A semigroup $S$ is defined to be *squared* if there exists a subset $A\subseteq S$ such that the function $A\times A\to S$, $(x,y)\mapsto xy$, is bijective.

Problem:Is each squared finite group trivial?

**Remarks (corrected in an Edit).**

I learned this problem from my former Ph.D. student Volodymyr Gavrylkiv.

It can be shown that a group with two generators $a,b$ and relation $a^2=1$ is squared. So the adjective

*finite*is essential in the above problem.Computer calculations show that no group of order $<64$ is squared.

For any set $X$ the rectangular semigroup $S=X\times X$ endowed with the binary operation $(x,y)*(a,b)=(x,b)$ is squared. This follows from the observation that for the diagonal $D=\{(x,x):x\in X\}$ of $X\times X$, the map $D\times D\to S$, $(x,y)\mapsto xy$, is bijective. So restriction to

*groups*in the formulation of the Problem is also essential.