Existence of m infinite subsets in an arbitrary group such that all products of one element from each (in order) are distinct

Question

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 $A_i$)

Answer 1

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), 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.

Answer 2

Define the sets $A_i$ inductively. At each stage $mq+r$ (with $0\leq r<m$) put one new element into $A_r$, being careful to keep all the products $a_0\cdots a_{r-1}$ with all $a_i\in A_i$ distinct. The distinctness requirement prohibits, at any stage, only finitely many elements from being put into the desired $A_r$, so, since $G$ is infinite, the inductive construction can proceed.

EDIT: As e1c25ec7 points out in a comment, this argument is incomplete. It can fail if a single distinctness requirement prohibits infinitely many elements $x$. That issue amounts to some product of previously added elements of the $A_i$'s and inverses thereof has an infinite centralizer. So one would need to either avoid producing such elements or exploit them (perhaps by producing an infinite abelian subgroup, as mentioned in the question).