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Let $G$ be a group, and for each ordered $n$ tuple $(g_1,...g_n)$ of elements of $G$, consider the function $f_n$ that outputs the number of permutations $\sigma\in S_n$ for which $g_{\sigma(1)}g_{\sigma(2)}...g_{\sigma(n)}=e$.

Is the group determined up to isomorphism by these functions $f_n$?

More precisely, if we have a set $G$, and two collections of functions $f^1_n,f^2_n$ on $n$ tuples of elements of $G$, such that there exists group structures $\mu_1,\mu_2:G\times G\rightarrow G$, $\epsilon_1,\epsilon_2:\ast\rightarrow G$ on $G$ giving rise to $f^i_n$ in the manner described above, is there a bijection $X\rightarrow X$ which identifies the groups $(\mu_1,\epsilon_1)$, and $(\mu_2,\epsilon_2)$?

Note that we can recover identity from $f_1$, inverses from $f_2$, and by looking at $3$ tuples, we can partially deduce the group law, in that we can say when $gh=k$ or $hg=k$ holds for any $g,h,k\in G$. So in particular, this has an affirmative answer when $G$ is abelian. For the general case, I wasn't able to find a similarly elementary argument.

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    $\begingroup$ In the title "multiset" sounds weird, as for a multiset in a non-abelian group, to "multiply to 1" doesn't make sense. The first question is a bit unclear: is $n$ fixed? it seems not, after reading the sequel. Also in the second question you mean "unique up to isomorphism", or maybe up to permutation of $X$ preserving $f$. $\endgroup$
    – YCor
    Commented Jun 17, 2020 at 13:32
  • $\begingroup$ Edited for clarity, I wasn't quite sure how to phrase the title, if you can think of better wording that would be very helpful. $\endgroup$
    – Chris H
    Commented Jun 17, 2020 at 13:49
  • $\begingroup$ Actually what you count is more precise, since you retain the number of permutations. $\endgroup$
    – YCor
    Commented Jun 17, 2020 at 13:51
  • $\begingroup$ It's equivalent to know, for each $g$ and $h$, the multiset $\{gh, hg\}$. (And knowing that multiset is equivalent to knowing the set.) You'e shown that counting gives us this information. Conversely, if we have $n$ elements, and count all the ways to choose two of them, multiply them in any order, then choose another, and multiply it in any order, and so on, until all are chosen, we will count each ordering the same number $2^{n-2}$ of times. So to recover your information, we can sum over the possible pairwise products and then divide by $2^{n-2}$. $\endgroup$
    – Will Sawin
    Commented Jun 17, 2020 at 14:38
  • $\begingroup$ Some more things you can recover from the $f_n$: orders of elements, subgroups, when two elements commute, centralisers, centre, conjugacy classes, normal subgroups, .... I don't know enough group theory to know if a subset of these properties recovers the group (maybe assuming $G$ is finite), but this is all pretty strong evidence that at least $(G,\mu_1)$ and $(G,\mu_2)$ are abstractly isomorphic. $\endgroup$ Commented Jun 18, 2020 at 4:14

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