Fact: There exists a finitely generated non-Hopf group. (An example is the Baumslag-Solitar group , although we cannot prove it yet.) So, by Lemma 5, there is a finitely generated non-residually finite group. Thus, free groups are not ERF: if is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection is not separable in . However, finitely generated subgroups of free groups are separable:
Marshall Hall’s Theorem (1949): is LERF.
This proof is associated with Stallings.
Proof: As usual, let where is a rose. Let be a covering map with finitely generated. Let be compact. We need to embed in an intermediate finite-sheeted covering.
Enlarging if necessary, we may assume that is connected and that . Note that we have . By Theorem 5 (see below), the immersion extends to a covering . Then . So lifts to a map .
The main tool in the proof above is this:
Theorem 5: The immersion can be completed to a finite-sheeted covering into which embeds:
Proof: Color and orient the edges of . Any combinatorial map of graphs corresponds uniquely to a coloring and orientation on the edges of . A combinatorial map is an immersion if and only if at every vertex of , we see each color arriving at most once and leaving at most once. Likewise, it’s a covering map if and only if at each vertex, we see each color arriving exactly once and leaving exactly once.
Let be the number of vertices of . For each color , let be the number of edges of colored . Then there are vertices of missing “arriving” edges colored , and there are vertices of missing “leaving” edges colored . Choose any bijection between these two sets and use this to glue in edges colored . When this is done for all colors, the resulting map is clearly a covering.
Note that the proof in fact gives us more. For instance:
Exercise 6: If is a finitely generated subgroup of , then is a free factor of a finite-index subgroup of .
Exercise 7 (Greenberg’s Theorem): If and is finitely generated, then is of finite index in .