Free Groups in Central Simple Algebras without Tits' Theorem

Abstract

Let R be a noncommutative central simple algebra, the center k of which is not absolutely algebraic, and consider units a b of R such that a a b freely generate a free group. It is shown that such b can be chosen from suitable Zariski dense open subsets of R, while the a can be chosen from a set of cardinality k (which need not be open).

Constructing free groups in a normal subgroup of the multiplicative group of division rings

“…However, if k has "sufficiently big" transcendence degree over the prime field, then this can be done, see [2,Theorem 7.3]. In all other cases we have only existential proofs, either invoking Tits' Alternative, see for example [3], or some density argument in the Zariski topology, see [8] Brought to you by | New York University Bobst Library Technical Services Authenticated Download Date | 6/8/15 12:34 PM We also briefly consider the existence of free symmetric pairs in D, whenever it is possible to extend the involution x D x 1 for all x 2 G, to a k-involution of D. Let us recall that an element x of D is symmetric if x D x, and a pair .u; v/ of elements of D is free if it generates a free noncyclic subgroup. Observe that when G is abelian-by-finite and D D k.G/, that is, D is generated by G over k, then D is finite-dimensional over its center.…”