x1 and x2 are units with order e1, prove that (x1*x2) has order e1.
[/QUOTE]

[QUOTE=R.D. Silverman;217383]Actually, let's skip this last question. It gets us too deep into group
theory and away from number theory.[/QUOTE]

Not to mention that it's false. What did you really mean here?

[QUOTE=jyb;217742]Not to mention that it's false. What did you really mean here?[/QUOTE]

You can prove the result by showing that the product of the units in a
cyclic group is -1, using the cyclic decomposition theorem to decompose
the unit group as a product of cyclic sub-groups, then use the Sylow
theorem to determine whether the number of such sub-groups is even or
odd (the parity determines whether the product is -1 or +1).


Yech.

[QUOTE=R.D. Silverman;217746]You can prove the result by showing that the product of the units in a
cyclic group is -1, using the cyclic decomposition theorem to decompose
the unit group as a product of cyclic sub-groups, then use the Sylow
theorem to determine whether the number of such sub-groups is even or
odd (the parity determines whether the product is -1 or +1).


Yech.[/QUOTE]

Note that the "last question" was not:
If x1 and x2 are units with order e1, prove that (x1*x2) has order e1.

The "last question" was to discover what determines when the product
is +1 and when it is -1. However, it *can* be done without the heavy
group-theory machinery. I am trying to lead Tomer through that now.