I've been corresponding via email with the OP about this, and he asked me to post an answer summarizing what I told him. I apologize for the length of this answer -- this is really quite a long story. I also apologize for sometimes butchering people's names. I am copying this answer from a long document I sent the OP, and unfortunately autocorrect screws things up without telling me (eg it wants to call Steinberg "Sternberg", though I hope caught most of those occurrences). $\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}}$ For $\Sp_{2g}(\Z/2)$, the correct theorem is that $\HH_2(\Sp_{2g}(\Z/2)) = 0$ for $g \geq 4$. This should be attributed to Steinberg and is contained in the paper cited by the OP. More generally, Steinberg showed that a similar theorem holds for $\Sp_{2g}(\Field_q)$. What Stein did in the cited paper was show how to extend what Steinberg did to $\Sp_{2g}(\Z/k)$ where $k$ is not prime. They both in fact dealt not just with the symplectic group, but also with more general finite Chevalley groups. Another good reference for Steinberg's work is sections 6 and 7 of Steinberg's Yale lecture notes, which were never published but which are available <a href="http://www.ms.unimelb.edu.au/\~{}ram/Resources/YaleNotes.pdf">here</a>. It is not easy to extract the above homological statement from Steinberg and Stein's papers since they are written in the language of algebraic k-theory. What they are imitating is the calculation of $\HH_2(\SL_n(\Field_q))$ that is described in Milnor's book on algebraic k-theory, which I highly recommend reading. To help you understand these papers, below I have written a guide to the calculation of $\HH_2(\SL_n(\Field_q))$ from Milnor's book. (nb: since this is long, I'm going to post it in stages so that my work isn't lost if things crash) ------------- I will begin by recalling the theory of universal central extensions. Let $G$ be a group. A **central extension** of $G$ is a group $\tG$ together with a short exact sequence $$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$ such that $C$ is contained in the center of $\tG$. This central extension is a **universal central extension** if for any other central extension $$1 \longrightarrow C' \longrightarrow \tG' \longrightarrow G \longrightarrow 1,$$ there exists a unique homomorphism $\tG \rightarrow \tG'$ such that the diagram $\require{AMScd}$ $$\begin{CD} 1 @>>> C @>>> \tG @>>> G @>>> 1 \\ @. @VVV @VVV @VV{=}V @. \\ 1 @>>> C' @>>> \tG' @>>> G @>>> 1 \end{CD}$$ commutes. The usual argument shows that universal central extensions are unique if they exist, but they might not exist. The following theorem summarizes their properties. A proof of it can be found in Theorem 5.7 and Corollary 5.8 of Milnor's book **Theorem 1:** Let $G$ be a group. Then $G$ has a universal central extension $$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$ if and only if $\HH_1(G;\Z) = 0$, in which case we have $C \cong \HH_2(G;\Z)$. For perfect groups, this reduces the computation of $\HH_2(G;\Z)$ to the construction of the universal central extension of $G$. -----------