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 elements
$$
ts\overline{ts}^{-1}
\qquad(t\in T, s\in S).
$$
There 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>
$$
\boxed{d(H)\leq1+d\bigl(d(G)-1\bigr).}
$$

Solution

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

{c}

If $H$ has finite index $d$ in a <finitely generated group> $G$, then <Schreier's lemma> gives
$$
\boxed{d(H)\leq 1+d\bigl(d(G)-1\bigr).}
$$
Equality holds when $G$ is a <free group> of finite rank.

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