Induced group homomorphism $\text{SL}_n(\mathbb{Z}) \twoheadrightarrow \text{SL}_n(\mathbb{Z}/m\mathbb{Z})$ surjective? Let $n, m &gt; 1$. The map $\mathbb{Z} \twoheadrightarrow \mathbb{Z}/m\mathbb{Z}$, of reduction mod $m$, induces a group homomorphism $F: \text{SL}_n(\mathbb{Z}) \to \text{SL}_n(\mathbb{Z}/m\mathbb{Z})$. My question is, is $F$ surjective or not?
This turned out to be a non-example.  I'm leaving it up for posterity as a CW answer.



As it turns out, the answer is no.  The smallest counterexample I can think of is as follows: we note that
$$
\pmatrix{2&amp;1&amp;1\\1&amp;2&amp;1\\1&amp;1&amp;2} \in SL_n(\Bbb Z/3\Bbb Z)
$$
However, this matrix is not the reduction mod $3$ of any element of $SL_n(\Bbb Z)$.
This turned out to be a non-example.  I'm leaving it up for posterity as a CW answer.



As it turns out, the answer is no.  The smallest counterexample I can think of is as follows: we note that
$$
\pmatrix{2&amp;1&amp;1\\1&amp;2&amp;1\\1&amp;1&amp;2} \in SL_n(\Bbb Z/3\Bbb Z)
$$
However, this matrix is not the reduction mod $3$ of any element of $SL_n(\Bbb Z)$.