Sylow Subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:21:18Z http://mathoverflow.net/feeds/question/18716 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18716/sylow-subgroups Sylow Subgroups Victor Miller 2010-03-19T05:15:31Z 2010-03-27T20:29:37Z <p>I had been looking lately at Sylow subgroups of some specific groups and it got me to wondering about <em>why</em> Sylow subgroups exist. I'm very familiar with the proof of the theorems (something that everyone learns at the beginning of their abstract algebra course) -- incidentally my favorite proof is the one by Wielandt -- but the statement of the three Sylow theorems still seems somewhat miraculous. What got Sylow to imagine that they were true (especially the first -- the existence of a sylow subgroup)? Even the simpler case of Cauchy's theorem about the existence of an element of order $p$ in a finite subgroup whose order is a multiple of $p$ although easy to prove (with the right trick) also seems a bit amazing. I believe that sometimes the hardest part of a proving a theorem is believing that it might be true. So what can buttress the belief for the existence of Sylow subgroups?</p> http://mathoverflow.net/questions/18716/sylow-subgroups/18718#18718 Answer by H. Hasson for Sylow Subgroups H. Hasson 2010-03-19T07:20:37Z 2010-03-19T07:20:37Z <p>I don't know if this was the original motivation, but this has some interesting motivating ideas: <a href="http://mathoverflow.net/questions/5659/abstract-nonsense-versions-of-combinatorial-group-theory-questions" rel="nofollow">http://mathoverflow.net/questions/5659/abstract-nonsense-versions-of-combinatorial-group-theory-questions</a> </p> http://mathoverflow.net/questions/18716/sylow-subgroups/18788#18788 Answer by KConrad for Sylow Subgroups KConrad 2010-03-19T19:52:19Z 2010-03-19T19:52:19Z <p>Victor, you should check out Sylow's paper. It's in Math. Annalen 5 (1872), 584--594. I am looking at it as I write this. He states Cauchy's theorem in the first sentence and then says "This important theorem is contained in another more general theorem: if the order is divisible by a prime power then the group contains a subgroup of that size." (In particular, notice Sylow's literal first theorem is more general than the traditional formulation.) Thus he was perhaps in part inspired by knowledge of Cauchy's theorem. </p> <p>Sylow also includes in his paper a theorem of Mathieu on transitive groups acting on sets of prime-power order (see p. 590), which is given a new proof by the work in this paper. Theorems like Mathieu's may have led him to investigate subgroups of prime-power order in a general finite group (of substitutions). </p> http://mathoverflow.net/questions/18716/sylow-subgroups/19378#19378 Answer by Vipul Naik for Sylow Subgroups Vipul Naik 2010-03-26T01:43:51Z 2010-03-26T13:29:47Z <p>The Sylow theorems are finite group analogues of a bunch of results about "maximal unipotent subgroups" in algebraic groups. Basically, the Sylow subgroups play a role analogous to the role played by the maximal unipotent subgroups. </p> <p>In the case where the group is the general linear group, the maximal unipotent subgroup can be taken as the group of upper triangular matrices with 1s on the diagonal, for instance. There are existence, conjugacy, and domination results for these analogous to the existence, conjugacy, and domination part of Sylow's theorems: maximal unipotents exist, every unipotent is contained in a maximal unipotent, all maximal unipotents are conjugate. The role analogous to "order" is now played by "dimension".</p> <p>The normalizer of the Sylow subgroup plays the role of the maximal connected solvable subgroup, also called the Borel subgroup (see <a href="http://en.wikipedia.org/wiki/Borel_fixed-point_theorem" rel="nofollow">Borel fixed-point theorem</a> and <a href="http://en.wikipedia.org/wiki/Lie-Kolchin_theorem" rel="nofollow">Lie-Kolchin theorem</a>). In the case of the general linear group, this is the group of upper triangular invertible matrices.</p> <p>There are similar results for Lie algebras too, basically arising from <a href="http://en.wikipedia.org/wiki/Engel_theorem" rel="nofollow">Engel's theorem</a> and <a href="http://en.wikipedia.org/wiki/Lie%27s_theorem" rel="nofollow">Lie's theorem</a>.</p> <p>In fact, much of the study of simple groups and their geometry relies on this geometric interpretation of Sylow subgroups, <em>p</em>-subgroups, and their normalizers. This deeper study of the geometry/combinatorics of simple groups is called local analysis in group theory and is closely related to the recently popular topic of "fusion systems" which are essentially studying the conjugation action of a group on subgroups of a particular Sylow subgroup.</p> <p>ADDED BASED ON COMMENT BELOW: For a finite field $F_q$ where <em>q</em> is a power of <em>p</em>, the maximal unipotent subgroup of $GL_n(F_q)$ is the $p$-Sylow subgroup. I had originally intended to mention this, but forgot.</p> http://mathoverflow.net/questions/18716/sylow-subgroups/19543#19543 Answer by jp for Sylow Subgroups jp 2010-03-27T20:29:37Z 2010-03-27T20:29:37Z <p>An extension of the Vipul's ideas can be found in the article (couldn't find a link to the pdf with google)</p> <p><strong>Subgroup complexes</strong> by <em>Peter Webb</em>, pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987).</p> <p>But as Mariano already commented, the analogy to the maximal unipotent subgroups of the general linear group was probably not Sylow's motivation. As commented before, he was maybe looking for maximal $p$-subgroups (i.e., maximal with respect to be a $p$-subgroup).</p> <p>This is also the leitmotif of my favorite proof of the Sylow theorems given by Michael Aschbacher in his book <strong>Finite Group Theory</strong>. It is based on Cauchy's theorem (best proved using J.H.McKay's trick to let $Z_p$ act on the set of all $(x_1, \dots, x_p) \in G^p$ whose product is $1$ by rotating the entries) and goes essentially like this:</p> <p>The group $G$ acts on the set $\mathrm{Syl}_p(G)$ of its maximal $p$-subgroups by conjugation. Let $\Omega$ be a (nontrivial) orbit with $S\in\Omega$. If $P$ is a fixed point of the action restricted to $S$ then $S$ normalizes $P$ and $PS=SP$ is a $p$-group. Hence $P=S$ by maximality of both $P$ and $S$, and $S$ has a <strong>unique</strong> fixed point. As $S$ is a $p$-group, all its orbits have order $1$ or a multiple of $p$, in particular $|\mathrm{Syl}_p(G)| = 1 \bmod p$. All orbits of $G$ are disjoint unions of orbits of $S$ proving $\Omega = 1 \bmod p$ and $\Omega' = 0 \bmod p$ for all other orbits $\Omega'$ of $G$. This implies that $\Omega = \mathrm{Syl}_p(G)$, as $\Omega$ was an arbitrary nontrivial orbit of $G$, showing that the action of $G$ is transitive. The stabilizer of $S$ in $G$ is its normalizer $N_G(S)$, and as the action is transitive $|G:N_G(S)| = |\mathrm{Syl}_p(G)| = 1 \bmod p$. It remains to show that $p$ does not divide $|N_G(S):S|=|N_G(S)/S|$. Otherwise, by Cauchy's theorem there exists a nontrivial $p$-subgroup of $N_G(S)/S$ whose preimage under the projection $N_G(S) \to N_G(S)/S$ is $p$-subgroup properly containing $S$ contradicting the maximality of $S$.</p>