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Math Help - Show that C^x is isomorphic to GL2(R)

  1. #1
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    Show that C^x is isomorphic to GL2(R)

    Show that C^x is isomorphic to the subgroup of GL2(R) consisting of all matrices of the form:
    a b
    -b a such that a^2+b^2 is not 0.

    I'm guessing I need to show C^x--->G given by c(a+bi)=
    a b
    -b a is an isomorphism.


    I need to show 1-1, onto, and homomorphism.
    1-1:
    Assume c(a+bi)=c(c+di). I need to show this implies a+bi=c+di
    a b c d
    -b a = -d a
    I'm not sure how this implies 1-1



    onto:Not sure how to show this




    homomorphism:
    c(ab)=c(a)c(b)
    Let a=a+bi, b=c+di
    a b c d
    -b a * -d c

    ac-db ad+bc
    -bc-ad -bd+ac =c(ab)
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  2. #2
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    i wasn't really sure how to write the matrices so I was looking for in the first
    first row: a b
    second row : -b a
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Show that C^x is isomorphic to the subgroup of GL2(R) consisting of all matrices of the form:
    a b
    -b a such that a^2+b^2 is not 0.

    I'm guessing I need to show C^x--->G given by c(a+bi)=
    a b
    -b a is an isomorphism.


    I need to show 1-1, onto, and homomorphism.
    1-1:
    Assume c(a+bi)=c(c+di). I need to show this implies a+bi=c+di
    a b c d
    -b a = -d a
    I'm not sure how this implies 1-1



    onto:Not sure how to show this




    homomorphism:
    c(ab)=c(a)c(b)
    Let a=a+bi, b=c+di
    a b c d
    -b a * -d c

    ac-db ad+bc
    -bc-ad -bd+ac =c(ab)
    Your homomorphism part looks right.

    For injectivity remember that two matrices are equal if and only if their entries are equal.

    For surjectivity it's much, much easier than you may think. Try it one more time.
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  4. #4
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    so the matrices are equal if a=c, b=d

    surjective we want y=c(a+bi)=first row a b
    second row -b a
    want to solve for a+bi
    Do I let y= a b
    c d
    and solve that way
    So c=-b, d=a
    a+bi=d-ci
    I feel like that was completely wrong
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    so the matrices are equal if a=c, b=d

    surjective we want y=c(a+bi)=first row a b
    second row -b a
    want to solve for a+bi
    Do I let y= a b
    c d
    and solve that way
    So c=-b, d=a
    a+bi=d-ci
    I feel like that was completely wrong
    What about the fact that c\left(a+bi\right)=\begin{bmatirx}a & b\\\ -b & a\end{bmatrix}
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  6. #6
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    I don't know where that comes from I have c(a+bi)=
    first row : a b
    second row -b a
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