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Consider an n-ary (polyadic) group, which is a generalization of the "usual" ("binary") groups to n-tuples. The definition requires appropriate standard generalizations of associativity and existence of inverse. Can this (or has this?) been generalized to arbitrary tuples? In other words, can the number of factors in the product be made arbitrary (but finite)? What "mathematical object" would this be? Does it have to be induced by some n-ary group or are there examples that are not?

( A useful classical paper with the relevant definitions is Polyadic Groups by Emil Leon Post from 1935: https://www.ams.org/journals/tran/1940-048-02/S0002-9947-1940-0002894-7/S0002-9947-1940-0002894-7.pdf )

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  • $\begingroup$ What is an "inverse" supposed to be if you are starting with a tenary operation $G\times G \times G \to G$? $\endgroup$
    – Yemon Choi
    Commented Feb 22, 2023 at 2:25
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    $\begingroup$ "Existence of inverse" in this case means that the equations xbc = d, axc = d, abx = d are uniquely solvable for all a, b, c, and d. $\endgroup$
    – P. Trinli
    Commented Feb 22, 2023 at 2:54
  • $\begingroup$ (Bad notation on my part, each of the equations should have its own quantifiers - i.e. me using the same variable names in all three is of no significance). $\endgroup$
    – P. Trinli
    Commented Feb 22, 2023 at 3:02
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    $\begingroup$ I have spent few years of my life for study of polyadic groups but honestly I couldn't understand your question. Do you mean an algebraic system $(G, f_1, f_2, f_3, \ldots)$ where each $f_n$ is an $n$-ary operation and every $(G, f_n)$ is a polyadic group, plus some relations among different $f_n$s (like $f_3(x_1, x_2, x_3)=f_2(f_2(x_1, x_2), x_3)$)? $\endgroup$
    – Sh.M1972
    Commented Feb 26, 2023 at 9:36
  • $\begingroup$ Sh.M1972, yes, exactly. $\endgroup$
    – P. Trinli
    Commented Feb 27, 2023 at 2:19

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Ok, I guess if associative law generalizes in the obvious way, the answer is trivial (unless I am missing something).

Denote group multiplication with brackets, such that e.g. in a binary group $a \cdot b \equiv [ab]$. In the case of an n-ary (polyadic) group, the associative law reads $$ [[a_1...a_n]a_{n+1}...a_{2n–1}] = [a_1...a_{i}[a_{i+1}...a_{i+n}]a_{i+n+1}...a_{2n–1}] \quad . $$ If one instead allows products of arbitrary arity, it seems that the associative law would generalize to e.g. imply $$ [abc]=[a[bc]] \quad . $$ Hence, binary multiplication determines multiplication of any arity and every object described in the question would simply be determined by a "usual" binary group.

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  • $\begingroup$ Is the answer more interesting if we only require arities strictly more than 2 (or any other fixed large number N)? $\endgroup$
    – Asvin
    Commented Mar 20 at 8:45
  • $\begingroup$ That's possible, have not thought about it. This would not define the "ordinary" polyadic group though. $\endgroup$
    – P. Trinli
    Commented Mar 26 at 16:02

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