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Math Help - Group theory

  1. #1
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    Jan 2009
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    Group theory

    Hey!

    For any subset of the affine plane R^2, let G() denote the group of all affine transformations f of R2 such that f() = .

    The elements of the group G = G(), where is the hyperbola xy = 1
    in R^2; i think are: (ax|dy) & (by|cx) where ad=bc=1

    Show that the elements of G preserving each of the two asymptotes of form a normal subgroup N of index 2 in G, isomorphic to the multiplicative group R* of non-zero real numbers, which acts regularly on .

    Do all the elements of G then preserve the two asymptotes? Any hints how to show the rest please?!

    No idea what i'm doin really i'm afraid, but here goes nothing:

    The index of a subgroup N of G is the number of distinct left
    (or right) cosets of N in G. Even if N is not a normal subgroup, the corre-
    spondence xN -> Nx is a bijection between the set of left cosets and the
    set of right cosets. If the index of N in G is equal to 2, then N is a normal subgroup of G:

    Showing that Nx = xN for each x in G. If x in N there is nothing
    to prove. Hence, let x is not in N. Since the index is two, the disjoint union of N and xN make up all of G. Analogously the disjoint union of N & Nx = G.
    Since N does not equal Nx as well as N does not equal xN, it follows that Nx = xN.
    Multiplying this with x^-1 gives that N^x = N and hence N is a normal
    subgroup.

    ???
    Thanks, any help would be much appreciated. x
    Last edited by dazed; May 10th 2009 at 03:46 PM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by dazed View Post
    Hey!

    For any subset of the affine plane R^2, let G() denote the group of all affine transformations f of R2 such that f() = .

    The elements of the group G = G(), where is the hyperbola xy = 1
    in R^2; i think are: (ax|dy) & (by|cx) where ad=bc=1

    Show that the elements of G preserving each of the two asymptotes of form a normal subgroup N of index 2 in G, isomorphic to the multiplicative group R* of non-zero real numbers, which acts regularly on .

    Do all the elements of G then preserve the two asymptotes? Any hints how to show the rest please?!

    No idea what i'm doin really i'm afraid, but here goes nothing:

    The index of a subgroup N of G is the number of distinct left
    (or right) cosets of N in G. Even if N is not a normal subgroup, the corre-
    spondence xN -> Nx is a bijection between the set of left cosets and the
    set of right cosets. If the index of N in G is equal to 2, then N is a normal subgroup of G:

    Showing that Nx = xN for each x in G. If x in N there is nothing
    to prove. Hence, let x is not in N. Since the index is two, the disjoint union of N and xN make up all of G. Analogously the disjoint union of N & Nx = G.
    Since N does not equal Nx as well as N does not equal xN, it follows that Nx = xN.
    Multiplying this with x^-1 gives that N^x = N and hence N is a normal
    subgroup.

    ???
    Thanks, any help would be much appreciated. x
    identify G with \left \{\begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \} \bigcup \left \{\begin{pmatrix}0 & a \\ \frac{1}{a} & \ 0 \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \} and N by \left \{\begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix}: \ a \in \mathbb{R}^{\times} \right \}. looking at H=\{1,-1 \} as a multiplicative subgroup of \mathbb{R}, we define f: G \longrightarrow H

    by f(A)= \det A. see that f is a surjective group homomorphism and \ker f = N. thus N is a normal subgroup of G and since G/N \cong H and |H|=2, the index of N in G is 2.

    the isomorphism between N and \mathbb{R}^{\times} is just the obvious one: \begin{pmatrix}a & 0 \\ 0 & \frac{1}{a} \end{pmatrix} \mapsto a.
    Last edited by NonCommAlg; May 10th 2009 at 07:46 PM.
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