# $Sp_{2g}(\Bbb Z) \rightarrow Sp_{2g}(\Bbb Z/m)$ is surjective?

How to show that the mod $m$ map $Sp_{2g}(\Bbb Z) \rightarrow Sp_{2g}(\Bbb Z/m)$ is surjective without using some deep structure theorem (like strong aprroximation)? Where $Sp_{2g}$ means the symplectic group.

Motivation: for $SL_n$ one can prove the surjectivity by noticing that $SL_n(A)$ is generated by elementary matrices for every local ring $A$. This answer gives a proof using elementary generators in the case $m$ is a prime. However, he didn't give any reference for such calculations.

• The link seems to be broken. – Stephen Apr 24 '18 at 12:37

• first use that $Sp_{2g}(\mathbb{Z})$ acts transitively on the decompositions of $\mathbb{Z}^{2g}$ as an orthogonal sum of $\mathbb{Z}^2$ endowed with the form $(x_1,y_1),(x_2,y_2)\mapsto x_1y_2-x_2y_1$.
• and then use that $Sp_2(\mathbb{Z})$ is surjective onto $Sp_2(\mathbb{Z}/m)$ (remark that $Sp_2=SL_2$).