How to make a function depending on some operation? Let $S$ be an infinite set and $f:S\times S\rightarrow\mathbb R$ be bounded.
Question: Are there simple hypotheses on $f$ such that there is a commutative and associative operation $\cdot$ on $S$ such that $f(x,y)=\phi(x\cdot y)$, for some $\phi$?
Example: if $f(x,y)=f(y,x)$, we can construct a commutative $\cdot$ as follows: since $S$ is infinite, the image of $f$ has cardinality at most $|S|$ and then there is an injective mapping $\Psi:Im(f)\rightarrow S$. Define $x\cdot y=\Psi(f(x,y))$. This is a commutative operation which verifies the required property, but, of course, in general it is not associative.
And it is not clear to me how to find a simple enough condition that guarantees associativity.
Thanks in advance for any help,
Valerio
 A: I do not think there is a nice solution. But the problem can be simplified a bit. First the condition that $f$ is bounded is irrelevant since you can always compose with $\arctan$. Second, the condition that $f$ is a function into $\mathbb{R}$ is not important, and in fact the only thing you need from $f$ is its kernel, i.e. the equivalence relation on $S\times S$:
$(a,b)\sim (c,d)$ iff $f(a,b)=f(c,d)$. Thus the problem is this: suppose that we are given an equivalence relation $\sim$ on $S\times S$ such that $(a,b)\sim (c,d)\to (b,a)\sim (d,c)$. When will there exists an associative commutative operation $\cdot$ such that $(a,b)\sim (c,d)$ if $a\cdot b=c\cdot d$. One can of course use the structure theory of commutative semigroups (these are more complicated than commutative groups but still manageable) to obtain 
more information but I am not sure very much can be achieved.  
A: Here is an approach you might take.  Consider the class of , oh, lets call them left-sections of f(x,y): fix x and define g_x(y) = f(x,y). Consider the equivalence classes induced by level sets of g_x.  Associativity combined with the shape of the level sets of g_x, g_z, and g_w will determine whether it is feasible to have w = x.z .  At the very least you should have some interesting conditions on f which will be necessary for the representation.  It may be that Green's relations will prove useful here.
Gerhard "Not an Expert in Semigroups" Paseman,  2012.01.11
