TITLE: Is there a formula for the size of Symplectic group defined over a ring $Z/p^k Z$?
QUESTION [6 upvotes]: Is there a formula for the size of Symplectic group defined over a finite ring $Z/p^k Z$?

REPLY [15 votes]: 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.