Proof of proposition 1

Submitted by asger on Tue, 01/22/2019 - 16:49

Proposition 1:

If G is a finite non-abelian simple group, then Fu(G) is the set of all functions on G.

In order to prove the proposition, we need a few lemmas. In the following it is implicitly assumed that G refers to a finite non-abelian simple group.

Lemma 1:

Let h, b be elements in G with b not equal to 1. Then there exist a positive integer r and elements c1, ..., cr in G, such that bc1 ... bcr = h. (The notation bc means b conjugated by c, i.e. bc = c b c-1).

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 is not 1 and G is simple, we must have N = G. This means that any element of G can be represented as a product of conjugates of b. This proves Lemma 1.

Lemma 2:

Let a, b be elements in G with b not equal to 1. If Fu(G) contains a function f, such that f(a) = b and f(x) = 1 for all x different from a, then Fu(G) is the set of all functions on G.

Proof.

Let g and h be arbitrary elements in G. By Lemma 1 we can find c1, ..., cr such that bc1 ... bcr = h. Define the function f[g, h] by:

 f[g,h](x) = f(a g-1x)c1 ... f(a g-1x)cr.

The function f[g, h] is in Fu(G) and has f[g, h](g) = h and f[g, h](x) = 1 for x different from g. Let F be any function on G. From the equation

  F(x) = y∈G f[y, F(y)](x)

we see that F is in Fu(G).  Proof of Lemma 2 complete.

Lemma 3:

For any four elements a, b, c, d in G, with a different from 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-1b. Furthermore we can assume that c = 1, by substituting the function x → c-1f(x)  for f , and replacing d by c-1d. Using Lemma 1 we can find elements c1, ..., cr in G such that bc1 ... bcr = d, so if we define the function f as:

f(x) = xc1 ... xcr,

we have f(a) = f(1) = 1 = c, and f(b) = d. Proof of Lemma 3 complete.

Proof of Proposition 1.

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 Lemma 2 ensures the conclusion of the proposition. Assume therefore that there exist two different elements a and b in G such that f(a) and f(b) are not equal to 1.

Since G is non-abelian simple and therefore has trivial center, we can find an element c in G, such that c f(b) c-1f(b)-1 is not 1. By Lemma 3 we can find a function e in Fu(G) such that e(a) = 1 and e(b) = c. Consider the function F in Fu(G) given by:

F(x) = e(x) f(x) e(x)-1f(x)-1.

We have F(a) equal to 1,  F(b) not equal to 1 and F(x) = 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.