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 $\prod_{j\in J,1<|J|<n}a_j$ 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.]
 A: 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 $1\leq k<n$ 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.
A: $n$-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.]
