5
$\begingroup$

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.

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ For Abelian groups, I get about half that size. How do you get larger? Gerhard "Maybe I Am Multiplying Wrong?" Paseman, 2017.10.21. $\endgroup$ – Gerhard Paseman Oct 21 '17 at 21:02
  • $\begingroup$ 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. $\endgroup$ – Dillon M Oct 21 '17 at 21:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Oct 21 '17 at 21:15
  • 1
    $\begingroup$ @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 $|A| \le 1$). $\endgroup$ – Salvo Tringali Oct 21 '17 at 21:54
  • 1
    $\begingroup$ 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 $|A^2|\le 1+n(n-1)/2$. $\endgroup$ – YCor Oct 21 '17 at 22:08
6
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
1
$\begingroup$

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?

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.