# Isomorphism and direct product of groups

1. Oct 11, 2007

### teleport

Just wondering if there is a general way of showing that

(Z, .)n isomorphic to Zm X Zp with the obvious requirement that both groups have the

same order?

2. Oct 11, 2007

### cogito²

Well considering they're not isomorphic in general, I really don't know.

3. Oct 11, 2007

### teleport

Oh didn't know this. Why not in general? Forgot to mention that (Z, .)n are the remainders mod n that are relatively prime to n, and exclude 0.

4. Oct 12, 2007

### matt grime

What are m and p, would b cogito's main point.

5. Oct 12, 2007

### teleport

What would m, p have to be? Are the possible values of m, p not just dependent of the order of (Zn, .), but also related in another way to n? Is there a requirement of some individual relation between n and m, and n and p?

6. Oct 12, 2007

### matt grime

From what you wrote originally, I presumed you'd just forgotten to say what m,p are. But I think you're now claiming that any two groups

Z_m x Z_p and Z_w x Z_q are isomorphic if mp=qw.

Is that what you meant?

I am now wondering what you meant.

7. Oct 12, 2007

### teleport

For example is the map f: (Zn, .) -> Zm X Zp (|(Zn, .)| = mp) with

f(a mod mp) = (a mod m, a mod p)

an isomorphism? This sort of implies your interpretation of my question, I guess.

8. Oct 12, 2007

### matt grime

Why haven't you tried to work it out for yourself?

9. Oct 12, 2007

### teleport

Ah, sorry for that. Yes I have now worked out a couple of examples of my own. The map that works to show isomorphism between (Zn, .) and Zm X Zp (when they are indeed isomorphic) is f(x^i) = (i mod m, i mod p), where x is a generator of (Zn, .). However from the result of my work it seems that this map only "works" when gcd(m,p) = 1. Since I know that isomorphism is shown with the use of the generators, then maybe Zm X Zp fails to have a generator when m and p are not relatively prime. Then I gave it a try and showed (I hope) that for Zm X Zp to be cyclic, gcd(m,p) must be 1, like this:

Because <1 mod m> = Zm and <1 mod p> = Zp, if Zm X Zp is cyclic, then its generator must be (1 mod m, 1 mod p). Take any
(i mod m, j mod p) in Zm X Zp. Then there must exist some k in Z s.t. k(1 mod m, 1 mod p) = (k mod m, k mod p) = (i mod m, j mod p). We must then have

k = i mod m

k = j mod p.

But then by the Chinese Remainder Theorem, k exists iff gcd(m,p) = 1, which implies
Zm X Zp is cyclic iff gcd(m,p) = 1.

Now definitely one requirement for Z_m x Z_p and Z_w x Z_q to be isomorphic (with mp=qw) is that they are both cyclic which is true if gcd(m,p) = gcd(w,q) = 1.

Thanks for not showing the ans :). This was good for me, since my test is coming up.

Last edited: Oct 12, 2007
10. Oct 12, 2007

### teleport

Wow I forgot the actual Chinese Thm statement and there is no iff. But the solution k to

k = i mod m

k = j mod p

exists only when i = j mod gcd(m,p). However no two Za, Zb s.t. gcd(a,b) =/= 1, have i = j mod gcd(a,b) for all i in Za and all j in Zb. Hence the same result follows that Zm X Zp is cyclic iff gcd(m,p) = 1.

Last edited: Oct 12, 2007
11. Oct 13, 2007

### teleport

I can't stop thinking about this. Now I tried out (Z_8, .) with Z_2 X Z_2 and there is the isomorphism (3^i)(5^j) = (i mod 2, j mod 2). The interesting thing is that neither of those groups are cyclic. All elements of (Z_8, .) except 1 have order 2 =/= |(Z_8, .)|, so the group can't be cyclic because there is no generator. If I was right, Z_2 X Z_2 is also not cyclic. So it seems that for the groups to be isomorphic either they must be both cyclic, or neither of them must be cyclic.

12. Oct 13, 2007

### matt grime

Are you getting at: if G is cyclic and H is isomorphic to G then H is cyclic? Of course that is true. Isomorphisms amongst other things preserve the order of elements.

13. Oct 13, 2007

### teleport

Makes sense. So is the conclusion that if both groups (Z_n, .) and Zm X Zp are cyclic with the same order, or if both are non-cyclic with the same order, then they are isomorphic, correct? The first case can be shown by using f as above. For the 2nd case I also tried (Z_15, .) with Z_2 X Z_4 and it also gives me isomorphic. The thing I noticed when showing they are isomorphic is that the map to be used is f(x^i y^j) = (i mod 2, j mod 4) with x, y not only coprime with |x| = 2 and |y| = 4, but x and y both had to be prime, otherwise the map would not show the isomorphism. Is this a casuality or a general thing? I will try to look for some counter example later. But I don't think there will be a computationally easier one. :)

14. Oct 13, 2007

### matt grime

No, absolutely not. Exercise find lots of counter examples. Surely the fact that there are non-abelian groups should make you nervous.

15. Oct 13, 2007

### teleport

I will try to find some time for the counter examples. I'm having my midterms next week and I have diverted my attention for some period now. Is there some formula that determines which n, p, m would make the isomorphism. If you know of any link with this info that would be great. I'm interested in seeing the proof. Hopefully won't be that complex. Thank you for the remarks.

16. Oct 14, 2007

### matt grime

There is an entire structure theorem for abelian groups that answers all of your questions. I'd google for it if I were you.