From: Robin Chapman Subject: Re: Finite group theory questions Date: Fri, 14 Jul 2000 07:17:37 GMT Newsgroups: sci.math Summary: Hall's theorem counting group elements dividing a conjugacy class In article <8kl98o$189$1@nnrp1.deja.com>, achava@hotmail.com wrote: > Hi all, > > I believe it was proved by Burnside that if G is a finite group and > if n divides the order of G, then the number of solutions to x^n = 1 is > a multiple of n. I also believe that the only known proof of this > result uses group representation theory and that no "elementary" proof > of this theorem is known (at least as of about 1975). Are these beliefs > actually true? No. There's an elementary proof in Marshall Hall's book. He attributes the result to Frobenius. > Also, is it true that the number of solutions to x^n = > g, where again n divides the order of G, but g is any element of G also > must be a multiple of n. Here is Theorem 9.1.1. in Hall: Let G be a group of order g and let C be a class of h conjugate elements. The number of solutions of x^n = c, where c ranges over C is a multiple of gcd(hn,g). -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Robin Chapman Subject: Re: Finite group theory questions Date: Fri, 14 Jul 2000 07:55:19 GMT Newsgroups: sci.math In article <8kmeq9$rcv$1@nnrp1.deja.com>, Robin Chapman wrote: > > Here is Theorem 9.1.1. in Hall: > Let G be a group of order g and let C be a class of h conjugate > elements. The number of solutions of x^n = c, where c ranges over C > is a multiple of gcd(hn,g). > For convenience I summarize Hall's proof. The result is trivial when g = 1 or n = 1 so use induction, assuming that the result is true for g' < g and for g' = g and n' < n. For a subset K of G let A(K,n) be the set of x in G with x^n in K and a(K,n) = |A(K,n)|. Case (i) h > 1. In this case a(C,n) = h a(c,n) for any c in H. If x^n = c, then x lies in C_G(c), the centralizer of c in G, which has order g/h < g. By the inductive hypothesis a(c,n) is a multiple of gcd(n,g/h) so a(C,n) is a multiple of gcd(hn,g). Case(ii) h = 1 and n is not a prime power. Write n = n_1 n_2 with n_1 and n_2 coprime and n_1, n_2 > 1. Then A(C,n) = A(D,n_1) where D = A(C,n_2). As D is a union of conjugacy classes then gcd(n_2,g) divides a(C,n). Similarly gcd(n_1,g) divides a(C,n). As n_1 and n_2 are coprime so are gcd(n_1,g) and gcd(n_2,g) and so thei product is gcd(n,g) and that divides a(C,n). Case(iii) h = 1, n = p^e with p prime and p divides the order of c in G (where C = {c}). Let u be the order of c. Each element x of A(c,n) has order nu and exactly n elements in the cyclic subgroup generated by x lie in A(n,c). Thus n divides a(n,c). Case(iv) h = 1, n = p^e with p prime and p is coprime to the order of c in G (where C = {c}). Note that c is in the centre Z of G. Let B be the subset of Z consisting of those elements with orders not divisible by p. As Z is Abelian then B is a subgroup of order b not divisible by p. Then c is in B. If c_1 and c_2 are in B then c_1 = c_2 y^n for some y in B. The map x |--> xy is a bijection from A(c_2,n) to A(c_1,n) so that a(c,n) is independent of the choice of c in B. We now have the sum g = b a(c,n) + sum_C a(C,n) where C runs through the conjugacy classes of G disjoint from B. All of these classes fall under an earlier case so we conclude that b a(c,n) is divisible by gcd(n,g). As gcd(n,g) is coprime to b then a(c,n) is divisible by gcd(n,g). -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy.