**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.

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