Let us denote by the set of -matrices whose determinant is , and by the quotient . Our aim here is to prove that is a free product:
As noticed in the previous note Some SQ-universal groups, as a corollary we get that is SQ-universal, that is every countable group is embeddable into a quotient of .
Usually, the theorem above is proved thanks to a natural action of on the hyperbolic plane by Möbius transformations. However, as noticed by Roger Alperin in his article published in The American Mathematical Monthly, it is sufficient to make act on the set of irrational numbers via:
Notice that for all precisely because is irrational. The argument below may be thought of as an application of ping-pong lemma (that we already met in the note Free groups acting on the circle in order to find free subgroups).
First, it is a classical exercice to prove that is generated by the two following matrices:
For convenience, let . Then and also generate , and consequently their images in generate it; furthermore, (resp. ) has order two (resp. order three) in . Notice that
, and .
Now, let and denote the set of positive and negative irrationals respectively. Clearly,
To conclude, we want to prove that, for any alternating word from and , in .
Case 1: has odd length. Then either begins and ends with a , hence , or begins and ends with a , hence . In particular, we deduce that in .
Case 2: has even length. Without loss of generality, we may suppose that begins with a ; otherwise, just conjugate by . Then either begins with a and ends with a , hence
or begins with a and ends with a , hence
In either case, we deduce that , so in . The proof is complete.