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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 $PGL_2(R)$?

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  • $\begingroup$ I guess by "$H$ is a subgroup of $PGL_2(R)$" you mean that there is a subgroup of $PGL_2(R)$ such that $H$ is the isomorphic image of this subgroup under the natural map $PGL_2(R) \to PGL_2(k)$? And what "non-elementary" way of showing this do you know? $\endgroup$ Jun 30, 2015 at 14:19
  • $\begingroup$ @FriederLadisch: "And what 'non-elementary' way of showing this do you know?" Perhaps your question was rhetorical. If not, you can deduce the result from SGA1, Exp. XIII, Cor. 2.12. Since $\text{PGL}_2(k)$ is the automorphism group of $\mathbb{P}^1_k$, you can form the quotient, which will again be $\mathbb{P}^1_k$, then lift the branched cover over $R$. $\endgroup$ Jul 1, 2015 at 11:11
  • $\begingroup$ @JasonStarr: I think I have a proof (or at least an idea of a proof), and compared to your proof sketch, it would perhaps qualify as "elementary". But I wanted to know what OP meant by "elementary", if he wants to avoid specific techniques/results that are used in the proof known to him, etc. $\endgroup$ Jul 2, 2015 at 9:53
  • $\begingroup$ @FriederLadisch: I cannot speak for the OP, but I would like to see an elementary proof. Using a result of de Jong, He and myself, I can say something about subgroups of the form $\mathbb{Z}/a\mathbb{Z}\times \mathbb{Z}/b\mathbb{Z}$ inside adjoint group via specialization over DVRs, and I would like to see a more elementary argument for that. Maybe your elementary argument would also give what I want. $\endgroup$ Jul 2, 2015 at 13:35

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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.

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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.

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    $\begingroup$ My first idea when seeing this question was also "This should be clear from modular representation theory". I think the arguments in your second paragraph all go through when $k$ is arbitrary, and $R$ complete. Only lifting $H$ to some finite subgroup of $\operatorname{GL}(d,k)$ may not be possible, if $k$ contains some transcendental elements, but not their roots. Of course, when $k$ is finite or algebraic over the prime field, then this is no problem. $\endgroup$ Jul 7, 2015 at 12:04
  • $\begingroup$ @Frieder : I suppose I was implicitly assuming algebraic over the ground field. $\endgroup$ Jul 7, 2015 at 15:57
  • $\begingroup$ @FriederLadisch : Well, I suppose that by introducing some scalars of finite order, you can keep all relevant groups finite, but in some of the situations you suggest, there might be enough scalars to do this. $\endgroup$ Jul 8, 2015 at 17:49

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