Corollary 2.5: two elements whose translates of a set are disjoint and stay inside it generate a free subsemigroup

Let $G$ be a group, let $a, b \in G$, and let $A \subseteq G$ be a nonempty subset such that

• $aA \cup bA \subseteq A$ — for every $x \in A$, both $ax \in A$ and $bx \in A$; and
• $aA$ and $bA$ are disjoint — for all $x, y \in A$, $ax \neq by$.

Then $a$ and $b$ generate a free subsemigroup: distinct words in two letters take distinct values when the letters are read as $a$ and $b$.

The conclusion, precisely. Let $W$ be the free monoid on two letters — all finite words, including the empty one, under concatenation — and let $\varphi : W \to G$ send a word $\ell_1 \cdots \ell_k$ to the product of the corresponding elements, multiplied left to right in the order the letters occur, with the empty word sent to $1$. Only positive words occur: no letter stands for an inverse and no cancellation is performed. The assertion is that $\varphi$ is injective. Since the empty word is included, this says in particular that no nonempty positive word in $a$ and $b$ equals $1$, and it gives $\exists a, b$ with $\varphi$ injective — the published predicate Chou.HasFreeSubsemigroupOfRankTwo — for the group $G$. It does not assert anything about inverses, so it is not a statement about a free subgroup.

What the hypotheses force. They are not vacuous, and they are satisfiable: one witness is the group of permutations of $\mathbb{Q}$ with $a : t \mapsto 2t$, $b : t \mapsto 2t + 1$ and $A$ the set of values of nonempty positive words. Two consequences are worth naming because they show the hypotheses are strong. Taking $y = x$ in the disjointness condition forces $a \neq b$. And the two conditions together force $A$, hence $G$, to be infinite: a finite $A$ would have to contain the two disjoint sets $aA$ and $bA$, each of the same size as $A$. So every instance with $G$ finite is vacuous.

The nonemptiness of $A$ is a genuine hypothesis and is not recorded in the declaration's name; without it the disjointness condition holds trivially for any $a, b$ and the conclusion fails.