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Apr 11, 2010 at 16:15 comment added Ian Agol I don't know a good reference for strong approximation, but in this context there's a very elementary argument one may make - see: ams.org/mathscinet-getitem?mr=1459136 Their argument shows that the group maps onto all but finitely many $SL_2(Z/p)$. This then implies that it maps onto all but finitely many $SL_2(Z/p^k)$, and therefore onto a finite-index subgroup of the pro-congruence completion (I don't know if this is standard terminology).
Apr 11, 2010 at 8:16 comment added Robin Chapman Thanks Ian, that's very interesting; it's a shame I can only tick one response.
Apr 11, 2010 at 0:54 history answered Ian Agol CC BY-SA 2.5