Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 1

The Nielsen–Schreier theorem says that every subgroup of a free group is free. Realize $F_n$ as the fundamental group of the rose $R_n$, the graph with one vertex and $n$ oriented loops. A subgroup $H\le F_n$ of finite index of a subgroup $d$ corresponds to a connected $d$-sheeted covering graph $X\to R_n$. The graph $X$ has $d$ vertices and $dn$ unoriented edges. Choosing a spanning tree leavesedges outside the tree, and these freely generate ${\pi }_1(X)\cong H$. Thus the Nielsen–Schreier formula is

Take the free abelian group $G={\mathbb{Z}}^2$, whose rank of a group is two, and its finite-index subgroup $H=2\mathbb{Z}\times \mathbb{Z}$. The subgroup has index two and is again isomorphic to ${\mathbb{Z}}^2$, so its rank is two. The Nielsen–Schreier formula would instead give $1+2(2-1)=3$. Henceis a counterexample.

Let $r=d(G)$ be the rank of a group, let $S$ be a generating set of size $r$, and suppose $[G:H]=d$. Choose a Schreier transversal adapted to a spanning tree in the Schreier coset graph. By Schreier's lemma, $H$ is generated by the elementsThere are $dr$ candidates. The $d-1$ oriented edges in the spanning tree give trivial candidates, leaving at most $dr-(d-1)$. This proves the Schreier index-rank inequality

Let $G$ have a generating set of a group with $r$ elements. Every subgroup $H$ of index $n$ gives a transitive group action of $G$ on the $n$ left cosets of $H$, hence a group homomorphism ${\rho }_H:G\to S_n$ after the cosets are labelled. Conversely, $H$ is the stabilizer subgroup of one point in this action.

A homomorphism $G\to S_n$ is determined by the images of the $r$ generators, so there are at most $(n!{)}^r$ such homomorphisms and at most $n(n!{)}^r$ point stabilizers. Therefore the finiteness of subgroups of a fixed finite index gives only finitely many subgroups of index $n$.