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The free product $G_1\ast G_2$ is a group with homomorphisms from $G_1,G_2$ such that any pair of homomorphisms $G_i\to K$ extends uniquely to a homomorphism $G_1\ast G_2\to K$. Concretely its elements are reduced words with nonidentity syllables alternating between the factors. Multiply by concatenating and then combining adjacent syllables from the same factor, deleting identities. The empty word is the identity, and this normal form theorem for a free product also makes the factor maps injective.

The Klein combination theorem, in its ping-pong form, is as follows. Suppose two nontrivial subgroups $A,B$ act on a set $X$, with disjoint nonempty subsets $X_A,X_B$, such thatIf at least one factor has at least three elements, then $⟨A,B⟩\cong A\ast B$ through the natural factor maps. The cardinality condition is needed in this version: two order-two groups can have the same image swapping the two domains.

Here is the proof. Assume $|A|\ge 3$. A reduced word beginning and ending with an $A$-syllable sends $X_B$ into $X_A$ by successive application of the displayed inclusions, so cannot act as the identity. A word beginning and ending in $B$ becomes one of the preceding type on conjugating by any nonidentity $a\in A$. If it begins in $A$ and ends in $B$, choose $a\in A$ different from both $1$ and its first syllable. The word $a^{-1}wa$ remains reduced after combining its first two syllables and begins and ends in $A$, so again it is nonidentity. If it begins in $B$ and ends in $A$, apply this last argument to its inverse. Every nonempty reduced word is covered, so the natural homomorphism $A\ast B\to ⟨A,B⟩$ has trivial kernel. This proves the Klein combination theorem.

Apply it to the two cyclic groups generated by the given Möbius transformations. On $X=\mathbb{R}\cup \{\infty \}$ takeTheir powers areFor $m\ne 0$ and $|x|<1$, $|x+2m|\ge 2|m|-|x|>1$, so $f^m{X}_B\subseteq X_A$. For $x\in X_A$, put $1/\infty =0$; thenso $g^m{X}_A\subseteq X_B$. The powers displayed also show that both transformations have infinite order. The Klein combination theorem consequently gives

For the finite-index formula, realize the free group $F_n$ as the fundamental group of the rose with $n$ edges. A subgroup $H$ of index $i$ corresponds to a connected $i$-sheeted covering graph. Explicitly, its vertices are right cosets $Hx$ and for each basis generator $s_j$ there is a directed edge $Hx\to Hx{s}_j$ projecting to the corresponding rose edge. This graph is connected because basis words reach every coset, and the based loops correspond precisely to words in $H$.

The covering graph has $i$ vertices and $ni$ geometric edges: each positively oriented rose edge has one lift from each vertex. A maximal tree has $i-1$ edges, and the complementary edges freely generate its fundamental group. Thus $H$ is a free group with the Nielsen–Schreier formulaThis uses the connected cover associated with $H$; no claim that arbitrary disconnected covers are connected is needed.

Finally the specified basis of $F_n$ gives $d(F_n)\le n$. Its abelianization is ${\mathbb{Z}}^n$, by the universal property of the free group, and reduction modulo two gives a quotient $(\mathbb{Z}/2\mathbb{Z}{)}^n$. Images of any generating set must span this $n$-dimensional vector space, so it contains at least $n$ members. Therefore $\boxed{d(F_n)=n}$, and the same argument applies to the free group $H$ above.

Using exactly the unshifted definition printed in the paper, the unshifted rank gradients are determined byThe supremum is $n$, attained for $H=F_n$. For $n\ge 1$, mapping one basis generator onto a generator of $\mathbb{Z}/i\mathbb{Z}$ and the others to zero provides subgroups of every positive index $i$. Hence the infimum is $n-1$, approached as $i\to \infty$:For the trivial free group $F_0$, both values are zero. Subtracting one from $d(H)$ would change the upper answer; that is a different convention from the one requested here.