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

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


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


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


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

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







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