# Maximum-sized product sets in infinite groups

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.

• For Abelian groups, I get about half that size. How do you get larger? Gerhard "Maybe I Am Multiplying Wrong?" Paseman, 2017.10.21. – Gerhard Paseman Oct 21 '17 at 21:02
• Now that I think about it further, I am only confident in the case of finitely-generated abelian groups. Since any such abelian group that is infinite must have an element of infinite order, the construction becomes easy. So I retract my claim about general abelian groups, for now. – Dillon M Oct 21 '17 at 21:07
• I am pretty sure at least one of us is confused. There are torsion infinite Abelian groups, and since ab=ba, I don't see how you can get an image that is much more than half the number of products you take. Gerhard "Maybe Abelian Means Something Else?" Paseman, 2017.10.21. – Gerhard Paseman Oct 21 '17 at 21:15
• @DillonM: Gerhard Paseman's observation applies to any (multiplicatively written) abelian group $G$, no matter whether $G$ is finitely generated or not (given a finite set $A \subseteq G$, it's easily seen that $|A^2| \le \frac{1}{2}|A|(1+|A|)$, so that $|A^2| = |A|^2$ iff $|A| \le 1$). – Salvo Tringali Oct 21 '17 at 21:54
• Any infinite group has a subset $A$ of each size $n\ge 1$ with $|A^2|\ge 1+n(n-1)/2$, by induction on $n$. Indeed, this is clear for $n=1$; if proved for $n$, choose $x_1,\dots,x_n$ with at least $1+n(n-1)/2$ products, and choose $x_{n+1}$ distinct from $x_i^{-1}x_jx_k$ for all $i,j,k$. Then the $x_{n+1}x_i$ are pairwise distinct, and distinct from all $x_jx_k$, so we get $1+n(n-1)/2+n=1+n(n+1)/2$. This is optimal since in a group of exponent two every $A$ of size $n$ satisfies $|A^2|\le 1+n(n-1)/2$. – YCor Oct 21 '17 at 22:08

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.

We can prove a more stronger result as follows.

Theorem. Suppose $G$ is an infinite group with no nontrivial finite conjugacy classes. Then there exists a subset $X$ of $G$ with $|X|=n$ such that $|X^2|=n^2$ for every $n\geq0$.

Proof. We construct $X$ by induction. Clearly, the result holds for $n=0$. Suppose $n\geq1$ and the result holds for $n-1$. Let $X\subseteq G\setminus\{1\}$ be such that $|X|=n-1$ and $|X^2|=(n-1)^2$. An element $x\in G\setminus X$ satisfies $|(X\cup\{x\})^2|=n^2$ if and only if

$$x\notin X^{-1}X^2\cup X^2X^{-1}\cup\bigcup_{\substack{a,b\in X\\a^G=b^G}}C_G(a)g_{a,b},$$ where $g_{a,b}\in G$ satisfies $a^{g_{a,b}}=b$ whenever $a,b\in X$ and $a^G=b^G$. If $$G=X^{-1}X^2\cup X^2X^{-1}\cup\bigcup_{\substack{a,b\in X\\a^G=b^G}}C_G(a)g_{a,b},$$ then $$G=\bigcup_{g\in X^{-1}X^2\cup X^2X^{-1}}\langle g\rangle \cup\bigcup_{\substack{a,b\in X\\a^G=b^G}}C_G(a)g_{a,b}.$$ Now, as $\langle g\rangle$ and $C_G(a)$ have infinite indices in $G$ for all $g\in X^{-1}X^2\cup X^2X^{-1}$ and $a\in X$, we get a contradiction by Theorem 4.4 of Neumann in Groups covered by permutable subsets. This contradiction guarantees the existence of $x$ and proves the theorem.

A similar argument shows that

Theorem. Let $G$ be an infinite group possessing an element $x$ with finite centralizer. Then there exists a subset $X$ of $x^G$ with $|X|=n$ such that $|X^2|=n^2$ for every $n\geq0$.

A natural question to ask is:

Question. Is the same result holds for any infinite group that is not an FC-group?