Subgroup of a (free) group. Let $F$ be a group, generated by $x_1,...,x_m$ and $H$ be its subgroup such that $|F:H|=n < \infty$. How to prove that $H$ can be generated by $n(m-1)+1$ elements?
 A: An argument I find really nice uses algebraic topology and finite graphs. More information may be found in Hatcher's or Massey's book, or on my blog and its references.
Let $F$ be a free group and $H$ be a subgroup. We view $F$ as the fundamental group of a graph $X$ (for instance, a bouquet of $\mathrm{rk}(F)$ circles). Let $Y \twoheadrightarrow X$ be a covering satisfying $\pi_1(Y) \simeq H$. In particular, the covering induces a structure of graph on $Y$ making the covering cellular. 
Furthermore, if $n=[F:H] < + \infty$, the covering $Y \twoheadrightarrow X$ is $n$-sheeted, hence $\chi(Y)=n \cdot \chi(X)$. If $T \subset X$ is a maximal subtree, then $ X$ is the disjoint union of $T$ and $\mathrm{rk}(F)$ edges, hence
$$\chi(X)= \chi(T)- \mathrm{rk}(F)=1- \mathrm{rk}(F);$$
in the same way, $\chi(Y)= 1- \mathrm{rk}(H)$. Therefore, $\mathrm{rk}(H)=1+n \cdot (\mathrm{rk}(F)-1)$. 
A: This question has been answered in a comment:

This is a standard result of Schreier, and it would be more sensible to look it up in a book. –  Derek Holt Feb 3 at 9:29 

