From: "Asger Grunnet" <asger@adslhome.dk>
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" <cbrown@cbrownsystems.com> wrote in message news:3BF3145D.3DFEAF6E@cbrownsystems.com...
>
>
> Asger Grunnet wrote:
> >
>
> <snip>
>
> > 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)) ).
> >
>
> <snip>
>
> > 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.