Subgroup of Projective general linear group on complete discrete valuation ring 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 $PGL_2(R)$?
 A: 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.
A: I think that an alternate strategy (which also works, as in Frieder Ladisch's answer, for general $d$, at least when $k$ and $R$ are large enough) is to "lift" finite $H \leq {\rm PGL}(d,k)$ to a central extension which is a finite $p^{\prime}$-subgroup ${\tilde H}$ of ${\rm GL}(d,k)$. Then ${\tilde H}$ is completely reducible. Then ${\tilde H}$ lifts to an isomorphic subgroup ${\tilde L}$ of ${\rm GL}(d,R)$ with scalar matrices in ${\rm GL}(d,k)$ lifting to scalar matrices in ${\rm GL}(d,R).$ Then ${\tilde L}$ with scalars factored out is isomorphic to $H$, and is isomorphic to a subgroup of ${\rm PGL}(d,R)$. This is fairly classical modular representation theory of finite groups ( dating back to R. Brauer and J.A. Green, etc.), and the underlying lifting techniques are not unrelated to those used in Frieder's answer.
Here is an outline, given that $R$ and $k$ are large enough: the irreducible $k{\tilde H}$-modules correspond to conjugacy classes (under the unit group of the group algebra $k{\tilde H}$) of primitive idempotents of $k{\tilde H}$. The primitive idempotents of $k{\tilde H}$ (up to the above conjugacy) in one-to-one fashion to primitive idempotents of $R{\tilde H}$ (up to the similar conjugacy). If ${\tilde Z}$ is the subgroup of scalar matrices in ${\tilde H}$ ( which is a finite $p^{\prime}$-group), then for each primitive idempotent $e$ of $k{\tilde H}$, there is a unique linear character $\lambda$ of ${\tilde Z}$ such that $e_{\lambda}e = e$, where $|{\tilde Z}| e_{\lambda} = \sum_{z \in {\tilde Z} } \lambda(z^{-1})z.$ There is a natural lift of $e_{\lambda}$ to an idempotent of  $R{\tilde Z}$, and the lifting process ( for all of $k{\tilde H}$) respects the correspondence between $e$ and $e_{\lambda}$. This translates to the fact that scalar matrices in ${\tilde H}$ lift to scalar matrices in ${\tilde L}$ when the representation is lifted.
