Is there a formula for the size of Symplectic group defined over a ring $Z/p^k Z$? Is there a formula for the size of Symplectic group defined over a finite ring $Z/p^k Z$?
 A: From L. C. Grove's Classical Groups and Geometric Algebra, page 27:
$$|\operatorname{SP}(2n,\mathbb{F}_q)|=q^{n^2}\prod_{i=1}^n(q^{2i}-1),$$
for some $q=p^k$. This works in the case of $\mathbb{F}_p\cong\mathbb{Z}/p\mathbb{Z}$, but I don't think it's helpful for the general case $\mathbb{Z}/q\mathbb{Z}$.
A: Hi.
EDIT: As Joe Silverman pointed out, this approach doesn't work as simple as I imagined it. Sorry for that. I leave the attempted proof here in case someone has an idea how to fix it.
Yes, there is such a formula. It works for in a similar fashion all of the classical algebraic group. Consider the projection $\mathbb{Z}/p^k\to\mathbb{Z}/p$. It induces a surjection $Sp_{2n}(\mathbb{Z}/p^k)\to Sp_{2n}(\mathbb{Z}/p)$ because every transvection has a preimage and the transvections generate $Sp_{2n}(F)$ for every field $F$. The kernel of the homomorphism consists obviously of those matrices $M$ with $M-I \in (p\mathbb{Z}/p^k\mathbb{Z})^{n\times n}$. There are $p^{(k-1)n^2}$ such matrices.
Therefore $|Sp_{2n}(\mathbb{Z}/p^k)| = p^{(k-1)n^2} |Sp_{2n}(\mathbb{Z}/p)|$. The order of the symplectic group over $\mathbb{Z}/p$ is known.
A: From Christopher Perez's answer, we have $|Sp_{2n}(\mathbb{Z}/p\mathbb{Z})| = p^{n^2} \prod_{i=1}^n (p^{2i}-1)$.  Following Johannes Hahn, we wish to determine the size of the kernel of the homomorphism  $Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p\mathbb{Z})$.
To compute this, we may induct on $k$: For $k \geq 1$, elements in the kernel of the homomorphism $Sp_{2n}(\mathbb{Z}/p^{k+1}\mathbb{Z}) \to Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})$ have the form $I + p^k A$ for a matrix $A = \left( \begin{smallmatrix} E & F \\ G & H \end{smallmatrix} \right)$ with values in $\mathbb{Z}/p\mathbb{Z}$.  The symplectic condition (given in Wikipedia) is equivalent to the conditions that $F^t = F$, $G^t = G$, and $E^t + H = 0$.  $F$ and $G$ are therefore symmetric matrices, while $H$ and $E$ determine each other, with no further conditions.  The kernel of one-step reduction therefore has size $p^{n(n+1)/2} \cdot p^{n(n+1)/2} \cdot p^{n^2}$, or $p^{2n^2 + n}$.
The final answer is therefore: $|Sp_{2n}(\mathbb{Z}/p^k\mathbb{Z})| = p^{(2k-1)n^2 + (k-1)n} \prod_{i=1}^n (p^{2i}-1)$
Edit: To address kassabov's complaint, I'll explain the calculation in a bit more detail.  The symplectic condition on $I + p^k A$ is that $(I + p^k A)^t \Omega (I + p^k A) \equiv \Omega \pmod {p^{k+1}}$, where $\Omega = \left( \begin{smallmatrix} 0 & I_n \\ -I_n & 0 \end{smallmatrix} \right)$.  Because $k \geq 1$, we may eliminate the $p^{2k} A^t \Omega A$ term when expanding, to get
$$ \Omega + p^k A^t \Omega + p^k \Omega A \equiv \Omega \pmod {p^{k+1}}$$
By subtracting $\Omega$ from both sides, we get the conditions I mentioned on the blocks in $A$.  As you mentioned, this coincides with the symplectic Lie algebra condition.  George McNinch gave an elegant explanation in the comments, but a possibly more pedestrian reason is that $(p^k)$ is a square zero ideal in $\mathbb{Z}/p^{k+1}\mathbb{Z}$, so one has a canonical isomorphism between the kernel of reduction and the Lie algebra tensored with the quotient ring.
