# 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?

• You mean $|G:H| = N$, no? – user61527 Feb 3 '14 at 4:49
• @T.Bongers Sorry, I've fixed it. Yes, I meant that $H$ is a subgroup of finite index in $F$. – user112072 Feb 3 '14 at 4:52
• This is a standard result of Schreier, and it would be more sensible to look it up in a book. – Derek Holt Feb 3 '14 at 9:29

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, $\mathrm{rk}(H)=1+n \cdot (\mathrm{rk}(F)-1)$.

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