From: "Asger Grunnet" Subject: Re: Groups with the same normal structure Date: Thu, 15 Nov 2001 22:11:07 +0100 Newsgroups: sci.math Summary: All functions on finite simple groups have "polynomial" form "Chas F Brown" wrote in message news:3BF3145D.3DFEAF6E@cbrownsystems.com... > > > Asger Grunnet wrote: > > > > > > > DEFINITION 2. For a finite group G, the set Fu(G) of functions G -> G > > is defined as all functions of the form > > > > x -> a_1 * x * a_2 * x * ... * x * a_n > > > > for group elements a_1, ..., a_n and a positive integer n. > > (Here * denotes the group multiplication.) > > > > Notice that Fu(G) is a semigroup with respect to functional > > composition ( (f, g) -> fog, where (fog)(x) = f(g(x)) ). > > > > > > > Finally, when G and H are non-isomorphic non-abelian simple groups > > of the same order (IIRC, this can happen for some of the orthogonal > > simple groups), we also have that Fu(G) and Fu(H) are isomorphic. > > This last fact follows from a slightly surprising proposition: > > > > PROPOSITION. If G is a non-abelian simple group, then Fu(G) > > is the set of all functions G -> G. > > > > This is more than just slightly suprising - can you show a proof? > > Cheers - Chas > > --------------------------------------------------- > C Brown Systems Designs > Multimedia Environments for Museums and Theme Parks > --------------------------------------------------- In the following I assume that G is a finite non-abelian simple group. LEMMA 1: Let a, b be elements in G with b =/= 1. If Fu(G) contains a function f, such that f(a) = b and f(x) = 1 for x =/= a, then Fu(G) is the set of all functions on G. PROOF: Let C denote the conjugacy class in G containing b. The subgroup N generated by C will be a normal subgroup of G. Since b =/= 1 and G is simple, we have that N = G. Let g and h be arbitrary elements in G. Using that N = G, we can find elements a_1, a_2, ..., a_r in G such that (a_1 b a_1^-1)(a_2 b a_2^-1) ... (a_r b a_r^-1) = h. The function f[g, h] defined by f[g,h](x) = (a_1 f(a g^-1 x) a_1^-1) ... (a_r f(a g^-1 x) a_r^-1) is in Fu(G) and has f[g,h](g) = h and f[g,h](x) = 1 for x =/= g. Let f' be any function on G. From the equation f'(x) = Prod( f[y, f'(y)](x), y in G) we see that f' is in Fu(G). Proof of lemma 1 complete. LEMMA 2: For any four elements a, b, c, d in G with a =/= b, there is a function f in Fu(G) such that f(a) = c and f(b) = d. PROOF: Without loss of generality we can assume that a = 1, by substituting the function x -> f(ax) for f, and replacing b by a^-1 b. Furthermore we can assume that c = 1, by substituting the function x -> c^-1 f(x) for f , and replacing d by c^-1 d. As in lemma 1 we can find elements a_1, ..., a_r in G such that (a_1 b a_1^-1) ... (a_r b a_r^-1) = d. So if we define the function f as f(x) = (a_1 x a_1^-1) ... (a_r x a_r^-1), we have that f(a) = f(1) = 1 = c, and f(b) = d. Proof of lemma complete. PROOF OF THE PROPOSITION: For any function f in Fu(G), let V(f) denote the set {x in G | f(x) = 1}. Choose a non-constant function f in Fu(G), such that |V(f)| is maximal. (Since f is not constant, we obviously have |V(f)| < |G|). If |V(f)| = |G| - 1, then we are done by lemma 1. Assume therefore that there exist two different elements a and b in G such that f(a), f(b) =/= 1. Since G is non-abelian simple, we can find an element c in G, such that c f(b) c^-1 f(b)^-1 =/= 1. By lemma 2 we can find a function f' in Fu(G) such that f'(a) = 1 and f'(b) = c. Consider the function F in Fu(G) given by: F(x) = f'(x) f(x) f'(x)^-1 f(x)^-1. We have F(a) = f'(a) f(a) f'(a)^-1 f(a)^-1 = 1 f(a) 1^-1 f(a)^-1 = 1, F(b) = f'(b) f(b) f'(b)^-1 f(b)^-1 = c f(b) c^-1 f(b)^-1 =/= 1, and F(x) = f'(x) f(x) f'(x)^-1 f(x)^-1 = f'(x) f'(x)^-1 = 1 for x in V(f). From this we see that |V(F)| > |V(f)|, which violates the maximality of |V(f)|. This contradiction completes the proof. I hope I haven't left out too many details. I have a bad tendency to do that sometimes. Feel free to ask if there is anything you don't understand. Asger.