# Isomorphism of relatively prime groups

• PennState666
In summary: That is, let \phi(g) be the function that takes an element a of Z_6 and returns the corresponding element of Z_2 \times Z_3, where the first argument is taken from the set Z_2 and the second argument is taken from the set Z_3.In summary, the group Z(sub 2) x Z(sub 3) is generated by <[1],[1]>, and is cyclic and abelian.
PennState666

## Homework Statement

Allow m,n to be two relatively prime integers. You must prove that Z(sub mn) ≈ Z(sub m) x Z(sub n)

## Homework Equations

if two groups form an isomorphism they must be onto, 1-1, and preserve the operation.

## The Attempt at a Solution

since m and n are relatively prime, the gcd(m, n) = 1.
mhm, very stumped from the start.

What do you know about the group Z(sub mn)? What are some facts about it? Is it abelian, etc.?

Isomorphisms isn't my thing, I do not know many facts about Z(sub mn). I can say the contents of Z(sub mn) are {[0],[1], [2],... [mn]}. generated by a single element? I am not sure...Abelian? yes i believe so...

PennState666 said:
Isomorphisms isn't my thing, I do not know many facts about Z(sub mn). I can say the contents of Z(sub mn) are {[0],[1], [2],... [mn]}. generated by a single element? I am not sure...Abelian? yes i believe so...

Yes, generated by a single element is the key. In other words, it's a cyclic group. (Side note: cyclic groups are always abelian, but you don't need that fact for this problem.)

So if Z(sub mn) is isomorphic to Z(sub m) x Z(sub n), then that group would also have to be cyclic, right? Can you show that Z(sub m) x Z(sub n) is cyclic?

mhmm i believe so, but i don't know where to start since I am dealing with a cross product...

If you had to guess at which element of ZmxZn generated the whole group, what would it be?

If you're not sure, try some small examples like Z2xZ3 or Z3xZ4 and see what you can come up with

What would the elements of Z2 x Z3 look like again?

Ordered pairs (a, b) such that a is from Z2, and b is from Z3.

PennState666 said:
What would the elements of Z2 x Z3 look like again?

Well, if A and B are any sets (whether or not they have a group structure), then what is the definition of A x B? This is called the Cartesian product of A and B. The elements are simply ordered pairs of the form $(a,b)$, where $a \in A$ and $b \in B$.

And if A and B are groups, then there's a natural way to define a group operation on A x B.

Surely this is discussed in your textbook or lecture? There's no way you can proceed with this problem without first understanding what the group is.

yes of course I just wanted to be refreshed and reassured...
Z2 x Z3 = {([0],[0]), ([0],[1]), ([0],[2]), ([1],[0]), ([1],[1]), ([1],[2])}...
I cannot grasp how this could be generated by a single element though?

PennState666 said:
yes of course I just wanted to be refreshed and reassured...
Z2 x Z3 = {([0],[0]), ([0],[1]), ([0],[2]), ([1],[0]), ([1],[1]), ([1],[2])}...
I cannot grasp how this could be generated by a single element though?

Look at the six elements. In order for the group to be generated by one element, clearly that element can't have a [0] in either of the two positions of the ordered pair. So that only leaves two candidates: either ([1],[1]) or ([1],[2]). Why not try each and see if it works? Just add it to itself over and over and see if you can generate the group.

the generator for the group Z(sub 2) x Z(sub 3) = <[1],[1]> and since the operation is addition, the generator for a group in the form Zn x Zm = <[1], [1]>. so Zn x Zm is cyclic and abelian.

PennState666 said:
the generator for the group Z(sub 2) x Z(sub 3) = <[1],[1]> and since the operation is addition, the generator for a group in the form Zn x Zm = <[1], [1]>. so Zn x Zm is cyclic and abelian.

OK, good. Note that <[1],[2]> also generates the group.

So Z2 x Z3 is cyclic, and I assume you observed that it has order 6. That looks promising, because Z6 is also cyclic with order 6.

Now you need to define an isomorphism $\phi: Z_{6} \rightarrow Z_2 \times Z_3$ between the groups. As a first step, I suggest choosing a generator $g$ for $Z_6$ and defining $\phi(g)$.

## What is isomorphism of relatively prime groups?

Isomorphism is a mathematical concept that describes a relationship between two groups. Two groups are considered isomorphic if there exists a bijective function between them that preserves their group structure. This means that the groups have the same number of elements and the same operations, making them essentially identical.

## How can you prove that two groups are isomorphic?

To prove that two groups are isomorphic, you must show that there exists a bijective function between them that preserves the group structure. This can be done by demonstrating that the function is both one-to-one and onto, and that it preserves the group's operation (e.g. if the groups are addition and the function preserves addition).

## What is the significance of relatively prime groups in isomorphism?

Relatively prime groups play a crucial role in isomorphism because they allow for the existence of a bijective function between them. This is because relatively prime groups have no common divisors, meaning that their elements cannot be broken down into smaller groups. This allows for the bijective function to preserve the group structure without any interference from common divisors.

## Can non-relatively prime groups be isomorphic?

Yes, non-relatively prime groups can be isomorphic, but it is less common. In these cases, the groups must have some other property that allows for the existence of a bijective function between them that preserves their group structure. For example, they may have the same number of elements and operations, but have different prime factorizations.

## What are some applications of isomorphism of relatively prime groups?

Isomorphism of relatively prime groups has many applications in mathematics and other fields. It is commonly used in cryptography to create secure codes and in chemistry to model molecular structures. It also has applications in computer science, physics, and engineering. Additionally, the concept of isomorphism is used to simplify and solve complex mathematical problems.

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