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Math Help - Group theory

  1. #1
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    Group theory

    Having some trouble figuring this out - any help at all would be greatly appreciated! Thanks

    S the group of bijective applications R - > R. x, y elements of S. For every real number t, x(t) = t +1, y(t) = 2t.
    G is a sub-group of S generated by {x,y}.

    1. For every integer n>=1, x
    n = y^(-n)xy^(n).
    Calculate x
    n(t) and
    show that (x
    n)^2 = xn-1

    2. Let Hn be a subgroup of G generated by xn Show that Hn is a subgroup of Hn+1 and that Hn is not equal to Hn+1
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  2. #2
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    Quote Originally Posted by rmangan View Post

    S the group of bijective applications R - > R. x, y elements of S. For every real number t, x(t) = t +1, y(t) = 2t.
    G is a sub-group of S generated by {x,y}.

    1. For every integer n>=1, x
    n = y^(-n)xy^(n).
    Calculate x
    n(t) and
    show that (x
    n)^2 = xn-1

    2. Let Hn be a subgroup of G generated by xn Show that Hn is a subgroup of Hn+1 and that Hn is not equal to Hn+1
    The elements x,y of all bijections on \mathbb{R} are functions x,y:\mathbb{R}\to \mathbb{R} defined by x(t) = t+1 and y(t) = 2t. Now x^n(t) = x\circ x \circ ... \circ x (t) = (((t+1)+1)+1) + ... + 1 = t+n. Also, y^n (t) = y\circ ... \circ y(t) = 2(...(2(2t))) = 2^n t.
    Thus, x_n(t) = y^{-n} x y^{n} (t) = 2^{-n}(2^{n} t + 1) = t + 2^{-n}.
    We see that x_{n}^2 (t) = (t+2^{-n})+2^{-n} = t + 2\cdot 2^{-n} = t + 2^{1-n} and so x_n = x_{n-1}^2.

    Wii see that \left< x_n \right> \subseteq \left< x_{n+1} \right> because x_n = x^2_{n+1}.
    Thus, H_n \subseteq H_{n+1} and can you show now H_n\not = H_{n+1}?
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