1. ## Special Linear Group

Hy. Im looking for a proof that SL_n(IZ)->SL_n(IZ/mIZ) is onto. Can you help me?

2. Originally Posted by housi
Hy. Im looking for a proof that $\textsf{SL}_n(\mathbb{Z})\to \textsf{SL}_n(\mathbb{Z}/m\mathbb{Z})$ is onto. Can you help me?
Have I interpreted the question correctly? If so, then I think that the result is false.

For $x\in\mathbb{Z}$, denote by $\bar{x}$ its image in the quotient ring $\mathbb{Z}_m = \mathbb{Z}/m\mathbb{Z}.$ I assume that $\textsf{SL}_n(\mathbb{Z}_m)$ denotes the set of all matrices with entries in $\mathbb{Z}_m$ with determinant $\bar{1},$ and that the map from $\textsf{SL}_n(\mathbb{Z})$ to $\textsf{SL}_n(\mathbb{Z}_m)$ is the natural quotient map given by $(a_{ij})\mapsto(\bar{a}_{ij}).$

Now take m = 7 and let $A = \begin{bmatrix}\bar{2}&\bar{0}&\bar{0}\\ \bar{0}&\bar{2}&\bar{0}\\ \bar{0}&\bar{0}&\bar{2}\end{bmatrix}\in\textsf{SL} _3(\mathbb{Z}_7).$ Then $\det(A)=\bar{1},$ but any lifting of A to a matrix in $\textsf{GL}_3(\mathbb{Z})$ will have determinant of the form 8+28k (mod 49), which can never be equal to 1.

Edit. Stupid mistake: 8 + 28k can be equal to 1 (mod 49). In fact, 8 + 12&#215;28 = 7^3 + 1; and in fact $\begin{bmatrix}2&0&7\\-7&2&0\\0&7&86\end{bmatrix}$ is a lifting of that matrix A to an element of $\textsf{SL}_3(\mathbb{Z}).$ So it looks to me as though the result may be true after all. But it's not at all obvious!

3. Originally Posted by Opalg
Have I interpreted the question correctly? If so, then I think that the result is false.
Yes you have interpretet the question correctly! More precisely i think i need this result to prove $\widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\mathrm{SL }_n(\widehat{\mathbb{Z}})$ (pro-finite completion) in the case $n\geq3$ (this is actually what i want to prove). With the solution of the congruence subgroup problem one can get $\widehat{\mathrm{SL}_n(\mathbb{Z})}\cong\lim_{m>0} \mathrm{SL}_n(\mathbb{Z})/\Gamma(m)$ (projective limes) where $\Gamma(m)$ denotes the kernel of $\mathrm{SL}_n(\mathbb{Z})\to\mathrm{SL}_n(\mathbb{ Z}/ m \mathbb{Z})$. Because it shouldn't be difficult to prove $\mathrm{SL}_n(\widehat{\mathbb{Z}})\cong\lim_{m>0} \mathrm{SL}_n(\mathbb{Z}/ m\mathbb{Z})$ it remains to prove $\mathrm{SL}_n(\mathbb{Z}/ m\mathbb{Z})\cong\mathrm{SL}_n(\mathbb{Z})/\Gamma(m)$. Thus i need to show that this natural projection is onto. Maybe one can solve this using matrices of the type $\mathrm{id}_n+E_{ij}$ where $E_{ij}$ has 1 at the place $(i,j), i\neq j$ and 0 else. But have no idea how to show...