3
$\begingroup$

Let $(A,f)$ be an $n$-ary semigroup ($n \ge 2$). Then there exists a ($2$-ary) semigroup $(\overline A,*)$ with an inclusion homomorphism $A \hookrightarrow \overline A$ such that that the restriction of its $n$-ary derivation to $A$ coincides with $f$, i.e. $(*_n)|_A = f$.

Is this true?

One possible direction would be, $(A,f) \cong \mathcal F_n(A) / \sigma$, and then $(\overline A, *) = \mathcal F_2(A) / \langle\sigma\rangle$ with the usual inclusion $\mathcal F_n(A) \hookrightarrow \mathcal F_2(A)$ would satisfy the theorem, if one could show that $\langle\sigma\rangle \cap \mathcal F_n(A)^2 = \sigma$.

(Where, $\mathcal F_n(A)$ is the free $n$-ary semigroup of strings $s$ which have length $|s| \equiv 1 \bmod (n-1)$ with entries in $A$ and operation $n$-ary concatenation. And $\langle\sigma\rangle$ is the smallest congruence on $\mathcal F_2(A)$ containing $\sigma$.)

edit: The terms "$n$-ary semigroup" and "derived" are used in the sense of this paper: http://www.quasigroups.eu/contents/download/2006/14_14.pdf .

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ It is not clear what the source algebra is. What relations/equations do n-ary semigroups satisfy? What arity does f have? Do you want the inclusion i and the existence of a term t of * of the appropriate arity such that i(f(stuff_from_A...)) = t(i(stuff_from_A)...)? Your F_n(A) does not seem to be closed under the operation as I understand it. Gerhard "It Must Be Something Else" Paseman, 2015.09.22 $\endgroup$
    – Gerhard Paseman
    Commented Sep 22, 2015 at 17:14
  • $\begingroup$ I have added a reference for the terminology. $\mathcal F_n(A)$ is closed under the operation, as the concatenation of any $n$ strings $s_1, \cdots, s_n$ which have $|s_i| \equiv 1 \bmod (n-1)$ will again have length $|s_1\cdots s_n| \equiv 1 \bmod (n-1)$. $\endgroup$
    – JustAskin
    Commented Sep 22, 2015 at 17:26

1 Answer 1

Reset to default
1
$\begingroup$

Summary of what I have learned wrt this question. It turns out that the relevant construction is the free covering semigroup (first constructed by G. Cupona; thanks to W. Dudek for pointing this out to me), i.e. we have a functor $\mathsf{FreeCov}: \mathsf{nSgrp} \to \mathsf{Sgrp}$ which sends $\mathcal F_n(A) / \sigma \mapsto \mathcal F_2(A) / \sigma$ (as I suspected). We also have the n-ary derivation which is a functor $d_n: \mathsf{Sgrp} \to \mathsf{nSgrp}$. My question at hand (existence of an injection of any n-semigroup into a binary-derived n-semigroup) is satisfied by the evident injection $$\mathcal F_n(A) / \sigma \hookrightarrow \mathcal d_n(\mathsf{FreeCov}(\mathcal F_n(A) / \sigma))$$

Also, it just so happens that $\mathsf{FreeCov}$ is left adjoint to $d_n$. Drawing from the pattern of $\mathsf{Free} \dashv \mathsf{Forgetful}$ we might wonder if $d_n$ is in some sense "forgetful". Yes. If we think of $\mathcal F_2(A) / \sigma$ as "knowing the value of" each string $a$, $ab$, $abc$, $abcd$, $abcde$ etc. (in the sense that the quotient by $\sigma$ defines a valuation map (of sets) $|\mathcal F_2(A)| \to A$, where $|-|:\mathsf{nSgrp} \to \mathsf{Set}$ is the underlying-set forgetful functor), then (fixing $n=4$ for example) $d_4(\mathcal F_2(A) / \sigma) = \mathcal F_4(A) / (\sigma \cap \mathcal F_4(A)^2)$ only "knows the value of" the strings with length $|s|\equiv 1 \bmod 3$, like $a$, $abcd$, $abcdefg$, etc., so it has "forgotten" the value of the strings of the other lengths.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .