Lemma 4: Let be surjective. Then is a separable subgroup of if and only if is a separable subgroup of .
Proof. If , then . So there is of finite index such that , but . So is as required.
Suppose is separable. Let . There is so that . Therefore, there exists of finite index in , with . Now . Then , and you can check that . QED
This lemma gives us examples of some ERF groups. For example, any finitely generated abelian group is ERF.
The Hopf Property
Definition. A group is not Hopf if and only if there is an epimorphism with nontrivial kernel.
Example. Consider . Then the map defined by is an epimorphism with nontrivial kernel. Hence is not Hopf.
Lemma 5 (Malcev): If a finitely generated group is RF, then it is Hopf.
Let be finitely generated and let be a finite index subgroup. The characteristic core of is defined to be
The characteristic core is of finite index by Exercise 2. It is also a normal subgroup, and one can check that if is an epimorphism and is the characteristic core of , then .
Proof of Lemma 5. Let be a surjection with . So there exists a finite index subgroup such that . Let be the characteristic core of . Then . Now descends to a homomorphism . is a surjection, but kills . QED
Theorem 4: Finitely generated free groups are Hopf.
Proof. Follows from the fact that free groups are RF and Lemma 5.
Corollary: If , then .
Proof. There is a surjection with nontrivial kernel. If , then we have contradicted the Hopf property. QED
Exercise 5. Give a different proof of the corollary above using homology and the Hurewicz Theorem.