Isomorphism of relatively prime groups

[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.

 

[jbunniii]

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

 

[PennState666]

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...

 

[jbunniii]

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?

 

[PennState666]

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

 

[Office_Shredder]

If you had to guess at which element of Z_mxZ_n 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

 

[PennState666]

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

 

[kru_]

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

 

[jbunniii]

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.

 

[PennState666]

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?

 

[jbunniii]

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.

 

[PennState666]

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.

 

[jbunniii]

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 $\phi(g)$.