Schur multiplier of $Sp(2g, \mathbb{Z}/2)$ for $g \geq 3$ This question is about the computation of $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$, where $Sp(2g, \mathbb{Z}/2)$ is the group of symplectic $2g \times 2g$ matrices over $\mathbb{Z}/2$. 
With respect to this computation I have seen quotations to the following two papers:


*

*Stein "Surjective stability in dimension $0$ for $K_2$ and related functors" (https://www.dropbox.com/s/jcy7sep2hpxy7gt/Stein.pdf?dl=0) 

*Steinberg "Générateurs, relations et revêtements de groupes algébriques"
(https://www.dropbox.com/s/jidchwwjssnmmlm/Steinberg1.pdf?dl=0)


According to the quotations of Stein's and Steinberg's papers the result is that
            $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z}) = 0$ for $g \geq 3$. 
My two concerns are:
1) The papers by Stein and Steinberg do not directly give the computation of  $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$. So my first question is if it is really possible to extract this result from these papers?
2) There is a discrepancy in the case $g=3$ between the quotations of Stein's and Steinberg's papers - which say that $H_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) =0$ - and the computation of this group done with GAP (the computational system for discrete algebra). GAP gives $H_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) = \mathbb{Z}/2$.  Checking on GAP, the process that the computer goes through to get the computation looks correct. Is there an error in Stein's and/or Steinberg's paper or is it a misinterpretation of the case $g=3$ in the quotation?
 A: I've been corresponding via email with the OP about this (it is a paper of mine that she got these citations from), and she asked me to post an answer summarizing what I told her.  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).
$\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\St}{St}$ 
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 here.
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.

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 $G$.

I now describe what happens in "infinite rank".  Let $R$ be a ring.  Define
$$\GL(R) = \bigcup_{n=1}^{\infty} \GL_n(R)$$
and let $E(R)$ be the subgroup of $\GL(R)$ generated by elementary matrices.
The low-dimensional homology groups of $\GL(R)$ and $E(R)$
are closely connected to the algebraic K-theory of $R$.  In particular,
$K_1(R)$ is by definition equal to $\HH_1(\GL(R);\Z)$.
It also turns out that $K_2(R) = \HH_2(E(R);\Z)$, but this
is a theorem rather than a definition.  To understand what happens in finite rank, we'll have to understand
where this comes from.
There a group $\St(R)$ called the Steinberg group.
It is defined via generators and relations, and informally can be described as the group generated by
elementary matrices in $\GL(R)$ with the ``obvious relations'' between elementary matrices.  Its precise
definition is as follows.


*

*The generators are symbols $e_{ij}(r)$ with $i,j \geq 1$ distinct
and $r \in R$.  This corresponds to the elementary matrix obtained from the identity matrix by putting an $r$ in
position $(i,j)$.

*The relations are as follows.


*

*For distinct $i,j \geq 1$ and $r,r' \in R$, we have $e_{ij}(r)
   \cdot e_{ij}(r') = e_{ij}(r+r')$.

*For distinct $i,j,k \geq 1$ and $r,r' \in R$, we have
$[e_{ij}(r),e_{jk}(r')] = e_{ik}(r r')$.

*For distinct $i,j \geq 1$ and distinct $k,l \geq 1$ such that $j
    \neq k$ and $i \neq l$, we have $[e_{ij}(r),e_{kl}(r')] = 1$ for all
$r,r' \in R$.
Since all these relations hold in $\GL(R)$, there is a group homomorphism $\St(R) \rightarrow \GL(R)$ whose
image is $E(R)$.  By definition, $K_2(R)$ is the kernel of this homomorphism, so we have a short
exact sequence
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1.$$
The main theorem concerning the Steinberg group is as follows (see Theorem 5.10 of Milnor's book).
Theorem 2: Let $R$ be a ring.  Then the extension
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1$$
is the universal central extension of $E(R)$.  In particular, $K_2(R)$ is an abelian group
and $\HH_2(E(R);\Z) = K_2(R)$.
This allows many concrete calculations.  One very simple one is as follows.
Example: Since $\SL(\Z)$ is generated by elementary matrices, we have $E(\Z) = \SL(\Z)$.  The group
$\HH_2(\SL(\Z);\Z)$ is then cyclic of order $2$.  Identifying it with $K_2(\Z)$, the generator
is $(e_{12}(1) e_{21}(1)^{-1} e_{12}(1))^4$.  Here the matrix
$$e_{12}(1) e_{21}(1) e_{12}(1) = \left(\begin{matrix}
0 & 1 \\
-1 & 0 
\end{matrix}\right)$$
is the one that rotates the plane by $90$ degrees, and hence has order $4$.
Assume now that $R$ is a commutative ring and let $u,v \in R^{\ast}$ be units.  Define
$$D_u = \left(\begin{matrix}
u & 0 & 0 \\
0 & u^{-1} & 0 \\
0 & 0 & 1
\end{matrix}\right)$$
and
$$D_v' = \left(\begin{matrix}
v & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & v^{-1}
\end{matrix}\right).$$
It is not hard to see that diagonal matrices like these can be written as products of elementary matrices, so
$D_u, D_v' \in E(R)$.  What is more, the matrices $D_u$ and $D_v'$ commute.  Letting $\tD_u$ and $\tD_v'$ be
lifts of $D_u$ and $D_v'$ to $\St(R)$, we obtain an element
$[\tD_u,\tD_v'] \in K_2(R)$, which will be called a symbol and denoted $\{u,v\}$.  It is not hard to see
that this does not depend on the choice of lifts.
The following theorem says that for fields, the symbols generate $K_2(R)$; see Corollary 9.13 of Milnor's book.
Theorem 3: Let $\Field$ be a field.  Then $K_2(\Field)$ is generated by the set of symbols $\{u,v\}$ as $u$ and $v$
range over $\Field^{\ast}$.
For finite fields, the final piece of the puzzle is as follows; see Corollary 9.9 of Milnor's book.
Theorem 4: Let $\Field$ be a finite field.  Then $\{u,v\} = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above with the fact that $\SL(\Field)$ is generated by elementary matrices for a field $\Field$,
we deduce the following theorem.
Theorem 5: Let $\Field$ be a finite field.  Then $\HH_2(\SL(\Field);\Z) = 0$.

Of course, what we are really interested in is $\GL_n(R)$, not $\GL(R)$.  Define $E_n(R)$ and $\St_n(R)$ in the obvious
way.  There is still a surjection $\St_n(R) \rightarrow E_n(R)$; denote its kernel by $C_n(R)$, so we have a short
exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1.$$
Associated to the natural inclusion $\GL_n(R) \hookrightarrow \GL_{n+1}(R)$ used to define $\GL(R)$ are an inclusion
$E_n(R) \hookrightarrow E_{n+1}(R)$ and homomorphisms $\St_n(R) \rightarrow \St_{n+1}(R)$ and $C_n(R) \rightarrow C_{n+1}(R)$
that fit into a commutative diagram
$$\begin{CD}
1 @>>> C_n(R) @>>> \St_n(R) @>>> E_n(R) @>>> 1 \\
@. @VVV @VVV @VVV @. \\
1 @>>> C_{n+1}(R) @>>> \St_{n+1}(R) @>>> E_{n+1}(R) @>>> 1.
\end{CD}$$
It is clear that $\St_n(R)$ is the limit of
$$\St_1(R) \rightarrow \St_2(R) \rightarrow \St_3(R) \rightarrow \cdots$$
and that $K_2(R)$ is the limit of
$$C_1(R) \rightarrow C_2(R) \rightarrow C_3(R) \rightarrow \cdots.$$
In the ``ideal'' situation, we would have theorems of the following form.


*

*For $n$ sufficiently large, the exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$; and hence $C_n(R)$ is an abelian group and $\HH_2(E_n(R);\Z) = C_n(R)$.

*For $n$ sufficiently large, the map $C_n(R) \rightarrow C_{n+1}(R)$ is surjective (this is called surjective stability).

*For $n$ sufficiently large, the map $C_n(R) \rightarrow C_{n+1}(R)$ is injective (this is called injective stability).
If these conditions are satisfied, then we would have $\HH_2(E_n(R);\Z) = K_2(R)$ for $n$ sufficiently large.
Unfortunately, one can give examples where these fail.  However, they do hold for many rings; in particular, they
hold for fields.  Our goal is
to talk about finite fields, so we will not try to give particularly general statements.  We begin with the
bit about being a universal central extension.  The proof of Theorem 2 can
be followed to deduce the following.
Theorem 6: Let $R$ be a ring and let $n \geq 5$.  Assume that $C_n(R)$ is contained in the center of $\St_n(R)$.  Then the extension
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$.  In particular,
$\HH_2(E_n(R);\Z) = C_n(R)$.
As the following theorem shows, the condition in this theorem is satisfied for fields; see Theorem 9.12 of Milnor's book.
Theorem 7: Let $\Field$ be a field.  Then $C_n(\Field)$ is contained in the center of $\St_n(\Field)$ for $n \geq 3$, and hence
$\HH_2(\SL_n(\Field);\Z) = C_n(\Field)$ for $n \geq 5$
Remark: In fact, it turns out that
$$1 \longrightarrow C_n(\Field) \longrightarrow \St_n(\Field) \longrightarrow \SL_n(\Field) \longrightarrow 1$$
is the universal central extension of $\SL_n(\Field)$ for $n \geq 3$ except for the following exceptions:


*

*$n=3$ and $\Field = \Field_2$, and

*$n=3$ and $\Field = \Field_4$, and

*$n=4$ and $\Field = \Field_2$.
This was proved by Steinberg.
We now turn to injective and surjective stability.  The key is the notion of symbol.  If $R$ is a commutative ring
and $u,v \in R^{\ast}$ are units, then we can define a symbol $\{u,v\}_n \in C_n(R)$ for any $n \geq 3$ in the obvious
way.  We then have the following; see Theorem 9.11 of Milnor's book.
Theorem 8: Let $R$ be a commutative ring and let $n \geq 3$.  Assume that $C_n(R)$ is contained in the center of $\St_n(R)$.
Then $C_n(R)$ is generated by the set of symbols $\{u,v\}_n$ as $u$ and $v$ range over the units of $R$.
Remark: Theorem 7 and Theorem 8 combine to show that if $\Field$
is a field, then $K_2(\Field)$ is generated by symbols $\{u,v\}$ for $u,v \in \Field^{\ast}$.  This is
precisely Theorem 3 above, and in fact this is how Theorem
3 is proved.
The proof of Theorem 4 above also works for the symbols $\{u,v\}_n$ and
gives the following.
Theorem 9: Let $\Field$ be a finite field and $n \geq 3$.  Then $\{u,v\}_n = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above, we deduce the following.
Theorem 10: Let $\Field$ be a finite field.  Then $\HH_2(\SL_n(\Field);\Z) = 0$ for $n \geq 5$.
Remark: In fact, as in the remark after Theorem 7, one can show that
$\HH_2(\SL_n(\Field);\Z) = 0$ for $\Field$ a finite field and $n \geq 3$ except for the following exceptions:


*

*$n=3$ and $\Field = \Field_2$, and

*$n=3$ and $\Field = \Field_4$, and

*$n=4$ and $\Field = \Field_2$.
