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On connection between reducibility of an n-ary quasigroup and that of its retracts

An $n$-ary operation $Q:S^n\to S$ is called an $n$-ary quasigroup of order $|S|$ if in the equation $x_0=Q(x_1,...,x_n)$ knowledge of any $n$ elements of $x_0,...,x_n$ uniquely specifies the remaining one. An $n$-ary quasigroup $Q$ is (permutably) reducible if $Q(x_1,...,x_n)=P(R(x_{s(1)},...,x_{s(k)}),x_{s(k+1)},...,x_{s(n)})$ where $P$ and $R$ are $(n-k+1)$-ary and $k$-ary quasigroups, $s$ is a permutation, and $1<k<n$. An $m$-ary quasigroup $R$ is called a retract of $Q$ if it can be obtained from $Q$ or one of its inverses by fixing $n-m>0$ arguments. We show that every irreducible $n$-ary quasigroup has an irreducible $(n-1)$-ary or $(n-2)$-ary retract; moreover, if the order is finite and prime, then it has an irreducible $(n-1)$-ary retract. We apply this result to show that all $n$-ary quasigroups of order 5 or 7 whose all binary retracts are isotopic to $Z_5$ or $Z_7$ are reducible for $n>3$. Keywords: $n$-ary quasigroups, retracts, reducibility, latin hypercubes