tag:blogger.com,1999:blog-264226589944705290.post9028544002401122656..comments2022-05-19T07:55:48.367-07:00Comments on God Plays Dice: Are most groups solvable?Michael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-264226589944705290.post-55417165803792563192008-03-20T19:57:00.000-07:002008-03-20T19:57:00.000-07:00I am told by a local expert that "almost all" grou...I am told by a local expert that "almost all" groups are solvable, in fact, nilpotent of class at most two, meaning solvable in at most two steps. So depending on how exactly this density is counted (I can think of at least four ways), I would say that a "random" group has 100% probability of being solvable. JVUnknownhttps://www.blogger.com/profile/08030082338335679725noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-83323670620312289192008-03-20T10:53:00.000-07:002008-03-20T10:53:00.000-07:00Susan, here's the connection: consider what it loo...Susan, here's the connection: consider what it looks like to take an nth root in the complex plane. You divide the plane up into n slices, and rotating the plane one nth of a turn is a symmetry of the whole situation. But rotations by nths of turns make up the cyclic group Cn! So a step in a solution by radicals that involves taking an nth root is tied to building up the symmetry group of your polynomial by a cyclic group of order n.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-16653879735658269062008-03-19T23:21:00.000-07:002008-03-19T23:21:00.000-07:00Hmmmâ€¦ I'm not familiar enough with group theory to...Hmmmâ€¦ I'm not familiar enough with group theory to understand the definition of solvable group you've given here, but in my abstract algebra class we've briefly studied solvable groups in the context of Galois theory and the idea of polynomials being solvable by radicals. (I assume the definition I learned is equivalent.) In this sense it might make sense to talk about picking a group at random--if I pick a polynomial at random and want to know if it is solvable by radicals, I'll look at its Galois group and determine whether that is solvable.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-76996988930696553282008-03-19T20:24:00.000-07:002008-03-19T20:24:00.000-07:00A couple ways the statement could make sense. If w...A couple ways the statement could make sense. <BR/><BR/>If we let f(n) denote the number of solvable groups of order n, and g(n) denote the total number of groups of order n, the statement could mean:<BR/><BR/>1. f(n)/g(n) tends to 1 as n tends to infinity (a group chosen randomly among the groups of a fixed large order tends to be solvable).<BR/>2. (The sum of f(i) from i=1 to n)/(The sum of g(i) from i=1 to n) tends to 1 as n tends to infinity (a group chosen randomly among the groups of at most a given fixed large order tends to be solvable). <BR/>3.Anonymousnoreply@blogger.com