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Jul 30, 2012 at 18:30 history edited Wilberd van der Kallen CC BY-SA 3.0
improved phrasing
Jul 27, 2012 at 17:14 vote accept Igor Belegradek
Jul 27, 2012 at 17:14 vote accept Igor Belegradek
Jul 27, 2012 at 17:14
Jul 27, 2012 at 17:13 history edited Igor Belegradek CC BY-SA 3.0
details added
Jul 27, 2012 at 16:43 comment added user6976 @Igor: $e_{i,j}(m)=e_{i,j}(1)^m$.
Jul 27, 2012 at 15:22 comment added Igor Belegradek Oops. So how does one prove the claim in the first sentence of your answer?
Jul 27, 2012 at 15:19 comment added Wilberd van der Kallen That will not work. They do not lie in $GL_n(\mathbb Z)$.
Jul 27, 2012 at 14:57 comment added Igor Belegradek Thank you! In fact, it is enough to consider the diagonal matrices $E_i(m)$ which have $m$ at $(i,i)$ entry and $1$ on all other diagonal entries. I am still in the process of proving that the group generated by $E_1(m), \dots, E_n(m)$ sits in any given finite index subgroup for large $m$ but it does sound believable.
Jul 27, 2012 at 14:19 history answered Wilberd van der Kallen CC BY-SA 3.0