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In the introduction section of the paper Triples of $2\times 2$ matrices which generate free groups the authors mentioning some thing...

In my words:

The matrices $\begin{pmatrix}1 & 0 \\ 2 & 1\end{pmatrix}$ and $\begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}$ are generating the free group of two generators.

How to prove the above statement?

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  • $\begingroup$ Take a look at the references in the paper you are reading. $\endgroup$ – Moishe Kohan Aug 9 '13 at 22:48
  • $\begingroup$ Well, one of the references is in German, the other is in Russian. So the request is perhaps not unreasonable. $\endgroup$ – Chris Godsil Aug 9 '13 at 22:54
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    $\begingroup$ @ChrisGodsil: And another one (Lyndon and Ullman) is in English. $\endgroup$ – Moishe Kohan Aug 9 '13 at 22:59
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This is nothing more than an application of the Ping-Pong Lemma.

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  • $\begingroup$ John Meier gives a wonderful explanation of this in his book Groups, graphs and trees. $\endgroup$ – user1729 Aug 11 '13 at 15:15
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It can be proven using the Ping Pong lemma. You can see the proof in Pierre de la Harpe's book Topics in geometric group theory.

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Your question is a particular case of Theorem 14.2.1 in Kargapolov and Merzljakov, "Fundamentals of the Theory of Groups", but you can easily adapt their proof to your case.

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Answer is in the this paper of Lyndon and Ullman.

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  • $\begingroup$ Now comes the hardest part... $\endgroup$ – 17SI.34SA Aug 9 '13 at 23:12
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    $\begingroup$ Yes, you have to learn how to play ping-pong without a partner. Some people find it difficult, but you should try, you will find it very useful later on, if you want to do group theory. $\endgroup$ – Moishe Kohan Aug 10 '13 at 4:21

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