Free groups of rank $\geq 2$ are commensurable I am reading Bowditch's notes on Geometric Group Theory and I am trying to understand a proof. He proves that if $F_p$, $F_q$ are free groups of rank $p,q\geq 2$, then they are commensurable, i.e. there are subgroups $G, G^{'}$ of $F_p$ and $F_q$, which are isomorphic, and have finite index. In the proof, he defines $K_n$ as the graph obtained by taking the circle, $R/nZ$, and attaching a loop at each point of $Z/nZ$ — that is $n$ additional circles. He proves that $\pi_1(K_n)=F_{n+1}$ (this is clear), but then states that $K_{mn}$ covers $K_n$, which I don't understand. Then he claims that $K_r$ covers both $K_p$ and $K_q$, where $r=(p-1)(q-1)+1$ and because of compactness, $F_r$ is a finite index subgroup of $F_p$ and $F_q$. Is there anyone who can help me understand the proof?

To describe the covering map $K_{mn} \mapsto K_n$, first map the main circle $R / (mnZ) \subset K_{mn}$ around the main circle $R / nZ \subset K_n$ by a covering map of degree $m$ (taking the point $x + mnZ$ to the point $x + nZ$ for each $x \in R$).

Next, the loop of $K_{mn}$ attached to the main circle at a vertex $p$ in $Z / (mnZ) \subset K_{mn}$ is mapped homeomorphically to the loop of $K_n$ attached to the main circle at the image of $p$ in $Z / nZ \subset K_n$.

The rest of the proof is standard covering space theory, explained in most algebraic topology textbooks. I recommend Hatcher's "Algebraic Topology", which has a beautiful page very early in the book showing many covering spaces of the graph $K_1$ whose fundamental group is $F_2$.