Let $R$ be a complete dvr and $k$ its residue field of positive characteristic. 
Let $H$ be a finite subgroup of $PGL\_2(k)$ such that the order of $H$ is prime with $char(k)$.


Is there some elementary way to show that $H$ is a subgroup of $PGL\_2(R)$?



Subgroup of Projective general linear group on complete discrete valuation ring
I don't know if the following qualifies as "elementary", but here is how I would prove this (for $2$ replaced by some $d\in \mathbb{N}$):
$\DeclareMathOperator{\gl}{GL}$ 


First, choose some finitely generated subgroup 
$U \leq \gl(d, k)$ with $U/(U\cap k^\*) = H$.
Then $U\cap k^\*$ is finitely generated abelian and thus $U\cap k^\* = Z \times F$,
where $F$ is free abelian, and $Z$ is finite.
The order of $Z$ is not divisible by the characteristic $p$ of $k$.
Thus $U/F$ is finite of order prime to $p$, and $U$ contains no elements of order $p$.


I claim that for each $i \in \mathbb{N}$, there is a subgroup
$U\_i \leq \gl(d, R/\pi^i R)$,
such that $U\_i \cong U\_{i-1}$ under the natural map,
and also $U\_i \cap (R/\pi^i R)^\* \cong U\_{i-1} \cap (R/\pi^{i-1}R)^\*$ under the natural map.
Thus we have a projective system
$$ \dots \to U\_i \to U\_{i-1} \to \dots \to U\_1 = U $$
with all maps isomorphisms.
Then the inverse limit $\widehat{U}$ is also isomorphic to $U$,
and can be viewed as subgroup of $\gl(d,R)$.
Moreover, we also have $\widehat{U} \cap R^\* \cong U\cap k^\*$ by the natural map,
and thus $\widehat{H}:= \widehat{U}/\widehat{U}\cap R^\* \cong H$ naturally,
which is what we want to prove. 


The $U\_i$'s are constructed inductively, with $U\_1=U$ already given.
Suppose we have defined $U\_i$.
Let $G\_{i+1} \leq \gl(d,R/\pi^{i+1}R)$ be a finitely generated subgroup which maps onto $U\_i$.
We can choose $G\_{i+1}$ so that $G\_{i+1}\cap (R/\pi^{i+1}R)^\*$ maps *onto*
$U\_i \cap (R/\pi^i R)^\*$.
In particular, $G\_{i+1} \cap (R/\pi^{i+1}R)^\*$ contains a free abelian group $F\_{i+1}$ mapping onto the corresponding subgroup $F\_i \cong F$ of $U\_i$.  

Set $P = G\_{i+1}\cap \big( 1 + \mathbf{M}\_n(\pi^{i}R/\pi^{i+1}R) \big)$,
so $G\_{i+1}/P \cong U\_i$.
Then $P$ is finitely generated and thus a finite $p$-group (elementary abelian, in fact).
We have $F\_{i+1} \cap P = 1$, and we can apply the Schur-Zassenhaus theorem to the finite group $G\_{i+1}/F\_{i+1}$ to conclude that $PF\_{i+1}/F\_{i+1}$ has a complement $U\_{i+1}/F\_{i+1}$.
Then $U\_{i+1}$ has the desired properties.  

Of course, instead of defining the system of subgroups, one can also construct a sequence of maps $\mu\_i \colon U \to \gl(d,R)$ which are homomorphisms mod $\pi^i$, and which converge to a group homomorphism $\mu\colon U\to \gl(d,R)$. This can be done even without invoking Schur-Zassenhaus, but gets a little bit clumsy.


