The author calls a group G an \({\mathfrak X}\)-group of for every positive integer d there exists a subgroup H such that the index G:H is finite and divisible by d. Extending recent work of A. Lubotzky (which ultimately depends on the classification of the finite simple groups) the author proves the following Theorem: Let R be a commutative ring, M a finitely- generated R-module and G a subgroup of \(Aut_ RM\). Under either of the following conditions G is an \({\mathfrak X}\)-group: (a) G is infinite and finitely generated. (b) R is finitely generated as a ring and G is not unipotent-by-finite. The main case is that of \(G\leq GL_ n(R)\), where R is a finitely generated domain, and the connected component of G is not unipotent. By enlargening R we may assume that G contains a triangular element g with eigenvalue \(\gamma\) of infinite order. Then by the author's previous result [Proc. Lond. Math. Soc., III. Ser. 36, 448-479 (1978; Zbl 0374.20040)] the maximal ideal topology on R induces the profinite topology on the group U of units of R. So there exists a semisimple ideal \({\mathfrak a}\) of finite index such that \(\gamma^{{\mathbb{Z}}}\cap (1+{\mathfrak a})\leq \gamma^{d{\mathbb{Z}}}\). Then \(H=G\cap (1+{\mathfrak a}^{n\times n})\).