Let $A$ be a finite subset of the group $H$. I am interested in sets with the property that


(1)$\qquad\qquad |\{ab\ \colon\ (a,b)\in A\times A\}| = |A|^{2}$.


Thus $A$ has property (1) if the product set $A^{2}$ is as large as possible.


Is it true that any infinite group has arbitrarily large finite sets satisfying (1)?


I can construct arbitrarily large such sets if $H$ is abelian, or non-Noetherian, or has infinite exponent. So the only case I am not sure about is the case of nonabelian, Noetherian groups with finite exponent.


**EDIT.** As pointed out in the comments, the original question does not make sense for abelian groups. A modified question that does apply to all groups is as follows: can we find sets $A$ and $B$ such that $\min\{|A|,|B|\}$ is arbitrarily large, and $|\{ab\ \colon (a,b)\in A\times B\}|=|A|\cdot|B|$. The comment by @YCor provides us with a method for constructing such sets.



Maximum-sized product sets in infinite groups
Indeed we even have, in a infinite group: for every finite subset $S$ and $n$ there exist a finite subset $F$ of cardinal $n$ such that the multiplication is injective on $S\times F$. Indeed, first find $F'=\{x\_1,\dots,x\_{n-1}\}$ of cardinal $n-1$ (by induction). Choose $x\_n$ distinct from $s^{-1}tx\_i$ for all $s,t\in S$ and $i<n$. So $Sx\_n$ is disjoint from $SF'$ and thus, writing $F=F'\cup\{x\_n\}$, we have $|SF|=|S|n$. 


As I said in my original comment, the same argument yields the existence, for all $n\ge 1$, of a subset $A$ of cardinal $n$ such that $|A^2|=1+n(n-1)/2$; this is optimal as we can see in an elementary abelian 2-group. As observed in the comments, in an abelian clearly we cannot do better than $n(n+1)/2$; conversely this bound can probably be achieved in any group of infinite exponent (using the existence of cyclic subgroups of arbitrary order). Also $n^2$ can be achieved in any non-abelian free group, hence in any group containing a non-abelian free subgroup.


