Schur covers of affine 2-transitive groups I am interested in Schur covers of minimal 2-transitive groups. A theorem of Burnside gives that every finite 2-transitive group is either almost simple or affine. In the time since, these groups have been classified as a consequence of the classification of finite simple groups (see Section 2 of this paper, for example). Since Schur covers were already determined for the simple groups in order to perform their classification, the Schur covers of almost simple groups are relatively easy to obtain. By contrast, I am having difficulty finding the Schur covers in the affine case.
Question: Where can I find descriptions of the Schur covers of minimal affine 2-transitive groups?
A good answer looks like:

"The Schur cover of PSL(n,q) is SL(n,q), with a few exceptions. See
  Karpilovsky, The Schur Multiplier, p.246."

but with a minimal affine 2-transitive group in place of PSL(n,q).
The following appear related, but are not what I'm asking for:


*

*"What is the Schur multiplier of the affine linear group AGL(n,q)?" computes the Schur multiplier of AGL(n,q). However, AGL(n,q) is not a minimal 2-transitive group, and I want a Schur cover, not just the multiplier.

*"Have finite doubly transitive groups been classified?" discusses the classification of affine 2-transitive groups at length. It does not mention Schur multipliers or covering groups.
 A: Here is an approximate answer. I believe that it is substantially correct, but there might be some small exceptions which I have not thought of.
I think that all the examples that are subgroups of ${\rm A \Gamma L}(1,q)$ have trivial multiplier, with a similar argument to that for ${\rm A G L}(1,q)$.
The groups of the form $G=N \rtimes H$ with $H = {\rm SL}(n,q)$ and $n \ge 3$ have trivial Schur Multiplier, possibly with a few exceptions.
One exception is when $H = {\rm SL}(3,4)$, in which $G$ has nontrivial multiplier $C_4^2$, and the covering group of $G$ is $N \rtimes \hat{H}$, where $\hat{H}$ is the unique covering group of ${\rm SL}(3,4)$.
For $G=N \rtimes H$ with $H = {\rm Sp}(n,q)$ with $n \ge 2$ even and and $q$ odd,the multiplier is elementary abelian of order $q$, and the covering group has structure $\hat{N} \rtimes H$, where $\hat{N}$ is a special group of exponent $p$ (where $q$ is a power of the prime $p$) and order $q^{n+1}$. The commutator map in $\hat{N}$ is the symplectic form on $N$ that is preserved by $H$.
In a few small cases, probably just with $n=24$, these groups are not minimally $2$-transitive. For example $N \rtimes {\rm SL}(2,3) < N \rtimes {\rm Sp}(2,5)$, $N \rtimes {\rm GL}(2,3) < N \rtimes {\rm Sp}(2,7)$, and $N \rtimes {\rm SL}(2,5) < N \rtimes {\rm Sp}(2,11)$, but in those cases the covering group is still obtained by replacing $N$ by $\hat{N}$.
For the examples of form $G=N \rtimes H$ with $H = {\rm Sp}(n,q)$ and $q$ even, and also their subgroups with $H = G_2(q)'$ when $n=6$, what seems to happen is that the covering group again has the form $\hat{N} \rtimes H$  with $|N| = q^{n+1}$, but now $\hat{N}$ is elementary abelian, and is a nonsplit module extension of the trivial module with the natural module for $H$ over ${\mathbb F}_q$. Such module extensions correspond to $H^1(H,N)$, which appears to be $1$-dimensional in these cases. (The first cohomology groups of classical groups on their natural modules are all known - I can check this later.) 
Again there may be some small exceptions. For example ${\rm Sp}(4,2)$ and ${\rm Sp}(6,2)$ have Schur multipliers of order $2$, so $H$ has to be replaced by its covering group $\hat{H}$ in those cases in the covering group of $G$.
