Automorphism group of genus 2 curve Suppose C is a genus 2 curve over a field k such that char k is not 2. Is there an easy way to show that the automorphism group is finite?
If we assume k is algebraically closed then C is hyperelliptic, does this help?
 A: Yes, here is a relatively easy way to do it using a few facts about hyperelliptic curves (where here a curve is a projective, geometrically regular, geometrically integral variety of dimension $1$ over a field $k$). These facts can be found in Ravi Vakil's notes "The Rising Sea: Foundations of Algebraic Geometry" in chapter $19$ and are all relatively simple to work through. Moreover this question is exercise $19.6.B$ from these notes.
Proposition 1): Let $C$ be a curve of genus at least $1$ with a degree two line bundle $\mathcal{L}$ such that $h^0(C,\mathcal{L})\geq 2$. Then $h^0(C,\mathcal{L}) = 2$, $\mathcal{L}$ is basepoint free and the corresponding complete linear system is hyperelliptic.
$$ C \xrightarrow[]{|\mathcal{L}|} \mathbb{P}^1$$
Proposition 2): Let $C$ be a hyperelliptic curve of genus at least $2$ and $\pi, \pi'$ any two hyperelliptic maps. Then there is an automorphism of $\mathbb{P}^1$ such that the following diagram commutes.
$$\require{AMScd}
\begin{CD}
C @>{1}>> C\\
@VV{\pi}V  @VV{\pi'}V \\
\mathbb{P}^1 @>{\alpha}>>\mathbb{P}^1\end{CD}$$
Proposition 3): Let $\pi: C \rightarrow \mathbb{P}^1$ be a hyperelliptic map for $C$ a curve of genus $g$ over a field $k$ of characteristic not $2$. Then $\pi$ has $2g+2$ branch points.
Now to prove that any genus $2$ curve over any (not necessarily algebraically closed) field of characteristic not $2$ has finite automorphism group:
Let $C$ be such a curve. Then $\omega_C$ has degree $2g-2 = 2$ and $g = 2$ sections, whence induces a hyperelliptic map by proposition $1$. Thus any genus $2$ curve is hyperelliptic. Fix some hyperelliptic map $\pi$ and let $\varphi:C \rightarrow C$ be any automorphism. Then by proposition $2$ there is an automorphism $\alpha$ of $\mathbb{P}^1$ such that the following diagram commutes.
$$\require{AMScd}
\begin{CD}
C @>{\varphi}>> C\\
@VV{\pi}V  @VV{\pi}V \\
\mathbb{P}^1 @>{\alpha}>>\mathbb{P}^1\end{CD}$$
But since $\pi$ and $\varphi \circ \pi$ have the same branch points (of which there are $6$ by proposition $3$) $\alpha$ must simply permute them. Thus we get a mapt $\operatorname{Aut}(C) \rightarrow S_6$ that is quickly checked to be a group homomorphism. Finally, notice that any automorphism in the kernel is an automorphism that commutes with $\pi$, since if $\alpha$ fixes all $6$ branch points, it fixed all of $\mathbb{P}^1$ (since the automorphisms act sharply triply transitively). There are exactly two such automorphisms* which implies that the size of the automorphism group of $C$ is indeed finite.
*Which can be seen either by using an explicit description of what a hyperellptic curve looks like over the usual open affine cover of $\mathbb{P}^1$, or using that such a map corresponds to an automorphism of the fraction field $\kappa(C)$ that fixes $\kappa(\mathbb{P}^1) \subset \kappa(C)$ and that this field extension is degree two, of characteristic not $2$.
