# Unit-product sets in finite decomposable sets in groups

A non-empty subset $$D$$ of a group is called decomposable if each element $$x\in D$$ can be written as the product $$x=yz$$ for some $$y,z\in D$$.

Problem. Let $$D$$ be a finite decomposable subset of a group. Is there a sequence $$x_1,\dots,x_n$$ of pairwise distinct elements of $$D$$ such that $$x_1\cdots x_n=1$$?

Remark 1. This problem is a non-commutative version of this still open problem posed by Zaimi in 2010 on MO. So, maybe in the non-commutative case there is a counterexample?

Remark 2. For any finite decomposable set $$D$$ it is possible to find a sequence $$z_1,\dots,z_n\in D$$ of length $$n\le|D|$$ such that $$z_1\cdots z_n=1$$.

Indeed, take any $$y_0\in D$$ and by induction for every $$i\ge |D|$$ find elements $$x_i,y_i\in D$$ such that $$x_iy_i=y_{i-1}$$. By the Pigeonhole Principle, there are numbers $$0\le i such that $$y_i=y_j$$. Then $$y_j=y_i=x_{i+1}y_{i+1}=x_{i+1}x_{i+2}y_{i+2}=\dots=x_{i+1}x_{i+2}\cdots x_jy_{j}$$and hence $$x_{i+1}\cdots x_j=1$$. Then for $$n=j-i$$ and $$(z_k)_{k=1}^n=(x_{k+n})_{k=1}^n$$ we have $$z_1\cdots z_n=x_{i+1}\cdots x_j=1$$. But in general, the elements $$z_1,\dots,z_n$$ are not pairwise distinct.

• If no error (and allowing relations of the type $x=y^2$) it seems there's no solution with $|D|=3$, which you maybe already checked. For $|D|=4$ there are many more possibilities but it's maybe still doable. – YCor Mar 12 at 9:21
• @YCor Truly speaking I expect a counterexample, maybe constructed by tools of the small cancellation theory. – Taras Banakh Mar 12 at 9:41