The Schreier index-rank formula

By [L1], choose a free basis $X$ of $F$ with $\mathrm{\mid }X\mathrm{\mid }=n$. Let $\mathrm{\Gamma }={\mathrm{Sch}}_X(H)$, and choose a rooted spanning tree $T$ in $\mathrm{\Gamma }$. Since the vertices of $\mathrm{\Gamma }$ are the right cosets of $H$, there are $d$ vertices. For each vertex and each $x\in X$, the Schreier graph has exactly one outgoing $x$-edge, so $\mathrm{\Gamma }$ has $dn$ positive labeled edges.

The tree $T$ has exactly $d-1$ edges. An edge $Ht\stackrel{x}{\to }Htx$ lies in $T$ exactly when the chosen representative of $Htx$ is $tx$, and then the corresponding Schreier generator is $tx(tx{)}^{-1}=1$. Every positive edge outside $T$ gives one nontrivial Schreier generator, so the number of nontrivial generators is $dn-(d-1)=1+d(n-1)$.

By [L3], the nontrivial generators counted in step 2.1 form a free basis of $H$. Therefore $H$ has finite rank and $\mathrm{rank}(H)=1+d(n-1)$.