Simple groups with the same cardinality as PSL_2(Z/p) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:35:38Z http://mathoverflow.net/feeds/question/61348 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p Simple groups with the same cardinality as PSL_2(Z/p) paul Monsky 2011-04-11T23:44:23Z 2012-08-23T07:39:15Z <p>In an undergrad honors algebra course it's sometimes shown that when $p$ is prime and $>3$ then $PSL_2(Z/p)$ is simple of of order $p(p-1)(p+1)/2$. But that this is the "only" simple group having that order is seldom or never (except when $p=5$) proved.</p> <p>I've worked out a fairly simple proof, using the Burnside transfer theorem, of this last result, but it's perhaps a little too intricate to present in class.</p> <p>QUESTION: Are there proofs of this result, on-line or in texts, that are appropriate for an undergrad honors algebra class? (If not, I might post an argument on arXiv).</p> <p>EDIT: Since no simple available proof has yet been found, I'll sketch the argument that I culled from classification arguments for Zassenhaus groups.</p> <p>Suppose G is simple of order p(p-1)(p+1)/2. First one shows that G has p+1 p-Sylows. Let S be the union of Z/p and {infinity}. An easy study of the conjugation action of G on the p-Sylows allows one to identify G with a doubly transitive group of (even) permutations of S, containing the p-cycle z-->z+1. Then the subgroup of G fixing both 0 and infinity is cyclic generated by z-->cz for some c. Once this is done, the key is in showing:</p> <p>A.--- The subgroup of elements that either fix 0 and infinity or interchange them is dihedral.</p> <p>Once A is shown it's not hard to show that z-->-1/z is in G, thereby identifying G with a fractional linear group. The proof of A is a counting argument when p=1 mod 4. But when p=3 mod 4 the situation is more delicate, and one uses Burnside tranfer.</p> http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p/61424#61424 Answer by Jim Humphreys for Simple groups with the same cardinality as PSL_2(Z/p) Jim Humphreys 2011-04-12T16:56:59Z 2011-04-12T16:56:59Z <p>I think the comments here converge toward the answer "no" to your basic question. While some very special small primes are treated in a few textbooks (using what seem to be <em>ad hoc</em> methods), it's much more natural to treat the entire family of groups at once in the wider context of the classification of finite simple groups. Here there should be some general methods in play, e.g., BN-pair structure, structure of "local" subgroups such as the normalizers of Sylow subgroups, centralizers of involutions, etc. Not to mention many techniques from character theory. </p> <p>I'm not aware of any treatment of this special family of rank 1 groups in a form suitable even for an honors course. The two books Danny Gorenstein wrote on finite groups, especially the second one published in 1982 when he and others felt the classification was complete, illustrate the difficulty at that time in organizing the subject in textbook form. In that later book the special result you want is embedded in a substantial treatment involving doubly transitive permutation groups, Zassenhaus groups, etc. I don't think it's feasible to extract from that textbook version an efficient proof for your purpose without bypassing the important underlying general ideas, though it's always interesting to see what can be written down in a more elementary and self-contained way. </p> <p>Anyway, the problem here is to work out in a very limited case the "recognition" theorem for finite simple groups: how do you argue that an unknown simple group (here specified just by an order formula) is isomorphic to some known group? </p> http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p/62013#62013 Answer by Geoff Robinson for Simple groups with the same cardinality as PSL_2(Z/p) Geoff Robinson 2011-04-17T08:32:57Z 2011-04-17T08:32:57Z <p>I am not sure that I know a really elementary proof of this fact, though I can imagine ways to start reconstructing one (Sylow theory and, as you say, transfer seem good places to start). But, in the end, you need a way to recognise this simple group. Seeing later comments, five pages sounds reasonable. There is a Theorem of M. Herzog (circa 1970) which (roughly speaking) characterizes finite groups of order less than $p^{3a}$ with a cyclic Sylow $p$-subgroup of order $p^{a}$, but this uses fairly sophisticated character theory, and (including background) needs a lot more than five pages.</p> http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p/62284#62284 Answer by paul Monsky for Simple groups with the same cardinality as PSL_2(Z/p) paul Monsky 2011-04-19T15:26:26Z 2011-07-22T21:27:21Z <p>I seem to nave reinvented the wheel! I quote Gallian from his 1976 survey of the current state of the classification problem (Mathematics Magazine 1976):</p> <p>In 1902 Frobenius determined all transitive permutation groups of degree p+1 and order p(p^2-1)/2 where p is prime.... it follows ... that PSL(2,p) is the only simple group of order p(p^2-1)/2. This appears to be the first arithmetical characterization of an infinite family of simple groups.</p> <p>I haven't been able to access the Frobenius paper, but I expect that I've more or less duplicated his argument. Thanks Junkie for your comments which led me to the Frobenius paper and the Gallian article citing it.</p> <p>EDIT: My version of the Frobenius paper is now up on arXiv: (GR) 1107.4130, "Frobenius' result on simple groups of order (p^3-p)/2." I abandoned my Burnside transfer approach for p=3 mod 4, replacing it with a simplification of Frobenius' more elementary proof. I thank those here who led me to his elegant article. Those who would like to use my note in the classroom can of course do so, but I disclaim responsibility.</p> http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p/62308#62308 Answer by Jim Humphreys for Simple groups with the same cardinality as PSL_2(Z/p) Jim Humphreys 2011-04-19T18:02:47Z 2011-04-19T18:02:47Z <p>This being community wiki, I'll add another "answer" in response to Paul's recent post. It's clear that the desired uniqueness statement for simple groups of this special order is embedded in the CFSG, so the question is whether there is a way to isolate the result for an honors class in a self-contained treatment (modulo knowledge of Sylow theorems and the like). Probably what Paul has arrived at in his own way is about as efficient as possible. </p> <p>As he points out, one wants to show that a Sylow <code>$p$</code>-subgroup of the unknown <code>$G$</code> is of index <code>$(p-1)/2$</code> in its normalizer (which plays the role of Borel subgroup <code>$B$</code> in a BN-pair of rank 1). Then you are looking at a doubly transitive permutation group on the <code>$p+1$</code> elements of the coset space <code>$G/B$</code> (thought of as rational points of a projective line). </p> <p>I've just looked at the relevant Frobenius paper, being probably the only person in the U.S. with all three volumes of his collected papers on a shelf several feet from my home computer. (Not that I've read them all. It was an extra set owned by Wilhelm Magnus.) The 19 page 1902 paper <em>Uber Gruppen des Grades <code>$p$</code> oder <code>$p+1$</code></em> was published in the <em>Sitzungsberichte ...</em>; it appears as number 66 in the third volume of collected papers containing later work on finite groups and character theory.</p> <p>The paper is somewhat electic, but the early Satz II affirms for any prime (except 7) the existence of a unique transitive permutation group of degree <code>$p+1$</code> and order <code>$p(p^2-1)/2$</code>. This requires four pages or so of heavy proof, with reference back to Sylow and others. It's hard to see at a glance how this translates into current terminology, but for example he sees very soon that the group permutes <code>$p+1$</code> objects which he denotes <code>$\infty, 0, 1, \dots, p-1$</code>. Once all the work is done, it follows as Satz III that for <code>$p&gt;3$</code> there is a unique simple group of the given order. </p> <p>Is it true that no later textbook proof has actually been given? It's not a result to be taken up in a standard elementary course, or even a graduate course where there is too much else to cover, but it illustrates nicely the starting point for CFSG beyond alternating groups. </p>