how to know if my ideas didnt tought before?

[Death]

I read all this from 1st post.

EPIC!!!!

[blob100]

Here are some theorems I know:
If (a,m)=1, and a is of order e, then if
a^f=1(mod m), we have e/f. And so: e/phi(m)
(becuase a^(phi(m))=1(mod m).
If d/p-1,where p a prime, there are phi(d) residue classes of order d modulo p.
I know the definition of a primitive root.
For every prime p>2, 1 is of order 1 and p-1 is of order 2.
I know some theorems about primitive roots too, is it needed to be shown?

[R.D. Silverman]

Originally Posted by blob100

Here are some theorems I know:
If (a,m)=1, and a is of order e, then if
a^f=1(mod m), we have e/f. And so: e/phi(m)

If a is of order e, then a^e = 1 mod m and e is the smallest
such integer. I don't understand what you mean by "we have e/f".

I suspect you mean that e divides f. Correct?


Do you mean to use "|" rather than "/"?? I suspect so.
x | y means x divides y. x/y means x divided by y. They are not the
same.

Start by proving the following:

Let d = phi(m). If a^e = 1 mod m, and e is the smallest such
integer then e divides d.

Hint: proof by contradiction works easily.

If you know this theorem, then you have everything you need to solve
problem 2.
=====================================================
The integers in [1,.... m-1] that are co-prime to m are the units of
the ring Z/mZ. (i.e. the ring of integers taken mod m). Units are elements
that have a multiplicative inverse. There are clearly phi(m) units. Thus,
the size of the set of units is phi(m). This set is also know as the unit
group mod m, because this set forms a group under multiplication
mod m.
======================================================

If you are up to it, try proving the following: [Lagrange's Theorem]
Let G be a group and let #G be its order. If H is a subgroup of G, then
#H divides #G.

Hint. Let {h1, h2, ..... hk) be the subgroup. Take any element a of G
not in H and consider the set {ah1, ah2, ah3..... ahk). This is
known as the coset of H (by a).

This is a very important fundamental theorem in both group theory and
number theory.

[blob100]

Originally Posted by R.D. Silverman

If a is of order e, then a^e = 1 mod m and e is the smallest
such integer. I don't understand what you mean by "we have e/f".

I suspect you mean that e divides f. Correct?


Do you mean to use "|" rather than "/"?? I suspect so.
x | y means x divides y. x/y means x divided by y. They are not the
same.

Start by proving the following:

Let d = phi(m). If a^e = 1 mod m, and e is the smallest such
integer then e divides d.

Hint: proof by contradiction works easily.

If you know this theorem, then you have everything you need to solve
problem 2.
=====================================================
The integers in [1,.... m-1] that are co-prime to m are the units of
the ring Z/mZ. (i.e. the ring of integers taken mod m). Units are elements
that have a multiplicative inverse. There are clearly phi(m) units. Thus,
the size of the set of units is phi(m). This set is also know as the unit
group mod m, because this set forms a group under multiplication
mod m.
======================================================

If you are up to it, try proving the following: [Lagrange's Theorem]
Let G be a group and let #G be its order. If H is a subgroup of G, then
#H divides #G.

Hint. Let {h1, h2, ..... hk) be the subgroup. Take any element a of G
not in H and consider the set {ah1, ah2, ah3..... ahk). This is
known as the coset of H (by a).

This is a very important fundamental theorem in both group theory and
number theory.

I meant | (tought these are the same).
The theorem I showed proves:
Let d = phi(m). If a^e = 1 mod m, and e is the smallest such
integer then e divides d.
As I said myself.
I can't understand Langrange's theorem.
What is a group? I understood it must assume closure.
By closure for every a and b in G, ab is in G.
Why won't I say ab*a is in G too? so on there are infinity many numbers in G? how can it be a finite group of numbers?

[R.D. Silverman]

Originally Posted by blob100

I meant | (tought these are the same).
The theorem I showed proves:
Let d = phi(m). If a^e = 1 mod m, and e is the smallest such
integer then e divides d.
As I said myself.

This is correct, but it is not what you said.

I can't understand Langrange's theorem.
What is a group? I understood it must assume closure.
By closure for every a and b in G, ab is in G.
Why won't I say ab*a is in G too? so on there are infinity many numbers in G? how can it be a finite group of numbers?

I group is a set of elements {a1, a2, .....} and an operation '*' on that set
such that the following is true.

(1) There exists an element e, such that e*a_i = a_i for all a_i
(an "identity" element)
(2) For each a_i and each a_j, a_i * a_j is also in the group
(3) For each a_i, there exists and a_k such that a_i * a_k = e.
(each element has a multiplicative inverse)
(4) (a_i * a_j) * a_k = a_i * (a_j * a_k) (the operator '*' is associative)

G can be infinite, or G can be finite. An example of an infinite group is
the set of 2 x 2 matrices whose entries are in R and for which their
determinant is non-zero. An example of a finite group is the integers
under multiplication mod 5. Its order is (surprise! phi(5) = 4).
Another example of a finite group is the rotations of a square. Just Label
the corners 1,2,3,4. The elements are the 4 different squares you can get
with 90 degree rotations. Another example is the permutations of the
integers 1....N.

As for your question: "ab*a is in G too? ", the answer is yes. In fact,
aba is known as the conjugation of b by the element a. Conjugation
is a major concept in group theory.

[blob100]

Originally Posted by R.D. Silverman

This is correct, but it is not what you said.




I group is a set of elements {a1, a2, .....} and an operation '*' on that set
such that the following is true.

(1) There exists an element e, such that e*a_i = a_i for all a_i
(an "identity" element)
(2) For each a_i and each a_j, a_i * a_j is also in the group
(3) For each a_i, there exists and a_k such that a_i * a_k = e.
(each element has a multiplicative inverse)
(4) (a_i * a_j) * a_k = a_i * (a_j * a_k) (the operator '*' is associative)

G can be infinite, or G can be finite. An example of an infinite group is
the set of 2 x 2 matrices whose entries are in R and for which their
determinant is non-zero. An example of a finite group is the integers
under multiplication mod 5. Its order is (surprise! phi(5) = 4).
Another example of a finite group is the rotations of a square. Just Label
the corners 1,2,3,4. The elements are the 4 different squares you can get
with 90 degree rotations. Another example is the permutations of the
integers 1....N.

As for your question: "ab*a is in G too? ", the answer is yes. In fact,
aba is known as the conjugation of b by the element a. Conjugation
is a major concept in group theory.

You didn't finish to answer my question.

The question was:
How it is possible to accept a group to be:
1)finite.
2)closure.

As I understood, if G (a finite group) contains a,b, exists:
ab, aba, abab, ababababa... and (a^n)(b^n).
This group's order -->oo(infinity).
But we assumed G as finite..
I know that two infinity sets (one and it's subset), are each bigger and smaller than the other.
But I can't assume infinity equals to finity.
I understand I'm mistaken somewhere, but I can't find where.

[blob100]

Can you please give an example of a group which agrees langrange's theorem (a group and it's subgroup).

Thank's

[R.D. Silverman]

Originally Posted by blob100

You didn't finish to answer my question.

The question was:
How it is possible to accept a group to be:
1)finite.
2)closure.

I DID answer your question. I gave several examples of finite
groups. They are clearly closed. What makes you think that they are
not?

As I understood, if G (a finite group) contains a,b, exists:
ab, aba, abab, ababababa... and (a^n)(b^n).
This group's order -->oo(infinity).

Where did you pull this assertion from??? What makes you think that

ab, aba, abab.... are all DISTINCT??? And saying that the 'group's
order -->oo' is mathematical gibberish. Either the group's order EQUALs
infinity or it is finite. Saying that the 'group's order -->oo' is treating
the order as a VARIABLE, which it definitely is not.


Consider the units of Z/5Z. These are the integers 1,2,3,4. mod 5.

Let a = 2, b = 3. Then ab = 1, aba = 2, and since this group is
commutative, (ab)^n = a^n b^n = 1 for ALL n. The group is clearly
finite and is clearly closed under multiplication. Why is this a problem?

[R.D. Silverman]

Originally Posted by blob100

Can you please give an example of a group which agrees langrange's theorem (a group and it's subgroup).

Thank's

Firstly, ALL groups 'agree with' Lagrange's Theorem. It is, after all, a
THEOREM.

Consider the group of units mod 15. The order of this group
is phi(15) = 8. Its sub-groups all have order 2 or 4. You can
verify this by simple arithmetic. e.g. consider the subgroup whose
elements are {1, 14}. I will let you find the other subgroups.


Shanks' book does a SUPERB job of showing the relationship between
modular multiplication and groups. Which is one reason why I suggested it.

[blob100]

Originally Posted by R.D. Silverman

Firstly, ALL groups 'agree with' Lagrange's Theorem. It is, after all, a
THEOREM.

Consider the group of units mod 15. The order of this group
is phi(15) = 8. Its sub-groups all have order 2 or 4. You can
verify this by simple arithmetic. e.g. consider the subgroup whose
elements are {1, 14}. I will let you find the other subgroups.


Shanks' book does a SUPERB job of showing the relationship between
modular multiplication and groups. Which is one reason why I suggested it.

I understood well done.

Thanks.

[blob100]

Is there a structure in factors of the form p#+1 (prime factorial:
2*3*5*7...*p)? Was it found?
These numbers (the factors (F)) are quiet interesting.
These aren't devided by the prime factorial's primes and agree
F<(p#+1)^0.5.