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Two functions, $f$ and $g$, that satisfy the following identity: $$f(g(a_1,...,a_n), g(b_1,...,b_n),...) = g(f(a_1,b_1,....), f(a_2,b_2,...)...)$$ (notice the "transposition" of the arguments), do they possess a named and/or known property? Or as an alternative, where can I find out more about it?

(Apologies if I asked a duplicate question, but I didn't know how to search for it)

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    $\begingroup$ Maybe it would be helpful to write it $f\left(g(A^1),g(A^2),\ldots ,g(A^n)\right)=g\left(f(A_1),f(A_2),\ldots , f(A_n)\right)$ $\endgroup$ – Alexander Gruber Jul 17 '18 at 0:16
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Wikipedia calls it "$f$ and $g$ commute" for multivariate functions:

The notion of commutation also finds an interesting generalization in the multivariate case; a function $f$ of arity $n$ is said to commute with a function $g$ of arity $m$ if $f$ is a homomorphism preserving $g$, and vice versa i.e.: $${f\big(g(a_{11},\ldots ,a_{1m}),\ldots ,g(a_{n1},\ldots ,a_{nm})\big)=g\big(f(a_{11},\ldots ,a_{n1}),\ldots ,f(a_{1m},\ldots ,a_{nm})\big)} $$

The given reference is Universal Algebra: Fundamentals and Selected Topics by Clifford Bergman.

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    $\begingroup$ This was definitely reminding me of universal algebra stuff. shudders Good find. $\endgroup$ – Cameron Williams Jul 17 '18 at 1:02
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    $\begingroup$ It is called Bisymmetry in the context of functional equations. $\endgroup$ – Jens Schwaiger Jul 17 '18 at 1:47
  • $\begingroup$ @JensSchwaiger Thanks for that tidbit. You should post it as an answer, perhaps with some reference. $\endgroup$ – dxiv Jul 17 '18 at 2:27
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In section 5 of https://www.google.at/url?sa=t&source=web&rct=j&url=https://www.imbs.uci.edu/files/personnel/luce/2000/AczelFalmagneLuce_Mathematica%2520Japonica_2000.pdf&ved=2ahUKEwjS_cWRuqXcAhWNZ1AKHZ7BDGwQFjAEegQIAhAB&usg=AOvVaw3WLjcxovgzVUfeuIABNdBJ

you my find this equation of generalized bisymmetry.

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