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Math Help - isomorphism question

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by wutang View Post
    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
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    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}.
    Last edited by Drexel28; April 28th 2010 at 07:26 PM. Reason: Typo
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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, b : [a, b]>

    (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!
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Swlabr View Post
    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, b : [a, b]>

    (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!
    That isn't what I was saying. I'm saying that define 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 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"
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    That isn't what I was saying. I'm saying that define 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 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
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    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.
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  8. #8
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    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!!!
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  9. #9
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    Quote Originally Posted by wutang View Post
    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!!!
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    That isn't what I was saying. I'm saying that define 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 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.
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Swlabr View Post
    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.
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