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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 $0\le i<j\le n$ 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.

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  • $\begingroup$ 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. $\endgroup$ – YCor Mar 12 at 9:21
  • $\begingroup$ @YCor Truly speaking I expect a counterexample, maybe constructed by tools of the small cancellation theory. $\endgroup$ – Taras Banakh Mar 12 at 9:41

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