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Let $K$ be a field of characteristic $p$. What are the finite subgroups of $PGL_2(K)$ whose orders are divisible by $p$? And if $G$ and $H$ are two such subgroups that are isomorphic, can one say when they are conjugate inside $PGL_2(K)$?

While the finite subgroups of $PGL_2(\mathbb{C})$ are well understood from a variety of different viewpoints, the answer to the above question does not appear to be well known. See, e.g., this article of Beauville for a pleasant discussion of the case in which $p$ does not divide the order of the group.

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Welcome to MO, Professor Faber. –  Pete L. Clark Jun 6 '11 at 5:04
By the way, since you are not assuming $K$ to be algebraically closed, my wild guess is that to get from the classification up to $\overline{K}$-conjugacy to the classification up to $K$-conjugacy will involve consideration of some (nonabelian) flat cohomology groups. Whether that's good or bad news I couldn't say... –  Pete L. Clark Jun 6 '11 at 5:07
I would have thought that L.E. Dickson already had something to say about this question, but when I have more time, I'll try to answer more fully. –  Geoff Robinson Jun 6 '11 at 7:27
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4 Answers

I'm going to answer a weaker question with much of the same flavor in it: Up to abstract isomorphism, what are the possible finite groups that can occur as a subgroup of $\mathrm{PGL}_2(K)$ when $\mathrm{char}(K)$ is odd and $K$ is algebraically closed?

Suppose that $G$ is a finite subgroup of $\mathrm{PGL}_2(K)$, where $K$ is of odd characteristic $p$. Since $G$ is finite, it may be represented by finitely many elements $g_1, \ldots, g_s \in \mathrm{GL}_2(K')$, where $K' \subset K$ is finitely generated over the prime field $\mathbb{F}_p$. But $G$ is finite, and so any transcendental entry of $g_i$ may be replaced by an algebraic element in some $\mathbb{F}_q$ for $q$ a sufficiently large power of $p$. This alters the group $G$, but not its isomorphism class. (There must be a fancier way to say this ... something like "$G$ is a finite group scheme over $K'$, and hence isotrivial.") In this way, we reduce to considering finite subgroups of $\mathrm{PGL}_2(\overline{\mathbb{F}}_p) = \mathrm{PSL}_2(\overline{\mathbb{F}}_p)$. Since $\mathrm{PSL}_2(\overline{\mathbb{F}}_p)$ is the union of its subgroups $\mathrm{PSL}(2,p^\alpha)$, we need only consider subgroups of $\mathrm{PSL}(2,p^\alpha)$.

I chased down some of the references suggested by Geoff Robinson in his answer. Indeed, L.E. Dickson (following E.H. Moore and Wiman) worked out all of the subgroups of $\mathrm{PSL}(2,q)$ for arbitrary $q$ in his book "Linear Groups with an Exposition of the Galois Field Theory." People spoke about groups in a very different way a hundred years ago, so I didn't have much luck with Dickson's version of the story. But when $q$ is odd, David Bloom gave a modern summary of Dickson's work in his 1967 paper "The subgroups of $\mathrm{PSL}(3,q)$ for odd $q$."

Theorem. (Dickson) Write $q = p^\alpha$, and let $G$ be a subgroup of $\mathrm{PSL}(2,q)$. Then one of the following occurs:

(a) $G$ is isomorphic to (i) $A_5$ (with $p \neq 5$), or (ii) $S_4$ or $A_4$;

(b) $G$ is cyclic and $p$-regular (i.e., of order prime to $p$);

(c) $G$ is dihedral and $p$-regular;

(d) $G = \langle \Gamma, \zeta \rangle$, where $\Gamma$ is a $p$-group and $\zeta$ is a $p$-regular element in the normalizer of $\Gamma$ (In this case, $G$ is contained in a Borel subgroup of $\mathrm{PSL}(2,q)$);

(e) $G$ is conjugate in $\mathrm{GL}(2,q) / \langle -I \rangle$ to $\mathrm{PSL}(2,p^{\beta})$ for some $\beta \mid \alpha$;

(f) $q \equiv 1 \pmod 4$; up to conjugacy in $\mathrm{GL}(2,q) / \langle -I \rangle$, $G$ contains $H = \mathrm{PSL}(2, p^\beta)$ as an index 2 subgroup (here $2\beta \mid \alpha$), and $G$ is generated by $H$ and the diagonal matrix $\|c, c^{-1} \|$, where $c^2$ is a fixed generator of $\mathbb{F}_{p^\beta}^\times$.

Cases (a,b,c,e,f) are reasonably straightforward groups. It's worth noting that if $p^\beta = 3$ in case (e) or (f), then one obtains a group that is isomorphic to $A_4$ or $S_4$, respectively. Case (d) is also not too difficult to work out. After conjugation, we may assume $G$ lies in the standard Borel subgroup of $\mathrm{PSL}(2,q)$: $$ B = \left\{ \begin{pmatrix} a & b \\ & a^{-1} \end{pmatrix} \ : \ a \neq 0 \right\}. $$ The subgroup $\Gamma$ consists of the unipotent elements of $G$, which gives an exact sequence $$ 0 \longrightarrow \Gamma \longrightarrow G \stackrel{\pi}{\longrightarrow} \mathbb{F}_q^\times, $$ where the map $\pi$ sends a Borel element to its upper left entry. A direct calculation shows that $\Gamma$ is normal in $G$. If $\zeta$ is a generator for the image of $\pi$, then $\zeta$ must be $p$-regular (its order divides $q-1$), and thus $G$ is a semidirect product: $G = \Gamma \rtimes \langle \zeta \rangle$. As to the structure of $\Gamma$, it is an abelian subgroup, and there exist finitely many elements $\gamma_1, \ldots, \gamma_r \in \mathbb{F}_q$ such that $\Gamma \cong (C_p)^r$ with generators $$ \left\{ \begin{pmatrix} 1 & \gamma_i \\ & 1 \end{pmatrix} \ : \ i = 1, \ldots, r \right\}. $$

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Post deleted, since Xander Faber's answer now gives definitive account of Dickson's results.

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After thinking about this question for a few months, I've managed to give a complete answer. The article is posted on the arXiv (arXiv:1112.1999v1 [math.NT]). The short version is that a finite $p$-irregular subgroup of $\mathrm{PGL}_2(k)$ is isomorphic to $\mathrm{PSL}_2(\mathbb{F}_q)$ or $\mathrm{PGL}_2(\mathbb{F}_q)$ for $q$ a power of the characteristic of $k$, to a $p$-semi-elementary group (a semi-direct product of a $p$-group and a cyclic group), to a dihedral group, or to $\mathfrak{A}_5$. Thanks to Geoff Robinson's suggestion above, I was able to modify Dickson's arguments to give a complete classification up to conjugacy over an an algebraically closed field. The same arguments go through for separably closed fields, except when the characteristic of $k$ is 2, where a little extra work is required. Finally, I use Galois descent (following Beauville) to pass to arbitrary fields.

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Since nontrivial unipotent elements have order $p$, you will have plenty of subgroups, doc. I don't think they are understood.

For instance, the answer to your second question is no as soon as $K$ contains at least $p^3$ elements. Just take two different $C_p^2$ in the additive group of $K$ and consider two upper triangular groups with off-diagonal elements in your $C_p^2$-s...

Having said that, it may be an interesting questions what finite simple groups sit there. Are they all $PSL(2,q)$?

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I think you mean PSL(2,q). And when p is odd they almost always will be, since they will have dihedral Sylow $2$-subgroups. I think the question should be asked about irreducible subgroups, which are pretty close to simple –  Geoff Robinson Jun 6 '11 at 7:59
Yes, sorry, correcting... –  Bugs Bunny Jun 6 '11 at 8:28
You are technically right: $p^0$ is possible as well... –  Bugs Bunny Jun 7 '11 at 9:23
Yes, but for 2x2 matrices higher powers are impossible: there are only two unipotent orbits. Or did I miss something? –  Bugs Bunny Jun 15 '11 at 21:08
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