Sylow subgroups of projective general linear groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:21:05Z http://mathoverflow.net/feeds/question/88017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups Sylow subgroups of projective general linear groups Alex Bartel 2012-02-09T17:40:47Z 2012-02-10T13:25:13Z <p>What is known about the Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$, where $q$ is any prime power? For example, according to Theorem 7.9 in Isaacs's Character Theory, these Sylow 2-subgroups cannot be generalised quaternion. Is there a classification of all these Sylows available?</p> <p>Concretely, let me make the following semi-conjecture, which, if true, would make me very happy:</p> <blockquote> <p>The irreducible complex representations of Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$ have trivial Schur index.</p> </blockquote> <p>Is this known? Any references to literature that discusses such questions would be very welcome.</p> <p>If anyone has information on Sylow $l$-subgroups for arbitrary $l$, I would also be very interested.</p> http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88030#88030 Answer by Jim Humphreys for Sylow subgroups of projective general linear groups Jim Humphreys 2012-02-09T19:56:01Z 2012-02-09T19:56:01Z <p>To reinforce Geoff Robinson's cautious comment, I'd encourage you to dig into a volume of the ongoing treatise by Gorenstein-Lyons-Solomon <em>The Classification of the Finite Simple Groups</em>, Number 3 (Amer. Math. Soc., 1998): Section 3.3 on "Equicharacteristic Sylow Structure" and Section 4.10 on "Cross-characteristic Sylow Structure", supplemented by references to the literature. The question raised concerns just one of the adjoint-type groups of Lie type, but even here the difficulty is apparent in the treatment by G-L-S of standard structural matters. As their section headings indicate, there are fundamental differences depending on whether or not the prime <em>r</em> (such as 2) involved in the Sylow subgroup is the natural/defining prime for the Lie-type group (having <em>q</em> as a power). But in either case it's already nontrivial to compute the <em>r</em>-rank and some of the commutator structure. </p> <p>The references (as of 1998) seem to cover the essential literature, including the work of Aschbacher and the paper by Carter and Fong mentioned by Geoff. All of this is above my pay grade, as they say, so I can only wish you luck. </p> http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88046#88046 Answer by Derek Holt for Sylow subgroups of projective general linear groups Derek Holt 2012-02-09T22:38:09Z 2012-02-09T23:58:09Z <p>For $q$ odd, it is not difficult to describe the structure of a Sylow 2-subgroup of $G={\rm GL}(n,q)$, so let me do that.</p> <p>If $q \equiv 1 \bmod 4$, then the subgroup of monomial matrices of $G$ contains a Sylow $2$-subgroup of $G$, which is a wreath product of a cyclic group of order $t$ (where $t$ is the 2-part of $q-1$), and a Sylow 2-subgroup of the symmetric group $S_n$. The Sylow subgroups of $S_n$ are of course themselves built up as wreath products of cyclic groups.</p> <p>It is a little more complicated when $q \equiv 3 \bmod 4$. In that case, let $t$ be the 2-part of $q^2-1$. Then, if $n$ is even, a Sylow 2-subgroup of $G$ is a wreath product of a Sylow 2-subgroup of ${\rm GL(2,q)}$, which is semidihedral of order $2t$, with a Sylow 2-subgroup of $S_{n/2}$. If $n$ is odd, then it is a direct product of a Sylow 2-subgroup of ${\rm GL}(n-1,q)$ with a cyclic group of order 2.</p> <p>To get a Sylow 2-subgroup of ${\rm PGL}(n,q)$, you have to factor our the cyclic scalar subgroup, which is a diagonal of the base group of the wreath product.</p> <p>For $q$ even, the upper unitriangular matrices form a Sylow 2-subgroup of ${\rm PGL}(n,q)$.</p> <p>I don't know much about Schur indices but I did some quick computations in Magma, and I found that for Sylow 2-subgroups of ${\rm GL}(n,q)$ for small $n,q$, going up $(n,q)=(6,5)$ and $(8,3)$, the Schur indices of all irreducible representations are indeed 1, which seems to provide good experimental evidence for your conjecture. You might find it easier to try and prove it for ${\rm GL}(n,q)$, since the structure of the Sylow 2-subgroups can be described so explicitly, at least for odd $q$.</p> <p>Added later. For odd prime $r$ not dividing $q$, let $d$ be minimal such that $r$ divides $q^d-1$, let $t$ be the $r$-part of $q^d-1$, and let <code>$m = \lfloor n/d \rfloor$</code>. Then a Sylow $r$-subgroup of ${\rm GL}(n,q)$ is a wreath product of a cyclic group of order $t$ with a Sylow $r$-subgroup of $S_m$.</p> <p>For all of the classical groups, Sylow subgroups in coprime characteristic arise in a similar way, as wreath products of a group coming from a base case with a Sylow subgroup of a symmetric group. But of course there are lots of little complications for the individual types. I had a student a few years ago write Magma code to construct all of these, so I can produce very explicit descriptions!</p> http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88098#88098 Answer by F. Ladisch for Sylow subgroups of projective general linear groups F. Ladisch 2012-02-10T12:55:43Z 2012-02-10T13:25:13Z <p>The irreducibles of the Sylow 2-subgroups of ${\rm GL}(n,q)$, $q$ odd, have indeed trivial Schur indices: Let $S$ be such a Sylow 2-subgroup. First, observe that the natural module $(\mathbb{F}_q)^n$ splits into a sum of simple modules $U_1\oplus \dotsb \oplus U_l$, and the dimension of each simple module is a power of $2$. This shows that $S \cong S_1 \times \dotsb \times S_l$, where the $S_i$'s are Sylow 2-subgroups of a ${\rm GL}(2^k, q)$. Thus, w.l.o.g. we may assume that $n=2^k$. Then I use induction on $k$. First note that if $S$ is a Sylow 2-subgroup of ${\rm GL}(2^k, q)$, then <code>$$T=\lbrace \begin{pmatrix} s &amp; \\ &amp; t \end{pmatrix} \mid s, t\in S \rbrace \cup \lbrace \begin{pmatrix} &amp; s \\ t &amp; \end{pmatrix} \mid s,t\in S\rbrace \cong S\wr C_2$$</code> is a Sylow 2-subgroup of ${\rm GL}(2^{k+1}, q)$, except when $k=0$ and $q\equiv 3\mod 4$. This follows from the description in Derek Holt's answer, but it can also be seen directly by observing that $T$ has the right order.<br> Write $N=S\times S$, so that $T= C_2 N$, and let $\chi\in {\rm Irr} (T)$. Three cases have to be considered:<br> 1. We have $\chi_N \in {\rm Irr} (N)$. By induction, $\chi_N$ has trivial Schur index. By Lemma~10.4 in Isaacs' character theory book, for example, it follows that $\chi$ has trivial Schur index.<br> 2. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are not Galois conjugate. Then $\chi = \theta^T$ and $\theta$ have the same field of values and the same Schur index (again, see Lemma~10.4 in Isaacs).<br> 3. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are Galois conjugate. This means that $\chi=\theta^T$, but $|\mathbb{Q}(\theta):\mathbb{Q}(\chi)|=2$. Now a representation $R\colon N\to {\rm GL}(\chi(1), \mathbb{Q}(\chi) )$ affording the character $\theta+\theta^g$ can be extended to a representation over the same field, since $N$ has a complement (of order 2) in $T$.<br> To get the induction going when $q\equiv 3\mod 4$, one has to check that the Sylow 2-subgroup of ${\rm GL}(2,q)$ has trivial Schur indices, but that is clear since it's a semidihedral group.</p>