What is the normalizer of $SL(2,q)$ inside $SL(2,\overline{\mathbb{F}_q})$? I'd like to understand the normalizer $N$ of $SL(2,\mathbb{F}_q)$ inside $SL(2,\overline{\mathbb{F}_q})$.
Specifically, I'd like to understand the quotient $N/SL(2,q)$.  I'm guessing this quotient must essentially be generated by the diagonal automorphisms (hence it should be isomorphic to $\mathbb{F}_q^\times/(\mathbb{F}_q^\times)^2$).
Is this true?
Does anyone know of a reference for this?
 A: In fact, we have the following result.
Thm.  Let $\mu_{2q-2}=\{u\in\overline{\mathbb{F}}_q^\times\mid u^{2q-2}=1\}$. For any $u\in\mu_{2q-2}$, set $D_u=\begin{pmatrix} u & 0 \cr 0 &u^{-1}\end{pmatrix}$.
Then, we have a group isomorphism $$\mu_{2q-2}/\mathbb{F}_{q}^{\times}\simeq N/SL_2(\mathbb{F}_q),$$ sending $u\mathbb{F}_{q}^{\times}$ to $D_uSL_2(\mathbb{F}_q)$.
Proof. Let $\sigma:SL_2(\overline{\mathbb{F}}_q)\to SL_2(\overline{\mathbb{F}}_q)$ which send $\begin{pmatrix}a & b\cr c & d\end{pmatrix}$ to $\begin{pmatrix}a^q & b^q \cr c^q & d^q\end{pmatrix}$. This is a group automorphism, and $P\in SL_2(\overline{\mathbb{F}}_q)$ lies in $SL_2(\mathbb{F}_q)$ if and only if $\sigma(P)=P$.
Notice that for all $S\in SL_2(\overline{\mathbb{F}}_q)$, we have $PSP^{-1}\in SL_2(\overline{\mathbb{F}}_q)$, so $P$ lies in the normalizer of  $SL_2(\mathbb{F}_q)$ if and only if $\sigma(PSP^{-1})=\sigma(P) S\sigma(P)^{-1}=PSP^{-1}$, that is $P^{-1}\sigma(P)$ lies in the centralizer of  $SL_2(\mathbb{F}_q)$. This centralizer consists of scalar matrices, so $P$ lies in the normalizer if only if $\sigma(P)=\alpha P$ for some $\alpha\in\overline{\mathbb{F}}_q^\times$.
Setting $P=\begin{pmatrix}a & b\cr c & d\end{pmatrix}$ , we get $a^q=\alpha a, b^q=\alpha b, c^q=\alpha c,d^q=\alpha d$. One of the entries of $P$ is nonzero, say $a$. Then $(b/a)^q=b/a$, and same for the other coefficients.
It follows easily that $P=u Q$, where $u\in\overline{\mathbb{F}}_q^\times$ and $Q\in GL_2(\mathbb{F}_q)$. Note that $u^2\det(Q)=1$, so $u^2\in\mathbb{F}_q^\times$, that is $u\in\mu_{2q-2}$ (since $u^2\in\mathbb{F}_q^\times$ if and only if $(u^2)^{q-1}=1$).
Writing $Q=\begin{pmatrix}1 & 0 \cr 0 & \det(Q)^{-1}\end{pmatrix}Q'$, $Q'\in SL_2(\mathbb{F}_q)$, we get $P=\begin{pmatrix} u & 0 \cr 0 &u^{-1}\end{pmatrix}Q', u\in\mu_{2q-2},Q'\in SL_2(\mathbb{F}_q)$. 
Hence any element of  $N/SL_2(\mathbb{F}_q)$ can be represented by a matrix $D_u=  \begin{pmatrix} u & 0 \cr 0 &u^{-1}\end{pmatrix}, u\in\mu_{2q-2}$, so we have a surjective morphism $u\in \mu_{2q-2}\mapsto \overline{D}_u \in N/SL_2(\mathbb{F}_q)$.
Now $D_u$ lies in $SL_2(\mathbb{F}_q)$ if and only if $u\in\mathbb{F}_q^\times$, so the first isomorphism theorem yields the desired result.
Now we have to elucidate the structure of $\mu_{2q-2}/\mathbb{F}_{q}^{\times}$.
Prop. The group  $\mu_{2q-2}/\mathbb{F}_{q}^{\times}$ is trivial if $q$ is even, and has order to if $q$ is odd.
Notice that $\mathbb{F}_q^\times= \{u\in\overline{\mathbb{F}}_q^\times\mid u^{q-1}=1\}.$
If $q$ is even (that is a power of $2$), then $u^{2q-2}-1=(u^{q-1}-1)^2$, since we are in characteristic $2$, hence the first part of the proposition.
If $q$ is odd, then $2q-2$ is prime to $q$, and it is well-known that $\mu_{2q-2}$ is cyclic of order $2q-2$. The result follows in this case, since $\mathbb{F}_q^\times$ has order $q-1$.
Corollary. The group $N/SL_2(\mathbb{F}_q)$ is trivial if $q $ is even, and has order $2$ if $q$ is odd. In the latter case, the nontrivial class is represent by $D_u$, where $u\in\mathbb{F}_q^\times$ is any element satisfying $u^{q-1}=-1.$
