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Math Help - Subgroup Test

  1. #1
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    Subgroup Test



    I want to use the "one step subgroup test".

    The identity is

    e=I= \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}

    And

    det(I)=1-0=1=(\sqrt{2})^0

    and 0 \in \mathbb{Z}

    Therefore H is not empty since I \in H.

    Now I have to show that for any two elements x_1,x_2 \in H, x_1x_2^{-1} is in H.

    Let \begin{bmatrix}a & b \\c & d\end{bmatrix}, \begin{bmatrix}a' & b' \\c' & d'\end{bmatrix} \in H

    x_2^{-1}= \frac{1}{(\sqrt{2})^k} \begin{bmatrix}d' & -b' \\-c' & a'\end{bmatrix}

    So we have:

    x_1x_2^{-1}= \begin{bmatrix}a & b \\c & d\end{bmatrix} \frac{1}{(\sqrt{2})^k} \begin{bmatrix}d' & -b' \\-c' & a'\end{bmatrix}

    =\frac{1}{(\sqrt{2})^k} \begin{bmatrix}-d'a+-c'b & -b'a'+b'a' \\cd'+-c'd & -b'c+a'd\end{bmatrix}

    And det (x_1x_2^{-1})=\frac{1}{(\sqrt{2})^k}(-d'a+-c'b)(b'c+a'd)-(-b'a'+b'a')(cd'+-c'd)

    If my working is correct so far, could anyone please show me how to manipulate this determinant to show that it satisfies the given condition, and x_1x_2^{-1} \in H.
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  2. #2
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    You don't need any of that.

    det(AB) = det(A)det(B) and det(A^{-1}) = \frac{1}{det(A)} are all you need.
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  3. #3
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    Thank you!

    So we have: det(x_1x_2^{-1})= \frac{(ad-bc)}{(a'd'-c'b')}=\frac{(\sqrt{2})^{k_1}}{(\sqrt{2})^{k_2}}

    where k_1,k_2 \in \mathbb{Z}

    So how could I now simplify this to show that the whole thing can be written as (\sqrt{2})^k??
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  4. #4
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    det(x_1x_2^{-1}) = \frac{(\sqrt{2})^{k_1}}{(\sqrt{2})^{k_2}} = (\sqrt{2})^{k_1-k_2} and k_1 - k_2 \in \mathbb{Z}.
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