Is it true that for every infinite group $G$ and every $m\in\mathbb{N}$ there are infinite subsets $A\_0,\dots,A\_{m-1}$ such that all the products $a\_0\cdot\dots\cdot a\_{m-1}$ with $a\_i\in A\_i$ are distinct?


(This is true at least when $G$ is either uncountable or contains an infinite abelian subgroup, and for $m=2$ this is true for all $G$. Also, one can always obtain arbitrarily large finite $A\_i$)



Existence of m infinite subsets in an arbitrary group such that all products of one element from each (in order) are distinct
Yes, such infinite subsets $A\_0, \dots, A\_{m-1}$ of an infinite group $G$ always exist. We can show this by wedging the problem into one of two polar opposite cases: abelian and ICC (i.e. every non-trivial element has an infinite conjugacy class).


**Fact.** An infinite group either has an infinite abelian subgroup or an infinite ICC quotient.


*Proof*. Let $FC(G) := \{ g \in G : [G : Z\_G(g)] < \infty \}$ denote the FC-centre of $G$ containing those elements whose centralizer is of finite index, or equivalently whose conjugacy class is finite. For $g, h \in FC(G)$ we have that $Z\_G(gh) \geq Z\_G(g) \cap Z\_G(h)$ and is also finite index, so $FC(G)$ is a subgroup of $G$ and in fact a characteristic one. There are two possibilities:


* If $FC(G)$ is infinite, then being an infinite FC-group (i.e. all conjugacy classes are finite) it contains an infinite abelian subgroup $\langle g\_0, g\_1, \dots \rangle$ constructed inductively: at each step, if the subgroup $\langle g\_0, g\_1, \dots, g\_i \rangle$ is finite, then its centralizer is a finite intersection of finite-index subgroups hence finite index itself and thus infinite so we can find some $g\_{i+1}$ commuting with and not contained in $\langle g\_0, g\_1, \dots, g\_i \rangle$.
* If $FC(G)$ is finite, then $G / FC(G)$ is infinite and ICC since the conjugacy class of any non-trivial $g FC(G) \in G / FC(G)$ will contain the infinitely many images of the conjugacy class of $g$ in $G$.


$\square$


So it suffices to prove the existence of such $A\_0, \dots, A\_{m-1}$ with the additional assumption that $G$ is abelian or ICC (since after finding such sets in a quotient of $G$, any choice of lifts to $G$ will still have the desired property). We now show that we can choose inductively $g\_0, g\_1, g\_2, \dots$ to put into the sets $A\_0, \dots, A\_{m-1}$ (say, cycling through the $A\_i$) so that at each step the desired uniqueness holds.


Start by putting $g\_0 = \dots = g\_{m-1} = 1$ into $A\_0, \dots, A\_{m-1}$. Suppose we are choosing $g\_n$ to add to $A\_i$. There are two cases to consider for distinct words $a\_0 a\_1 \dots a\_{m-1}$ and $b\_0 b\_1 \dots b\_{m-1}$ that possibly give equal group elements that involve $g\_n$ and otherwise only $g\_j$ with $j < n$: (1) $a\_i = g\_n$ and $b\_i \neq g\_n$, and (2) $a\_i = b\_i = g\_n$.


(1) simply rules out finitely many possibilities for $g\_n$.


(2) is equivalent to the equation $g\_n^{-1} (u^{-1} u') g\_n = v (v')^{-1}$ that we get by rewriting $u g\_n v = u' g\_n v'$ for $u = a\_0 \dots a\_{i-1}, v = a\_{i+1} \dots a\_{m-1}, u' = b\_0 \dots b\_{i-1}, v' = b\_{i+1} \dots b\_{m-1}$. Note that $u^{-1} u' \neq v (v')^{-1}$ as otherwise picking $a\_i = b\_i = 1$ would then give distinct words giving equal group elements, contrary to our inductive hypothesis. If $G$ is abelian then there is no solution to this conjugacy equation so we have no additional constraints on $g\_n$. If $G$ is ICC then the set of solutions is either empty (if $u^{-1} u'$ and $v (v')^{-1}$ are not conjugate) or is a coset of $Z\_G(u^{-1} u')$. In the latter case this is a coset of an infinite-index subgroup, by ICC, because $u^{-1} u' \neq v (v')^{-1}$ so their being conjugate rules out $u^{-1} u' = 1$.


Both (1) and (2) only rule out finitely many cosets of infinite-index subgroups (in (1) we have cosets of the trivial subgroup!), so since we cannot write an infinite group as a finite union of such cosets by a result of B.H. Neumann (see [MO question](https://mathoverflow.net/q/17396)), there is some choice of $g\_n$ for which neither (1) nor (2) can give a failure of the desired uniqueness, which completes the proof.


