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Let $G$ be a finite group. I'm trying to understand the structure (if exists) of the set of functions $f:G\times G\to\mathbb{Z}_p$ ($p$ is a given prime) satisfying the condition:

$$f(a,b)+f(b,c)+f(c,a)=0$$

Where $a,b,c\in G$ and can be subjected to conditions if "all the elements of $G$" gives boring results (especially a condition like "$abc\in S$" for some subset $S\subseteq G$ is good for me).

Also, the most interesting thing for me are such functions that also satisfy the additional constraint $f(0,0)=1$.

Note that a basic structure always exists: such a function $f$ is given by a solution of a set of linear equations of the form $X_{a,b}+X_{b,c}+X_{c,a}=0$, the variables corresponding to the values of $f$. However, I'm afraid this does not give me sufficient insight.

A guiding example for me is group cohomology; the setting is almost the same but the equation defining 2-cocycles is slightly different ($f(a,bc)+f(b,c)-f(ab,c)-f(a,b)=0$ if I'm not mistaken)

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    I have a doubt that others may as well have: $f: G \to \mathbb Z_p$ . So, what is the meaning of $f(a,b)$. Do you mean that $f$ is function on the set product?2012-02-05
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    Yes, I meant $f:G\times G\to \mathbb{Z}_p$. Thank you.2012-02-05
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    I think, $f(a,b)=ab-ba$ is a good map to study and understand.2012-02-05
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    Did you mean $f(e,e)=0$ instead of $f(0,0)=1$? Assuming your $0$ denotes the identity of $G$, the condition $f(0,0)=1$ can only be satisfied if $p=3$.2012-02-05
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    Also note that your conditions do not involve the group structure of $G$ at all. You probably want to require a bit more than you did.2012-02-05
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    Marc, note that I allow you to limit the triples $a,b,c$ involved to those having the property $abc\in S$ for a given $S\subseteq G$. This solves the $f(0,0)=1$ problem if $0\notin S$, and involves the group operation on $G$.2012-02-05
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    @GadiA: But note that what you allow is _weakening_ the condition you impose; for an extreme case if $S=\emptyset$ you are no longer requiring anything for $f$. You cannot expect to find _more_ structure by weakening the constraint on $f$.2012-02-05
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    Marc, why not? A group has less conditions than a field, but no one will say they do not have an interesting structure - quite different than that of fields. Of course I don't want to drop ALL demands; in the setting the problem arose in I have a specific S of size 3.2012-02-05

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Not so much an answer as a brief remark. Your condition is cohomological in nature, but it might be better to think of singular cohomology rather than group cohomology (if this doesn't mean anything to you, then I would recommend Allan Hatchers (free!) text-book on algebraic topology - particularly the motivational subsection in the beginning of chapter 2...).

More specifically you could note that any function of the form $$ g(x,y) = h(x) - h(y)$$ where $$ h(x) = \sum_{g} a_{g}\delta_g(x) $$ where $\delta_g$ is the kronecker delta, and $a_n \in \mathbb{Z}_p$, will solve your functional equation. Hence the natural first step would be to mod out the $\mathbb{Z}_p$-module generated by your functions by this space of trivial solutions. Which will give you the first singular cohomolgy group. Of course the exact nature of this group will depend on the topological space, i.e. in your case presumably on the set $S$ in the condition $abc\in S$.

EDIT: Another usefull term in connection with your particular condition, might be the notion of a Cayley graph of a group (it seems to me that the case $S=1$ should correspond to the situation where the topological space whose first cohomology group we want, is the Cayley graph of the full group. Though I may very well be confusing things...).

EDIT 2: Hm.. Guess I missed the condition $f(0,0)=1$. To connect my idea with the problem as posed one would probably need to find a function $r : G \to \mathbb{Z}_p$ such that $$ r(0) =1 \ \text{ and } \ r(a)+r(b)+r(c) =0 \ \text{ whenever} \ abc\in S$$ and then define $$ f(x,y) = g(x,y) + \sum_g r(g) \delta_g(x)$$ (where $g$ is anti-symmetric, i.e. $g(x,y)=-g(y,x)$). As explained in the comments though, it seems hard to find such an $r$ in general - and for some choices of $S$ impossible.

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    Thank you, this is a very interesting answer.2012-02-05