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Math Help - Group ismorphism (Z2xZ4)

  1. #1
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    Group ismorphism (Z2xZ4)

    Hi,

    Can you show that (Z2xZ4,+)/N isomorphic to (Z4,+)?
    (N=<(1,2)>)
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  2. #2
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    Re: Group ismorphism (Z2xZ4)

    My solution is: N={(0,0),(1,2)} and f((a,b))->x (mod4) (a(mod2),b(mod4))

    But f(0,0)=f(2,4)=f(1,2)+f(1,2) so f(1,2)=2(mod4). Hence N is not kerf so i cannot use first ismorphism theorem. Am I wrong?
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  3. #3
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    Re: Group ismorphism (Z2xZ4)

    it suffices to find a homomorphism from Z2xZ4 onto Z4 whose kernel is N (do you understand why? this is "the hard part").

    let's look at the cosets of Z2xZ4/N:

    we have N = {(0,0),(1,2)}, the identity

    we have (0,1) + N = {(0,1),(1,3)} (this is also (1,3) + N)

    we have (0,2) + N = {(0,2), (1,0)} (which is (1,0) + N, as well)

    we have (0,3) + N = {(0,3), (1,1)} (which is (1,1) + N).

    so if we map (0,k) + N ---> k, it appears we will have an isomorphism.

    so it looks like we want a mapping that sends (0,k) ---> k and (1,k) ---> k+2 (mod 4) (there are other possible homomorphisms).

    can we combine these two into a single condition?

    how about sending (k,m) ---> m + 2k (mod 4)?

    now verify that f(k,m) = m + 2k (mod 4) is indeed an onto homomorphism, and that ker(f) = N.
    Last edited by Deveno; November 7th 2012 at 08:47 AM.
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  4. #4
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    Re: Group ismorphism (Z2xZ4)

    Quote Originally Posted by Deveno View Post
    it suffices to find a homomorphism from Z2xZ4 onto Z4 whose kernel is N (do you understand why? this is "the hard part").

    let's look at the cosets of Z2xZ4/N:

    we have N = {(0,0),(1,2)}, the identity

    we have (0,1) + N = {(0,1),(1,3)} (this is also (1,3) + N)

    we have (0,2) + N = {(0,2), (1,0)} (which is (1,0) + N, as well)

    we have (0,3) + N = {(0,3), (1,1)} (which is (1,1) + N).

    so if we map (0,k) + N ---> k, it appears we will have an isomorphism.

    so it looks like we want a mapping that sends (0,k) ---> k and (1,k) ---> k+2 (mod 4) (there are other possible homomorphisms).

    can we combine these two into a single condition?

    how about sending (k,m) ---> m + 2k (mod 4)?

    now verify that f(k,m) = m + 2k (mod 4) is indeed an onto homomorphism, and that ker(f) = N.
    Is my solution wrong?
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  5. #5
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    Re: Group ismorphism (Z2xZ4)

    you don't exactly say what your solution IS.

    you write f((a,b)) = x (mod 4) <---what is x????
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