3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @ChrisGodsil: And another one (Lyndon and Ullman) is in English. $\endgroup$ – Moishe Kohan Aug 9 '13 at 22:59
4
$\begingroup$

This is nothing more than an application of the Ping-Pong Lemma.

$\endgroup$
  • $\begingroup$ John Meier gives a wonderful explanation of this in his book Groups, graphs and trees. $\endgroup$ – user1729 Aug 11 '13 at 15:15
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Answer is in the this paper of Lyndon and Ullman.

$\endgroup$
  • $\begingroup$ Now comes the hardest part... $\endgroup$ – 17SI.34SA Aug 9 '13 at 23:12
  • 1
    $\begingroup$ 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

protected by user26857 Aug 25 '15 at 20:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.