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Math Help - Identity and Inverse of a Permutation set

  1. #1
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    Identity and Inverse of a Permutation set

    G = GL(2,R), H = {A 2 G : det(A) = 1}

    (where the set of all permutations which send 1 → 1)

    How could i possibly determine the identity (and consequently the inverse) of the above?
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  2. #2
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    Quote Originally Posted by Lightningepsilon View Post
    G = GL(2,R), H = {A 2 G : det(A) = 1}

    (where the set of all permutations which send 1 → 1)

    How could i possibly determine the identity (and consequently the inverse) of the above?

    Uuh?? First you have G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\} ...a group of matrices and a subgroup of this group.

    And then you write something about permutations which send 1 to 1, and you ask about the identity and inverse "of the above": what "above"?...What is going on??

    Tonio
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    Quote Originally Posted by tonio View Post
    Uuh?? First you have G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\} ...a group of matrices and a subgroup of this group.

    And then you write something about permutations which send 1 to 1, and you ask about the identity and inverse "of the above": what "above"?...What is going on??

    Tonio
    Sorry for the very badly worded question!

    What i meant was we have G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\} and i have to determine whether or not H is a subgroup of G.

    I've determined H is closed - now i have to find out whether or not H has an identity and an inverse. Neither or which i can find.

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    Quote Originally Posted by Lightningepsilon View Post
    Sorry for the very badly worded question!

    What i meant was we have G:=GL(2,\mathbb{R})\,,\,\,H:=\{A\in G\;;\;\det A=1\} and i have to determine whether or not H is a subgroup of G.

    I've determined H is closed - now i have to find out whether or not H has an identity and an inverse. Neither or which i can find.

    Ok, now I see: well, you have to answer the following questions:

    1) Is H non-empty? (almost trivial)

    2) If A\in H then A^{-1}\in H ?

    3) If A,B\in H then AB\in H ?

    Hint for (2)-(3): \det(AB)=\det A\cdot \det B

    Tonio
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    Quote Originally Posted by tonio View Post
    Ok, now I see: well, you have to answer the following questions:

    1) Is H non-empty? (almost trivial)

    2) If A\in H then A^{-1}\in H ?

    3) If A,B\in H then AB\in H ?

    Hint for (2)-(3): \det(AB)=\det A\cdot \det B

    Tonio
    (3)
    A,B\in H then
     det(AB) = det(A)det(B) = 1 * 1 = 1
    and so AB\in H

    (2)
    I suppose the inverse must be 'I' and so i have to find A^{-1}
    where A*A^{-1} = I.

    So as A\in H then  det(A) = 1

    1 = det(I) = det(AA^{-1}) = det(A)det(A^{-1})
    hence det(A^{-1}) = 1 so A^{-1}\in H.

    I can't thank you enough Tonio for helping me, I really appreciate it
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