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

**Also sprach Zarathustra** · August 28th 2013, 08:41 AM

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!

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