How do I do...

Z x Z/<(1,2)> isomorphic to Z

and

Z x Z x Z/<(1,1,1)> isomorphic to Z x Z

Thanks.

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- Dec 11th 2011, 06:37 PMJSB1917Direct Products
How do I do...

Z x Z/<(1,2)> isomorphic to Z

and

Z x Z x Z/<(1,1,1)> isomorphic to Z x Z

Thanks. - Dec 11th 2011, 08:30 PMDrexel28Re: Direct Products
What have you done my friend? Without any context, I'm not sure how you would even want to proceed by this? The obvious thing to do, for the first example you want a mapping $\displaystyle \mathbb{Z}^2\to\mathbb{Z}$ whose kernel is $\displaystyle a(1,2)$. How about something like $\displaystyle (1,0)\mapsto 2$, and $\displaystyle (0,1)\mapsto -1$. Then, $\displaystyle (a,b)\mapsto 2a-b$ which....

- Dec 11th 2011, 09:14 PMJSB1917Re: Direct Products
- Dec 11th 2011, 09:30 PMDrexel28Re: Direct Products
Clearly you are going to want to construct an epimorphism $\displaystyle \mathbb{Z}^2\twoheadrightarrow\mathbb{Z}$ with $\displaystyle \ker=\langle(1,2)$ and then apply the FIT. Now, since $\displaystyle \mathbb{Z}^2$ has basis $\displaystyle \{(1,0),(0,1)\}$ (i.e. it's free) you need only specify what you want your homomorphism to do to those guys, and then just extend by linearity. Now, how could you define a map on them, so that your kernel is what you want?

- Dec 12th 2011, 02:47 AMDevenoRe: Direct Products
Drexel28 has already given you a mapping which has the property that f(2,1) = 0. all you have to do is show it is an onto homomorphism....

- Dec 13th 2011, 05:46 PMJSB1917Re: Direct Products
I have f(a,b)=2a-b for the first showed it's a homomorphism, showed the kern f = <(1,2)>, just need the onto.

Similarly, for the other one I have f(a,b,c)=(a-c,b-c), showed it's a homomorphism, kern f = <(1,1,1)>, just need the onto.

I just need help with onto. - Dec 14th 2011, 01:37 AMDevenoRe: Direct Products
pick a k in Z. now we need to find a pair of integers (a,b) such that f(a,b) = 2a-b = k. there could be a LOT of pairs with this property, we only need to find ONE. so let's make it easy on ourselves, and see if we can find a pair of the form (0,b) such that f(0,b) = - b = k.

well, it appears that b = -k, will do the trick. so

(we could have decided to let (a,b) = (a,k) (that is, choose b = k). then, f(a,k) = 2a - k = k implies that 2a = 2k, and we could choose a = k. and sure enough f(2k,k) = 2k-k = k. there's literally infinite ways we could have gone about them, and all are equally good. the trick is to pick "some" value for either a, or b, and then find out how we have to choose "the other part of the pair". for example, suppose we choose a = m. then f(a,b) = 2m - b. if 2m - b = k, then b has to be 2m - k. and sure enough, f(m,2m-k) = 2m - (2m - k) = 2m - 2m + k = k. see how that works?).

a similar logic applies to the second problem. we want to pick a,b,c so that a-c = k, b-c = m, for any arbitrary pair (k,m). one possible choice is (k,m,0). do you see the similarity here with solving a system of linear equations?