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:

**Theorem:** .

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:

and .

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,

and .

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.

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