# relatively prime isomorphism groups

by dim&dimmer
Tags: groups, isomorphism, prime
 P: 22 1. The problem statement, all variables and given/known data Show that Z/mZ X Z/nZ isomorphic to Z/mnZ iff m and n are relatively prime. (Using first isomorphism theorem) 2. Relevant equations 3. The attempt at a solution Okay, first I want to construct a hom f : Z/mZ X Z/nZ ---> Z/mnZ I have f(1,0).m = 0(mod mn) = f(m,0) = f(e) = e and f(0,1).n = 0(mod mn) ..... Now f(1,0) = kn for some k because f(1,0).m has to be divisible by mn and f(0,1) = lm likewise My hom is f(a,b) = a.f(1,0) + b.f(0,1) Does this make sense??? If so, then I have to show that ker(f) is trivial to prove f is an iso. ker(f)(a,b) = (0,0) So f(a,b) = akn + blm and I have to show akn +blm = 0. I find this all very confusing so any help would be greatly appreciated. Dim
 P: 394 Since the hint was to use the 1st Isom Thm, why don't you define a homomorphism g from Z onto Z/mZ x Z/nZ, such that ker g is mnZ.
 P: 22 Ok, this sounds good but I haven't got a clue about how to start doing it. Would the isomorphism then end up looking something like f : Z/ker(g) ---> Z/mZ x Z/nZ ==> f : Z/mnZ --> Z/mZ x Z/nZ (as ker(g) = mnZ) Any tips on constructing the hom g:Z -->Z/mZ xZ/nZ would be appreciated, Ive got to the stage where I can't even think, maybe a break for a day or two. I know g(mn) = (00), but lost as to how to construct this.
P: 394

## relatively prime isomorphism groups

 Quote by dim&dimmer Any tips on constructing the hom g:Z -->Z/mZ xZ/nZ would be appreciated
Define g(i)=(thing1,thing2). Usually in these problems the thing1 and thing2 are very simple. By very simple, I mean that someone not even in your class could possibly guess them correctly. Try something, and see if it works.
 P: 22 Well, Im none the wiser, you'll have to dumb it down for me, for g(i) does i represent the identity or neutral element, would g(i) = (0,0) ? Still lost
 Emeritus Sci Advisor PF Gold P: 16,101 Z is a free abelian group with one generator. You should know what every homomorphism Z --> A looks like, no matter what abelian group A happens to be.
P: 394
 Quote by dim&dimmer Any tips on constructing the hom g:Z -->Z/mZ xZ/nZ would be appreciated
I'm just using i to denote an integer. The domain of g is Z, so for any integer i, you have to define g(i). Since g(i) is in Z/mZ xZ/nZ, you know g(i) is some ordered pair. The first component of that ordered pair will be some function of i. What's the easiest function you know? The output must be in Z/mZ, of course, but if you accidentally forget that, you might guess right anyway. The second component of that ordered pair will be some function of i. What's the easiest function you know? So define g(i) to be (easy1,easy2) and then see if it's a homomorphism. See what its kernel is. See if it is onto. If you are lucky, you will guess right and you'll be done. If not, try again. Post your attempts.
 P: 22 Attempt 1: Define g : Z --> Z/mZ x Z/nZ, z |---> (z (mod m), z (mod(n)) Since domain and range are abelian g is a homomorphism as g(ab) = (a+b mod(m), a+b mod(n)) = (a mod(m), a mod(n)) + (b mod(m), b mod(n)) = g(a) + g(b) ker (g) = mnZ ,as n(m.z mod(m)) = n.0 = 0 g is onto. I think this is right, so the first part is done. Now I have to show that f : Z/ker(g) ---> im(g) is an isomorphism. ==> f : Z/mnZ --> Z/mZ x Z/nZ I think that f(x) = (x mod(m), x mod(n)), for any x in Z/mnZ. Is this right?? So f is a homomorphism, using same argument as for g (abelian). Then, if ker(f) is trivial, f is an isomorphism. Suppose f(z) = (0,0), then g(z) = (0,0) So, if f maps any z in Z/mnZ to e, so that z is an element of ker(f), then the only z it can map is equal to e. Therefore, ker(f) = {e}, and f is an isomorphism. While this looks ok to me, I know it is incorrect as I have failed to address the "iff m and n are relatively prime" part. I'd really appreciate a bit more "concrete" help if possible, I know that you want me to think it out for myself, but I've already spent far too much time on this problem relative to its overall course mark. Dim.
P: 394
 Quote by dim&dimmer Attempt 1: Define g : Z --> Z/mZ x Z/nZ, z |---> (z (mod m), z (mod(n)) Since domain and range are abelian g is a homomorphism as g(ab) = (a+b mod(m), a+b mod(n)) = (a mod(m), a mod(n)) + (b mod(m), b mod(n)) = g(a) + g(b) ker (g) = mnZ ,as n(m.z mod(m)) = n.0 = 0 g is onto. I think this is right, so the first part is done. Now I have to show that f : Z/ker(g) ---> im(g) is an isomorphism.
It is an isomorphism by the First Isomorphism Theorem, so you are now done with one part.

 While this looks ok to me, I know it is incorrect as I have failed to address the "iff m and n are relatively prime" part.
Hmm, well, I guess you didn't completely show ker(g) = mnZ. You have one inclusion ("is a subset of," or "contains," which one?), but don't you need relatively prime to get the other inclusion?

As for the if and only if part, just assume m and n are not relatively prime, and show the two groups in question can not be isomorphic. You could consider the order of the identity element. Come to think of it, that might have been an easier way to do the whole darn thing (the word "cyclic" comes to mind).
 P: 394 OK, the first isom thm is a nice way to do it. To see how to finish it, just look at example. g: Z -> Z/6Z x Z/10Z given by g(i)=(i,i) where first component is mod 6 and second is mod 10. What is ker g? Well, i is in ker g iff i=0 mod 6 and i=0 mod 10. So i is divisible by 6 and by 10. Iff i is divisible by 30. Note Z/30Z has 30 elements but Z/6Z x Z/10Z has 60. Therefore not isomorphic. Replace 10 by 5 to get another example. Use m and n in general to write up your proof. Hope that's more concrete.
 P: 22 ok, Z/mZ is a cyclic group (iso to Z mod(m)) with order m, and Z/nZ has order n. Isomorphism preserves orders so ker(g) = mnZ iff m and n are coprime, so that the order of Z/mZ x Z/nZ is mn. Does this complete the proof

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