Nielsen–Schreier theorem

Remarks

A topological proof: $F$ is the fundamental group of a bouquet of circles, a graph. Then $H$ is the fundamental group of a covering space, which is again a graph. The fundamental group of a connected graph is free. For the rank formula, the covering graph of degree $e$ of a bouquet of $n$ circles has $e$ vertices and $en$ edges. So its Euler characteristic is $e(1 - n)$. A connected graph with Euler characteristic $\chi$ has free fundamental group of rank $1 - \chi$. An algebraic proof uses a Schreier transversal and gives explicit free generators, the Schreier generators. Example: in the free group $F(a, b)$, let $H$ be the kernel of the map to $\Z/2$ sending $a$ and $b$ to $1$. It has index $2$, so it is free of rank $1 + 2 \cdot 1 = 3$; a free basis is $a^2$, $ab$ and $ab^{-1}$. So a subgroup of a free group can have larger rank than the group. The free group of rank $2$ contains free subgroups of every finite rank, and of countably infinite rank, such as its commutator subgroup. The finite index hypothesis is needed in the rank formula: the commutator subgroup of $F(a, b)$ has infinite index and is not finitely generated.

Going deeper

For a student

The theorem is one of the starting points of combinatorial and geometric group theory. The covering space proof is the model for Bass–Serre theory, where groups acting on trees are studied, and the Kurosh subgroup theorem is the analogue for subgroups of free products. The rank formula is a form of the multiplicativity of the Euler characteristic under finite coverings.