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Math Help - Understanding a proof by R.C Lyndon and J.L Ullman.

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    Understanding a proof by R.C Lyndon and J.L Ullman.

    Here, http://projecteuclid.org/DPubS/Repos...mmj/1028999969 in this article I have difficulties understanding the theorem on page 162.




    Theorem. Let A, B and C=AB be an elements of group GL_2(\mathbb{Z}), all with real fixed points. Suppose that the fixed points of each of these matrices under action on the extended real axis {\mathbb{R}}^{*}, lie in an interval of {\mathbb{R}}^{*} containing no fixed points of order two.
    Then A and B generate a free group.


    Proof. First, we change the notation by replacing B by B^{-1}. The hypothesis now asserts that C=AB^{-1} has real fixed points, that is, points p such that pA=pB. If A has distinct fixed points, one is a source a^- and one is a sink a^+; if A is parabolic, we take a^-=a^+ to be its sole fixed point. We employ similar notation b^- and  b^+ for fixed points of B. The points a^-, \ a^+, \ b^- and b^+ divide {\mathbb{R}}^{*} into four open intervals(some possibly empty). The hypothesis requires that both fixed points(possibly coincident) of C lie in same one of these intervals, and not in one bounded by two fixed points of A or by two fixed points of B.


    If p\in (a^+,b^+), then pA\in (a^+,p ) and pB\in (p,b^+). Since (a^+,p )\cap (p,b^+)=\emptyset, it is impossible that pA=pB. Thus, no fixed point of C lies in an interval (a^+,b^+), or, similarly in an interval (b^+,a^+).
    (The proof continuous...)


    I don't understand the second paragraph in the proof.

    If p\in (a^+,b^+), then pA\in (a^+,p ) and pB\in (p,b^+).







    Thank you for any help!
    Last edited by Also sprach Zarathustra; August 28th 2013 at 07:48 AM.
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