This paper deals with the following general situation: we are given an algebraic group G defined over a number field K, and a subgroup F of the group G(K) of K-rational points of G. Then what should it mean for F to be a 'large' subgroup? We might require F to be a lattice in G, to be arithmetic, to contain many elements of a specific kind, to have a large closure in some natural topology, et cetera. There are many theorems proving implications between conditions of'size' of this kind. We shall consider the case of G a K-simple group, usually Q-simple. The Zariski topology of G, for which the closed sets are those defined by the vanishing of polynomial functions, is very coarse. On the other hand, the various valuations v of K each give rise to a 'strong' topology on G(K); if v is non-archimedean, and Kv is the completion of K with respect to v, then G(KV) is a non-archimedean Lie group, and if £>„ is the ring of integers of Kv, the open subgroup G(£5 J of G(KV) of integral points is defined for all but finitely many v. The Zariski closure F of a subgroup F of G(K) is a /C-algebraic subgroup of G, while the strong closure Fy is a Lie subgroup of G(KV). For elementary reasons the dimension of Tv is at most the dimension of F. Our results are in the opposite direction: we show, under suitable conditions on G, that if F = G then for almost all v we have that Tv contains G(OV). To illustrate this with a specific example, if Ml 5. . . , Mk are matrices in SL2(Q) generating a group F which is Zariskidense in SL2, then for all sufficiently large prime numbers p we have F p = SL2(Zp). Further, this result is effective, in the sense that we could, in principle, check the hypothesis for given Mx,..., Mk, and derive a bound on p beyond which the conclusion holds. (It is perhaps worth remarking that this special case, which is the simplest application of our results, may be proved with much less effort than the general theorem.)
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