The two cyclic basis factors of the rank-two free group are self-commensurating with trivial cross-conjugate intersections

For nonzero $k\in \mathbb{Z}$, the reduced word for $a^k$ has length $\mathrm{\mid }k\mathrm{\mid }$, so [F6] gives $a^k\mathrm{\ne }e$; the same holds for $b^k$. If $a^m=a^n$, then [F8] gives $a^{m-n}=e$, forcing $m=n$ by the preceding fact; likewise the powers of $b$ are distinct. By [F9], the power maps $\mathbb{Z}\to A$ and $\mathbb{Z}\to B$ are surjective, and they are injective by these distinctness arguments; [F8] makes them homomorphisms. Thus $A$ and $B$ are infinite cyclic. By [F5], their free product is free on the canonical copies of $a,b$. Let $\varphi :A\ast B\to F$ be induced by the factor inclusions using [F2], and let $\psi :F\to A\ast B$ send the free basis $a,b$ to those copies using [F1]. The composite $\varphi \psi$ fixes $a,b$, so it is ${\mathrm{i}\mathrm{d}}_F$ by [F1]; the composite $\psi \varphi$ restricts to the identity on each factor, so it is ${\mathrm{i}\mathrm{d}}_{A\ast B}$ by [F2]. Hence $\varphi$ is an isomorphism and we identify $F=A\ast B$. The factor maps are injective because each nonidentity factor element is a nonempty reduced word by [F3], which also gives the reduced syllable normal form.

Let $g\mathrm{\notin }A$. Its reduced syllable form, after removing an initial and terminal $A$-syllable when present, is $g=a_{0}w{a}_1$ with $a_0,a_1\in A$ and a nonempty reduced word $w$ beginning and ending in nonidentity $B$-syllables. For $x\in A\setminus \{e\}$, cyclicity of $A$ gives $g^{-1}xg=a_1^{-1}{w}^{-1}{a}_0^{-1}x{a}_{0}w{a}_1=a_1^{-1}{w}^{-1}xw{a}_1$. The middle word $w^{-1}xw$ is reduced and contains $B$-syllables on both ends; multiplication by the outer $A$-elements cannot cancel those syllables. By [F3] this element is not in $A$. Therefore $g^{-1}Ag\cap A=\{e\}$ for every $g\mathrm{\notin }A$. Interchanging $A$ and $B$ gives $g^{-1}Bg\cap B=\{e\}$ for every $g\mathrm{\notin }B$.

In the identification of step 1.1, let ${\rho }_A:F=A\ast B\to A$ be the retraction which is the identity on $A$ and trivial on $B$, supplied by the universal property [F2]. If $g^{-1}{b}^{k}g=a^{\mathrm{\ell }}$ lies in $g^{-1}Bg\cap A$, then applying ${\rho }_A$ gives $e=a^{\mathrm{\ell }}$, so the intersection element is $e$. Thus $g^{-1}Bg\cap A=\{e\}$ for every $g$. The retraction ${\rho }_B:F\to B$ proves $g^{-1}Ag\cap B=\{e\}$ for every $g$.

By [F10], every singleton in $F$ is open; by [F11], each singleton rectangle in $F\times F$ is open, so the product topology on $F\times F$ is discrete. For either multiplication or inversion, every open set containing the image of a point has an open preimage containing that point, because the domain is discrete; [F12] therefore gives continuity. Thus [F13] makes $F$ a topological group. Every subgroup of $F$ is open by [F10], so the commensurator definition [F4] applies to both $A$ and $B$. If $g\in A$, then $g^{-1}Ag=A$, so both indices in [F4] are $1$. If $g\mathrm{\notin }A$, step 2.1 gives $A\cap g^{-1}Ag=\{e\}$, whose index in the infinite cyclic group $A$ is infinite; hence $g\mathrm{\notin }{\mathrm{Comm}}_F(A)$. Therefore ${\mathrm{Comm}}_F(A)=A$. The same argument with $B$ gives ${\mathrm{Comm}}_F(B)=B$, and step 2.2 gives the two cross-factor intersections in the statement. AC is the stated inherited assumption [F7]; the proof steps use no further choice.