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I am looking whether there is any compact way to write the following: Suppose we have an abelian group $G$. For a subset $A\subset G$ let $S_A$ be the sum of its elements. I want to find the number of triples of $k$-element subsets $(A,B,C)$ with $S_A+S_B+S_C=0$.

There is an obvious way to do it by just summing over any triple $(x,y,z)\in G$ with $x+y+z=0$ and writing how many sets have $S_A=x$, $S_B=y$, $S_C=z$ (there are expressions for these counts in the literature), but this does not lead to a nice compact expression, or at least I do not see the simplifications.

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    $\begingroup$ Let $G$ have multiplicative notation, let $\chi$ run over all characters on $G$, then the number of your triples is $n^{-1}(\sum_{|A|=k}\chi(\prod(A))^3$. For computing the sum $\sum_{|A|=k} \chi(\prod(A))$, you may use inclusion-exclusion: say, for $k=3$ you write in the group algebra: $\sum_{|A|=3} \prod(A)=\frac16 (\sum_{g\in G} g)^3-\frac12\sum_{g\in G} g^2\cdot \sum_{h\in G} h+\frac13\sum_{g\in G} g^3$, and apply to this equation the character $\chi$ (extended to the group algebra). $\endgroup$
    – Fedor Petrov
    Commented Nov 21, 2021 at 11:11
  • $\begingroup$ (continuation) Usually everything cancels, you need only the unit character and the characters which are unit on the subgroup of cubes. $\endgroup$
    – Fedor Petrov
    Commented Nov 21, 2021 at 11:11
  • $\begingroup$ Hi Fedor! Great idea. To formalize what you wrote in general you need to write the elementary symmetric sum in terms of the power sum. We have the the green boxed formula ( for f_X(m)) math.stackexchange.com/questions/3983165/… $\endgroup$
    – Vlad Matei
    Commented Nov 21, 2021 at 14:16
  • $\begingroup$ But now it still seems delicate since we sum over partitions and we have to be careful which parts in our partition of $k$ share a common factor with $|G|$. $\endgroup$
    – Vlad Matei
    Commented Nov 21, 2021 at 14:19
  • $\begingroup$ Yes, but it seems to be unavoidable, because the answer really depends on arithmetic properties of $|G|$ $\endgroup$
    – Fedor Petrov
    Commented Nov 21, 2021 at 14:51

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