Abstract. It is well known that for the ring H(Z) of integral quaternions the unit group U(H(Z)) is finite. On the other hand, for the rational quaternion algebra H(Q), its unit group is infinite and even contains a nontrivial free subgroup. In this note (see Theorem 1.5 and Corollary 2.6) we find all intermediate rings Z ⊂ A ⊆ Q such that the group of units U(H(A)) of quaternions over A contains a nontrivial free subgroup. In each case we indicate such a subgroup explicitly. We do our best to keep the arguments as simple as possible.
Motivation and main result.In this paper N ⊂ Z ⊂ Q ⊂ R have standard meaning as subsets of the field C of complex numbers, U(R) denotes the group of units of any associative ring R with 1 = 0, F stands for a free nonabelian group with two generators, and SO n (R) ⊂ GL n (R) for the well known linear groups over a given ring R. We also apply some other standard notation and terminology (see for example [10,13,17]).In [4,8,9,11], and some other papers, for various rings R explicit copies of F ⊂ U(R) were found. We consider the same problem for orders, but not only Z-orders, in finite-dimensional, semisimple Q-algebras. An old and simple, but very useful and effective result in this direction, due to Sanov and Brenner (see [17]), is:As a consequence of this result we have Theorem 1.2. Let R be any order in a finite-dimensional semisimple Q-algebra. If R has a nonzero nilpotent element then there exists an effective way to construct a copy of F ⊆ U(R).