### 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 c_{1}, ..., c

_{r}in G, such that b

^{c1}... b

^{cr}= h. (The notation b

^{c}means b conjugated by c, i.e. b

^{c}= 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 c_{1}, ..., c

_{r}such that b

^{c1}... b

^{cr}= h. Define the function f[g, h] by:

f[g,h](x) = f(a g^{-1}x)^{c1} ... f(a g^{-1}x)^{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### 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. Using Lemma 1 we can find elements c

_{1}, ..., c

_{r}in G such that b

^{c1}... b

^{cr}= d, so if we define the function f as:

f(x) = x^{c1} ... x^{cr},

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 thatSince G is non-abelian simple and therefore has trivial center, we can find an element c in G, such that c f(b) c^{-1}f(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)^{-1}f(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.