Let $n = q(q^2-1)$ (the order of $SL(2,\mathbb{F}\_q)$ I think). How many groups of order $2n$ contain $SL(2,\mathbb{F}\_q)$ as a (necessarily normal) subgroup? Is this number known exactly? Seems like it would be something well known, but maybe I slept through Group Theory class that day?



Extensions of $SL(2,\mathbb{F}_q)$
The general classification of extensions $1 \to N \to G \to H \to 1$ with $N$ and $H$ fixed (which here are $N = \text{SL}\_2(\mathbb{F}\_q)$ and $H = \mathbb{Z}\_2$) is that they correspond to equivalence classes of homomorphisms $H \to \mathbf{Aut}(N)$, where $\mathbf{Aut}(N)$ denotes the [automorphism 2-group](http://ncatlab.org/nlab/show/automorphism+2-group) of $N$. 


This is the 2-group whose objects are automorphisms of $N$ and whose morphisms are natural transformations between these; explicitly, these are given by pointwise conjugation with elements of $N$. Its $\pi\_0$ is the outer automorphism group $\text{Out}(N)$ and its $\pi\_1$ is the center $Z(N)$. In particular: 


* If $Z(N) = 1$ then $\textbf{Aut}(N) \cong \text{Out}(N)$ and we get that extensions correspond to equivalence (which here means conjugacy) classes of homomorphisms $H \to \text{Out}(N)$.
* If $\text{Out}(N) = 1$ then $\textbf{Aut}(N) \cong B Z(N)$ and we get that extensions correspond to equivalence classes of homomorphisms $H \to BZ(N)$, which in turn are classified by second group cohomology $H^2(H, Z(N))$ with trivial action.


Unfortunately, when $N = \text{SL}\_2(\mathbb{F}\_q)$ neither of these useful simplifying assumptions hold in general:


* The center $Z(N)$ consists of scalar multiples of the identity, hence is isomorphic to the group $\mu\_2(\mathbb{F}\_q)$ of square roots of unity in $\mathbb{F}\_q^{\times}$. This is trivial if $q$ is even but has order $2$ if $q$ is odd.
* The "projective general semilinear group" $\text{PGL}\_2(\mathbb{F}\_q) \rtimes \text{Gal}(\mathbb{F}\_q/\mathbb{F}\_p)$, where $\mathbb{F}\_q$ has characteristic $p$, naturally acts on $N$, and some of this action survives to the outer automorphism group. (I believe, but am not sure how to prove, that in fact it surjects onto the outer automorphism group.)


So things look a bit delicate to me. In general, in order to classify homomorphisms $H \to \textbf{Aut}(N)$, you can do the following:


* First classify homomorphisms $H \to \text{Out}(N)$.
* If $f : H \to \text{Out}(N)$ is a homomorphism, the obstruction to lifting it to a homomorphism $H \to \textbf{Aut}(N)$ is a class in third group cohomology $H^3(H, Z(N))$ given by the pullback of a universal class in $H^3(\text{Out}(N), Z(N))$ (the [Postnikov invariant](http://ncatlab.org/nlab/show/Postnikov+system) of $\textbf{Aut}(N)$) along $f$. Next, determine when the obstruction vanishes.
* If the obstruction vanishes for some $f$, lifts of $f$ are classified by second group cohomology $H^2(H, Z(N))$ where the action of $H$ on $Z(N)$ is induced by $f$. Finally, compute all of these cohomology groups.


As an insufficiently delicate person I will refrain from doing this except in the special case that $q = 2^n$ is even, so that the center vanishes. In this case the projective general and special linear groups coincide and I believe, but am not sure how to prove, that the outer automorphism group of $N = \text{SL}\_2(\mathbb{F}\_{2^n})$ is precisely 


$$\text{Gal}(\mathbb{F}\_{2^n}/\mathbb{F}\_2) \cong \mathbb{Z}\_n.$$


If I'm right, then extensions are classified by homomorphisms $\mathbb{Z}\_2 \to \mathbb{Z}\_n$. There is one if $n$ is odd, corresponding to the trivial extension $\text{SL}\_2(\mathbb{F}\_{2^n}) \times \mathbb{Z}\_2$, and two if $n$ is even, corresponding to the trivial extension and the semidirect product 


$$\text{SL}\_2(\mathbb{F}\_{2^n}) \rtimes \mathbb{Z}\_2$$ 


where $\mathbb{Z}\_2$ acts by conjugation by the unique nontrivial element of the Galois group of order $2$. 


