Here, http://projecteuclid.org/DPubS/Repos...mmj/1028999969 in this article I have difficulties understanding the theorem on page 162.
Theorem. Let , and be an elements of group , all with real fixed points. Suppose that the fixed points of each of these matrices under action on the extended real axis , lie in an interval of containing no fixed points of order two.
Then and generate a free group.
Proof. First, we change the notation by replacing by . The hypothesis now asserts that has real fixed points, that is, points such that . If has distinct fixed points, one is a source and one is a sink ; if is parabolic, we take to be its sole fixed point. We employ similar notation and for fixed points of . The points and divide into four open intervals(some possibly empty). The hypothesis requires that both fixed points(possibly coincident) of lie in same one of these intervals, and not in one bounded by two fixed points of or by two fixed points of .
If , then and . Since , it is impossible that . Thus, no fixed point of lies in an interval , or, similarly in an interval .
(The proof continuous...)
I don't understand the second paragraph in the proof.
If , then and .
Thank you for any help!