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).
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I learned this problem from my former Ph.D. student Volodymyr Gavrylkiv.
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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.
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Computer calculations show that no group of order $<64$ is squared.
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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.
Best Answer
I think, as implicitly suggested by Yemon Choi, it is possible to explain the proof of the answer of user49822 by making more use of idempotents. Suppose that the finite group $G$ is squared via the subset $A$. The element $ e = \frac{1}{|G|}\sum_{g \in G} g$ is a primitive idempotent of $\mathbb{C}G.$
Let $ f= \frac{1}{|A|}\sum_{a \in A} a.$ Then we have $f^{2} = ef = fe = e = e^{2}$. Thus $(f-e)^{2} = 0 = e(f-e) = (f-e)e.$ Now $f = e +(f-e)$ is the sum of commuting matrices (in the regular representation of $\mathbb{C}G$, say) with the second matrix nilpotent.
Thus $f$ has trace $1$ in the regular representation of $\mathbb{C}G.$ This forces $1 \in A$ since all non-identitiy elements of $G$ have trace zero in the regular representation. But then $A = \{1 \}$, since (as in Jeremy Rickard's comment) if $a \neq 1 \in A$, then $a = 1a = a1$ gives two different expressions for $a$. Alternatively, (using traces) the fact that $1$ appears with coefficient $|G|^{-1}$ in $e$ tells us that $1$ appears with multiplicity $|G|^{-1}$ in $f$ as well, so that $\sqrt{|G|} = |A| = |G|$ and $|G| = 1.$