Finite subgroups of $PGL_2(K)$ in characteristic $p$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:17:02Z http://mathoverflow.net/feeds/question/67007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67007/finite-subgroups-of-pgl-2k-in-characteristic-p Finite subgroups of $PGL_2(K)$ in characteristic $p$ Xander Faber 2011-06-06T02:10:31Z 2011-12-12T06:57:17Z <p>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 $PGL_2(K)$? </p> <p>While the finite subgroups of $PGL_2(\mathbb{C})$ are <a href="http://mathoverflow.net/questions/16026" rel="nofollow">well understood from a variety of different viewpoints</a>, the answer to the above question does not appear to be well known. See, e.g., <a href="http://arxiv.org/abs/0909.3942" rel="nofollow">this article of Beauville</a> for a pleasant discussion of the case in which $p$ <i>does not</i> divide the order of the group.</p> http://mathoverflow.net/questions/67007/finite-subgroups-of-pgl-2k-in-characteristic-p/67014#67014 Answer by Bugs Bunny for Finite subgroups of $PGL_2(K)$ in characteristic $p$ Bugs Bunny 2011-06-06T07:32:08Z 2011-06-07T09:24:19Z <p>Since nontrivial unipotent elements have order $p$, you will have plenty of subgroups, doc. I don't think they are understood.</p> <p>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... </p> <p>Having said that, it may be an interesting questions what finite simple groups sit there. Are they all $PSL(2,q)$?</p> http://mathoverflow.net/questions/67007/finite-subgroups-of-pgl-2k-in-characteristic-p/67035#67035 Answer by Geoff Robinson for Finite subgroups of $PGL_2(K)$ in characteristic $p$ Geoff Robinson 2011-06-06T12:49:44Z 2011-06-10T08:21:52Z <p>Post deleted, since Xander Faber's answer now gives definitive account of Dickson's results.</p> http://mathoverflow.net/questions/67007/finite-subgroups-of-pgl-2k-in-characteristic-p/67403#67403 Answer by Xander Faber for Finite subgroups of $PGL_2(K)$ in characteristic $p$ Xander Faber 2011-06-10T02:33:18Z 2011-06-10T02:33:18Z <p>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? </p> <p>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 $\mathrm{PSL}(2,p^\alpha)$.</p> <p>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 $q$." </p> <p><b>Theorem</b>. (Dickson) Write $q = p^\alpha$, and let $G$ be a subgroup of $\mathrm{PSL}(2,q)$. Then one of the following occurs:</p> <p>(a) $G$ is isomorphic to (i) $A_5$ (with $p \neq 5$), or (ii) $S_4$ or $A_4$;</p> <p>(b) $G$ is cyclic and $p$-regular (i.e., of order prime to $p$);</p> <p>(c) $G$ is dihedral and $p$-regular;</p> <p>(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 $\mathrm{PSL}(2,q)$);</p> <p>(e) $G$ is conjugate in $\mathrm{GL}(2,q) / \langle -I \rangle$ to $\mathrm{PSL}(2,p^{\beta})$ for some $\beta \mid \alpha$;</p> <p>(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 $\mathbb{F}_{p^\beta}^\times$.</p> <hr> <p>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 $\mathrm{PSL}(2,q)$: <code>$$B = \left\{ \begin{pmatrix} a &amp; b \\ &amp; a^{-1} \end{pmatrix} \ : \ a \neq 0 \right\}.$$</code> 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 <code>$\gamma_1, \ldots, \gamma_r \in \mathbb{F}_q$</code> such that $\Gamma \cong (C_p)^r$ with generators <code>$$\left\{ \begin{pmatrix} 1 &amp; \gamma_i \\ &amp; 1 \end{pmatrix} \ : \ i = 1, \ldots, r \right\}.$$</code></p> http://mathoverflow.net/questions/67007/finite-subgroups-of-pgl-2k-in-characteristic-p/83226#83226 Answer by Xander Faber for Finite subgroups of $PGL_2(K)$ in characteristic $p$ Xander Faber 2011-12-12T06:57:17Z 2011-12-12T06:57:17Z <p>After thinking about this question for a few months, I've managed to give a complete answer. The article is posted on the arXiv <a href="http://arxiv.org/abs/1112.1999" rel="nofollow">(arXiv:1112.1999v1 [math.NT])</a>. 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. </p>