Determining if these surjections have sections Let $\pi:\ \operatorname{GL}(2,k)\ \longrightarrow\ \operatorname{PGL}(2,k)$ be the canonical homomorphism, and pick some finite subgroup $G\subset \operatorname{PGL}(2,k)$. Then we have an exact sequence
$$1\ \longrightarrow\ \{\alpha I\mid \alpha \in k\}\ \longrightarrow\ \pi^{-1}(G)\ \longrightarrow\ G\ \longrightarrow\ 1.$$
Specific question: Let $k$ be algebraically closed of characteristic zero and let $G$ be some subgroup of $\operatorname{PGL}(2,k)$ isomorphic to $A_5$. Does this sequence split in this case? (It seems to me it shouldn't, but I don't have a proof.)
General question: How would you go about determining, for any particular case of $k$ and $G$, if this sequence splits?
 A: The sequence does not split: Suppose toward a contradiction that there does exist a splitting $s$. Then there exists a subgroup of $\operatorname{GL}(2,k)$ isomorphic to $A_5$, which has a presentation
$$A_5=\langle a,b\vert\ a^2=b^3=(ab)^5=1\rangle.$$
In particular there exist $A,B\in\operatorname{GL}(2,k)$ satisfying
$$A^2=B^3=(AB)^5=I,$$
and hence $\det(A)$, $\det(B)$ and $\det(AB)$ are second, third and fifth roots of unity, respectively. Because the determinant is multiplicative, it follows that $\det(A)=\det(B)=\det(AB)=1$.
Note that $A$ is a root of its characteristic polynomial, which is given by
$$p_A(X)=X^2-\operatorname{Tr}(A)\cdot X+\det(A)\cdot I.$$
As $\det(A)=1$ and $A^2=I$ it follows that $-\operatorname{Tr}(A)\cdot A+2\cdot I=0$, and hence
$$A=\frac{2}{\operatorname{Tr}(A)}\cdot I,$$
which shows that $\pi(A)=I\in\operatorname{PGL}(2,k)$. Therefore $(\pi\circ s)(g)=I$ for any $g\in G$ of order $2$, contradicting the fact that $s$ is a splitting of the sequence.
Note that this works for any $k$ with $\operatorname{char}k\neq2$.
A: For the "specific question", inspired by Servaes' answer, I note that knowledge of $A_5$'s character table (for algebraically closed, characteristic zero $k$) allows us to rule out a splitting of the sequence, simply on the grounds that $GL(2,k)$ doesn't have a subgroup isomorphic to $A_5$ because $A_5$ does not have a 2-dimensional faithful representation.
$$\begin{array}{c | c c c c c} &\text{id}&\text{dbl transp}&\text{3-cycle}&\text{5-cycle}_A&\text{5-cycle}_{A^2} \\ \hline \text{Triv} &1&1&1&1&1\\ \text{3-dim}_A&3&-1&0&\varphi&\hat\varphi\\\text{3-dim}_B&3&-1&0&\hat\varphi&\varphi\\\text{4-dim}&4&0&1&-1&-1\\\text{5-dim}&5&1&-1&0&0\end{array}$$
where $\varphi,\hat\varphi$ are the roots of $x^2-x-1$. The smallest non-trivial representation of $A_5$ is dimension $3$, and any faithful representation of $A_5$ must be a direct sum containing at least one nontrivial representation. (In fact, since $A_5$ is simple, all nontrivial representations are faithful; but 2 dimensions just isn't enough...)
I am still very interested in your thoughts about the "general question".
