Let H={(n,n)|nis an element of Z (integers)}.Prove (Z+Z)/H is isomorphic to Z.

The hint I was given was that my mapping should involve subtraction. I am totally clueless on this one and any help would be of great benefit to me.

Thanks

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- Apr 28th 2010, 06:57 PMwutangisomorphism question
Let H={(n,n)|n

**i**s an element of Z (integers)}**.**Prove (Z+Z)/H is isomorphic to Z.

The hint I was given was that my mapping should involve subtraction. I am totally clueless on this one and any help would be of great benefit to me.

Thanks - Apr 28th 2010, 07:07 PMtonio
- Apr 28th 2010, 07:09 PMDrexel28
- Apr 29th 2010, 12:37 AMSwlabr
This does note work. You are (I believe) claiming, essentially, that $\displaystyle (A \times B)/C \cong B$ when $\displaystyle A \cong C$. It is (again, I believe) true when everything is finite and abelian. However, take $\displaystyle A=B=\mathbb{Z}$ and $\displaystyle C=2A$.

$\displaystyle A \times B = <a, b : [a, b]>$

$\displaystyle (A \times B)/2A = <a, b : [a, b], a^2> \cong \mathbb{Z} \times C_2$ and so these groups are clearly not isomorphic.

I know that what you claim is not true as I once claimed it was true too! - Apr 29th 2010, 08:56 AMDrexel28
That isn't what I was saying. I'm saying that define $\displaystyle \theta:H\to\mathbb{Z}\times\{0\}:(n,n)\mapsto (n,0)$

Clearly this is injective since $\displaystyle (n,0)=(n',0)\implies n=n'\implies (n,n)=(n',n')$ and surjective since if $\displaystyle (n,0)\in\mathbb{Z}\times\{0\}$ then $\displaystyle \theta\left((n,n)\right)=(n,0)$. To see it's a homomorphism we note that $\displaystyle \theta\left((n,n)(m,m)\right)=\theta\left((nm,nm)\ right)=\left(nm,0\right)=\left(n,0\right)(m,0)=\th eta((n,n))\theta((m,m))$.

So, $\displaystyle H\cong\mathbb{Z}\times\{0\}$. So now define $\displaystyle \phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}:(z,0)\maps to z$. This is clearly injective and surjective and $\displaystyle \phi\left((z,0)(z',0)\right)=\phi\left((zz',0)\rig ht)=zz'=\phi\left((z,0)\right)\phi\left((z',0)\rig ht)$. Thus, $\displaystyle \mathbb{Z}\times\{0\}\cong\mathbb{Z}$ and so the conclusion follows by the transitivity of "is isomorphic to" - Apr 29th 2010, 12:26 PMtonio
- Apr 29th 2010, 04:41 PMDrexel28
- Apr 29th 2010, 05:06 PMwutang
Here is my proof.

Let f: Z+Z to Z where f(a,b)=a-b. NTS that F is an epimorphism.

f is obviously a well defined function.

Operation preserving: let x,y be elemnts of Z+Z. The f(x,y)=x-y = f(x)-f(y). Thus f is operation preserving.

Onto- Let z-y be an element of Z and let z,y be elements of Z+Z. Then

f(z,y)=z-y. Thus f is onto.

show that H is contained in Ker f.

Let x be an element of H. Then x=(n,n) for some n that is an element of Z. Then (n,n) is an element of kerf since f(n,n)=0. Thus x is an element of ker f and H is contained in Ker f.

Show that Ker f is contained in H.

Let x be an element of Ker f. Then x=(n,n) for some n that is an element of Z since f(n,n)=0. Then (n,n) is an element of H by the definition of H. Thus x is an element of H. So ker f is contained in H.

Therefore by th 1st isomorphism theroem, (Z+Z)/H is isomorphic to Z.

P.S. I will be getting my labtop back soon, so I wil be able to write in LaTex again!!! - Apr 29th 2010, 05:38 PMDefunkt
This is not what you are required to show. In order to prove that f is operation conserving, prove that $\displaystyle f((x,y)+(z,w)) = f(x,y)+f(z,w) ~ \forall x,y,z,w \in \mathbb{Z}$

Quote:

Onto- Let z-y be an element of Z and let z,y be elements of Z+Z. Then

f(z,y)=z-y. Thus f is onto.

Quote:

show that H is contained in Ker f.

Let x be an element of H. Then x=(n,n) for some n that is an element of Z. Then (n,n) is an element of kerf since f(n,n)=0. Thus x is an element of ker f and H is contained in Ker f.

Quote:

Show that Ker f is contained in H.

Let x be an element of Ker f. Then x=(n,n) for some n that is an element of Z since f(n,n)=0. Then (n,n) is an element of H by the definition of H. Thus x is an element of H. So ker f is contained in H.

Quote:

Therefore by th 1st isomorphism theroem, (Z+Z)/H is isomorphic to Z.

P.S. I will be getting my labtop back soon, so I wil be able to write in LaTex again!!!

- Apr 30th 2010, 12:37 AMSwlabr
- Apr 30th 2010, 06:53 AMDrexel28