1. ## isomorphism question

Let H={(n,n)|n is 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

2. Originally Posted by wutang
Let H={(n,n)|n is 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

Show that $\phi(a,b):= a-b$ is an epimorphism from $\mathbb{Z}\times\mathbb{Z}$ onto $\mathbb{Z}$ and use then the first isomorphism theorem.

Tonio

3. Originally Posted by tonio
Show that $\phi(a,b):= a-b$ is an epimorphism from $\mathbb{Z}\times\mathbb{Z}$ onto $\mathbb{Z}$ and use then the first isomorphism theorem.

Tonio
Or construct the obvious isomorphism between $H$ and $\mathbb{Z}\times\{0\}$ which is trivially isomorphic to $\mathbb{Z}$.

4. Originally Posted by Drexel28
Or construct the obvious isomorphism between $H$ and $\mathbb{Z}\times\{0\}$ which is trivially isomorphic to $\mathbb{Z}$.
This does note work. You are (I believe) claiming, essentially, that $(A \times B)/C \cong B$ when $A \cong C$. It is (again, I believe) true when everything is finite and abelian. However, take $A=B=\mathbb{Z}$ and $C=2A$.

$A \times B = $

$(A \times B)/2A = \cong \mathbb{Z} \times C_2$ 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!

5. Originally Posted by Swlabr
This does note work. You are (I believe) claiming, essentially, that $(A \times B)/C \cong B$ when $A \cong C$. It is (again, I believe) true when everything is finite and abelian. However, take $A=B=\mathbb{Z}$ and $C=2A$.

$A \times B = $

$(A \times B)/2A = \cong \mathbb{Z} \times C_2$ 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!
That isn't what I was saying. I'm saying that define $\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" alt="\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" />

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

So, $H\cong\mathbb{Z}\times\{0\}$. So now define $\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" alt="\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" />. This is clearly injective and surjective and $\phi\left((z,0)(z',0)\right)=\phi\left((zz',0)\rig ht)=zz'=\phi\left((z,0)\right)\phi\left((z',0)\rig ht)$. Thus, $\mathbb{Z}\times\{0\}\cong\mathbb{Z}$ and so the conclusion follows by the transitivity of "is isomorphic to"

6. Originally Posted by Drexel28
That isn't what I was saying. I'm saying that define $\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" alt="\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" />

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

This shouldn't be confusing after some algebra is learnt: the group operation on $\mathbb{Z}\times\mathbb{Z}$ is coordinatewise addition, not multiplication! Fortunately enough, your proof works mutandis-mutandis if instead product we use addition .

So, $H\cong\mathbb{Z}\times\{0\}$. So now define $\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" alt="\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" />. This is clearly injective and surjective and $\phi\left((z,0)(z',0)\right)=\phi\left((zz',0)\rig ht)=zz'=\phi\left((z,0)\right)\phi\left((z',0)\rig ht)$. Thus, $\mathbb{Z}\times\{0\}\cong\mathbb{Z}$ and so the conclusion follows by the transitivity of "is isomorphic to"
I still can't see how $\mathbb{Z}\times\mathbb{Z}/H\cong \mathbb{Z}$ , which is what was wanted, follows from the above...

Tonio

7. Originally Posted by tonio
I still can't see how $\mathbb{Z}\times\mathbb{Z}/H\cong \mathbb{Z}$ , which is what was wanted, follows from the above...

Tonio
I just used "multiplication" because I was thinking about groups in general where you usually write multiplication. Of course it was meant to be taken as addition. Also, I misread the question.

8. 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!!!

9. Originally Posted by wutang
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.
This is not what you are required to show. In order to prove that f is operation conserving, prove that $f((x,y)+(z,w)) = f(x,y)+f(z,w) ~ \forall x,y,z,w \in \mathbb{Z}$

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.
This is not clear. First you should say that any element $x \in \mathbb{Z}$ can be written as $z-y$ for some $z,y \in \mathbb{Z}$, say, $z=x,y=0$

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.
This is good.

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.
You're sort of assuming what you are asked to prove here. You should let $x \in Ker(f)$, write $x=(x_1,x_2) \in \mathbb{Z+Z}$ and reach that $x_1=x_2$

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!!!

10. Originally Posted by Drexel28
That isn't what I was saying. I'm saying that define $\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" alt="\theta:H\to\mathbb{Z}\times\{0\}n,n)\mapsto (n,0)" />

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

So, $H\cong\mathbb{Z}\times\{0\}$. So now define $\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" alt="\phi:\mathbb{Z}\times\{0\}\to\mathbb{Z}z,0)\mapsto z" />. This is clearly injective and surjective and $\phi\left((z,0)(z',0)\right)=\phi\left((zz',0)\rig ht)=zz'=\phi\left((z,0)\right)\phi\left((z',0)\rig ht)$. Thus, $\mathbb{Z}\times\{0\}\cong\mathbb{Z}$ and so the conclusion follows by the transitivity of "is isomorphic to"
My counter-example still works here (I think you are just doing what I was saying you couldn't do in a more complicated way!). There exists a bijection between A and 2A, but 2A does not quotient out all of A.

11. Originally Posted by Swlabr
My counter-example still works here (I think you are just doing what I was saying you couldn't do in a more complicated way!). There exists a bijection between A and 2A, but 2A does not quotient out all of A.
I think you didn't read my other response. You disagree that $H\cong \mathbb{Z}$? That's what I thought the question was, I didn't realize he said anything about quotient groups.