# Name of a group-like structure

The late Vladimir Arnold, in

Arnold, V., Arithmetics of binary quadratic forms, symmetry of their continued fractions and geometry of their de Sitter world, Bull. Braz. Math. Soc. (N.S.) 34, No. 1, 1-42 (2003). ZBL1044.11016.

introduced (in the context of binary quadratic forms, but the concept is general) the following definition (in multiplicative notation): a subset $$\mathcal{A}$$ of a semigroup $$\mathcal{S}$$ has the trigroup property if for any triplet $$(a_1,a_2,a_3)\in\mathcal{A}^3$$ one always has $$a_1a_2a_3\in\mathcal{A}$$. Remark: of course $$a_1a_2$$, $$a_1a_3$$ or $$a_2a_3$$ might not be in $$\mathcal{A}$$ (that's the point).

I have been working on another topic where the following generalisation pops up naturally: a subset $$\mathcal{A}$$ of a semigroup $$\mathcal{S}$$ has the $$n$$-group property if for any $$n$$-tuple $$(a_1,\dots ,a_n)\in\mathcal{A}^n$$ one always has $$\prod_{i=1}^na_i\in\mathcal{A}$$. Remark: some of the smaller products $$\prod_{j\in J,1<|J| might not be in $$\mathcal{A}$$.

Question: (a) has that generalisation been already studied, perhaps with a different name ? As far as I can tell from papers citing Arnold's it's not the case, but maybe it came up before his paper. (b) in the event it has no name yet, is the name $$n$$-group property reasonable, or would it be confusing with something else ?

[Edited several times to take into account the comments.]

• What you call a "set" is a set equipped with a binary operation, right? – abx Dec 29 '18 at 7:40
• Arnold wouldn't make such an abstract setting :) He considers subsets of the set of integers, under multiplication. – YCor Dec 29 '18 at 7:47
• @abx: yes, you are correct, apologies. Now corrected. – Thomas Sauvaget Dec 29 '18 at 8:10
• And actually, it's not $\mathcal{A}$ that is equipped with a binary operation, but a larger set... so $\mathcal{A}$ should be assumed to be a subset of a semigroup. – YCor Dec 29 '18 at 9:21
• $n$-submonoid sounds fine and clear to me (with dash outside the math mode!), if it's assumed to contain the unit. Otherwise, $n$-subsemigroup. – YCor Dec 29 '18 at 19:02

$$n$$-group property is not a good idea, because it's a substructure (and also because groups have inverses). One option is $$n$$-subsemigroup (or any obvious variant such as $$n$$-fold subsemigroup, $$[n]$$-subsemigroup, if any reason to do so)...

As mentioned in the comments, an $$n$$-subsemigroup with unit is just the same as a submonoid, so no need to define $$n$$-submonoid.

[Note that it's the name of a substructure, not of a structure. Finding axioms for the structure itself is not obvious, and possibly not clearly defined (one way to define it, using the language of universal algeba, would be to characterize, in sets endowed with an $$n$$-ary law, the subvariety generated by semigroups with their $$n$$-ary law). For instance, there is a classical notion of Lie triple system.]

• Thank you for this. A substructure is fine for my purposes, I shall not venture into trying to to find the axioms for the structure (albeit this looks interesting for its own sake). – Thomas Sauvaget Jan 2 at 15:18

This isn't an answer but it's too long for a comment. I suggest that $$n-1$$ is more important than $$n$$ in this context, for the following reason. Suppose $$A$$ is an $$n$$-group in a semigroup $$S$$. Then, for any positive integer $$k$$, we can define $$A^k$$ to be the set of all products of $$k$$ factors from $$A$$. The definition of $$n$$-group says $$A^n\subseteq A=A^1$$, and it follows that $$A^k\subseteq A^r$$ where $$r$$ is the remainder when $$k$$ is divided by $$n-1$$ (I take remainders to be in the range $$1\leq r\leq n-1$$ rather than the customary $$0\leq r\leq n-2$$ because there is no $$A^0$$).

The union $$\bigcup_kA^k$$ is a subsemigroup $$S'$$ of $$S$$. If the $$A^k$$'s for $$1\leq k are pairwise disjoint, then we get a homomorphism from $$S'$$ to the additive group $$\mathbb Z/(n-1)$$ by sending all elements of $$A^k$$ to $$k$$. Conversely, any homomorphism $$h$$ from a subsemigroup of $$S$$ to $$\mathbb Z/(n-1)$$ gives an $$n$$-group, namely $$h^{-1}\{1\}$$. ("Conversely" may be an overstatement here, since the two processes are, in general, inverse to each other only on one side.) The situation where the $$A^k$$'s are not pairwise disjoint looks considerably more complicated, but maybe someone can provide some insight into it.

• @Qfwfq You're right, if there's an identity element present. I had assumed that there wouldn't be, because if $1\in A$ then the trigroup or $n$-group property for $A$ just says that $A$ is a semigroup (take all but 2 of the $n$ factors to be $1$). But I overlooked the possibility that there might be an identity element in $S$ but not in $A$. – Andreas Blass Jan 1 at 17:37
• Oh yeah, sure. I read "semigroup" but thought "monoid"... – Qfwfq Jan 1 at 19:23