# 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?

"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:

• The related post about the Schur multiplier of ${\rm AGL}(n,q)$ is relevant because ${\rm AGL}(1,q)$ is a minimal affine $2$-transitive group, and it is shown in the answer to that post that it has trivial Schur multiplier except for ${\rm AGL}(1,4)$. Do you know whether there is a classification of minimal finite affine $2$-transitive groups? – Derek Holt Jun 4 '18 at 7:52
• But there are other sharply $2$-transitive affine groups with nontrivial multiplier, such as $5^2:{\rm SL}(2,3)$ with multiplier of order $5$, and $11^2:{\rm SL}(2,5)$ with multiplier oforder $11$. – Derek Holt Jun 4 '18 at 9:11
• Ignoring sporadic cases, Kantor's statement of the classification (linked in my question) suggests that there are four families of minimal finite affine 2-transitive groups. Three have the form $N\rtimes H$, where $N$ is an elementary abelian group of order $v$, and $H$ is either $SL(n,q)$ with $q^n=v$, $Sp(n,q)$ with $q^n=v$, or $G_2(q)'$ with $q^6=v$ and $q$ even. The remaining family consists of the minimal 2-transitive subgroups of $A\Gamma L(1,v)$, e.g., the group you pointed out: $AGL(1,v)$. I don't know if it is easy to further classify the minimal 2-transitive groups of this form. – Dustin G. Mixon Jun 4 '18 at 16:50

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 $G$.