I read all this from 1st post.

EPIC!!!!

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?

[QUOTE=blob100;213611]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)
[/QUOTE]

If a is of order e, then a^e = 1 mod m [b]and[/b] 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
[b]group[/b] 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
[b]not[/b] 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.

[quote=R.D. Silverman;213619]If a is of order e, then a^e = 1 mod m [B]and[/B] 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
[B]group[/B] 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
[B]not[/B] 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.[/quote]
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?

[QUOTE=blob100;213622]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.
[/QUOTE]


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

[QUOTE]
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?[/QUOTE]


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.

[quote=R.D. Silverman;213623]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.[/quote]

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.

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

Thank's

[QUOTE=blob100;213626]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.

[/QUOTE]


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

[QUOTE]
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).
[/QUOTE]

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?

[QUOTE=blob100;213628]Can you please give an example of a group which agrees langrange's theorem (a group and it's subgroup).

Thank's[/QUOTE]

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.

[quote=R.D. Silverman;213635]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.[/quote]
I understood well done.

Thanks.

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.