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Math Help - Is this an isomoprphism of groups?

  1. #1
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    Is this an isomoprphism of groups?

    G is the group of 2x2 invertible upper triangular matrices with real entries. Let H be the normal subgroup such that a_{11}=a_{22}=1.

    If I have the following surjective homomorphism

    f:G\rightarrow (\mathbb{R}^*,\times)

    such that f(g)=a_{11}a_{22}.

    Then by the first isomorphism theorem I have the following isomorphism:

    f:G/H\rightarrow (\mathbb{R}^*,\times)

    such that f(\bar{g})=a_{11}a_{22}

    Is this correct?

    Thanks for any help
    Last edited by hmmmm; July 29th 2011 at 02:13 PM.
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  2. #2
    Super Member girdav's Avatar
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    Re: Is this an isomoprphism of groups?

    It's correct. M
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  3. #3
    Super Member girdav's Avatar
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    Re: Is this an isomoprphism of groups?

    It's correct. Maybe you should denote by \widetilde f the morphism between G/H (I guess H is the matrix of G which have determinant 1) and \mathbb R^*.
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  4. #4
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    Re: Is this an isomoprphism of groups?

    yeah thanks very much, I edited it. My notation is bad, how did you put the squiggle above the f?
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  5. #5
    Super Member girdav's Avatar
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    Re: Is this an isomoprphism of groups?

    Quote Originally Posted by hmmmm View Post
    yeah thanks very much, I edited it. My notation is bad, how did you put the squiggle above the f?
    \widetilde f
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  6. #6
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    Re: Is this an isomoprphism of groups?

    Uh,

    I don't think so. The kernel of that homomorphism is certainly strictly larger than H. What about the matrix A= \begin{pmatrix}-1&0\\0&-1\end{pmatrix}? Certainly A\in G\setminus H but \mathrm{det}A=1.
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  7. #7
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    Re: Is this an isomoprphism of groups?

    Ah of course, in fact if a_{22}=(a_{11})^{-1} then that maps to the identity.

    So i have to have another map f such that f(a_{11},a_{22})=1 and it has to be surjective right?
    Last edited by hmmmm; August 2nd 2011 at 08:59 AM.
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  8. #8
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    Re: Is this an isomoprphism of groups?

    What about:

    Let J be the subgroup of \mathcal{SL}_r(\mathbb{R}) such that a_{12}=a_{21}=0.

    Then we have a surjectrive homomorphism [\bar{f}:G\rightarrow J[/TEX] with kernel H.

    So from the first isomorphism theorem we have an isomorphism:

    \widetilde{f}:G/H\rightarrow J

    thanks for any help
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Re: Is this an isomoprphism of groups?

    Quote Originally Posted by hmmmm View Post
    What about:

    Let J be the subgroup of \mathcal{SL}_r(\mathbb{R}) such that a_{12}=a_{21}=0.

    Then we have a surjectrive homomorphism [\bar{f}:G\rightarrow J[/TEX] with kernel H.

    So from the first isomorphism theorem we have an isomorphism:

    \widetilde{f}:G/H\rightarrow J

    thanks for any help
    What precisely are you trying to do? Find a nicer description of G/H? Is this a different question?
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  10. #10
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    Re: Is this an isomoprphism of groups?

    Yeah that is what I am trying to do, any help would be apperciated
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