Lemma 4: Let \varphi: G_{1} \longrightarrow G_{2} be surjective.  Then H is a separable subgroup of G_{2} if and only if \varphi^{-1}(H) is a separable subgroup of G_{1}.

Proof. (\Longrightarrow) If g_{1} \in G_{1} \smallsetminus \varphi^{-1}(H), then \varphi(g_{1}) \notin H. So there is K_{2} < G of finite index such that H_{2} \subset K_{2}, but \varphi(g_{1}) \notin K_{2}.  So K_{1} = \varphi^{-1}(K_{2}) is as required.

(\Longleftarrow) Suppose \varphi^{-1}(H) < G_{1} is separable.  Let g_{2} \in G_{2} \smallsetminus H.  There is g_{1} \in G_{1} \smallsetminus \varphi^{-1}(H) so that \varphi(g_{1}) = g_{2}.  Therefore, there exists K_{1} \supset \varphi^{-1}(H) of finite index in G_{1}, with g_{1} \notin K_{1}.  Now \varphi(K_{1}) = K_{2}.  Then H \subset K_{2}, and you can check that  g_{2} \notin K_{2}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 G \longrightarrow G with nontrivial kernel.

Example. Consider G = \mathbb{Z}^{\infty}.  Then the map G \longrightarrow G defined by (n_{1}, n_{2}, \ldots) \longmapsto (n_{2}, n_{3}, \ldots) is an epimorphism with nontrivial kernel.  Hence G is not Hopf.

Lemma 5 (Malcev): If a finitely generated group G is RF, then it is Hopf.

Let G be finitely generated and let K be a finite index subgroup.  The characteristic core of K 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 \alpha is an epimorphism and H is the characteristic core of K, then \alpha(H) = H.

Proof of Lemma 5. Let f: G \longrightarrow G be a surjection with g \in \ker(f) \smallsetminus \left\{ 1 \right\}. So there exists a finite index subgroup K < G such that g \notin K. Let H be the characteristic core of K. Then g \notin H.  Now f descends to a homomorphism \widehat{f}: G/H \longrightarrow G/H\widehat{f} is a surjection, but kills gH \neq H. QED

Theorem 4: Finitely generated free groups are Hopf.

Proof. Follows from the fact that free groups are RF and Lemma 5.

Corollary: If k > 1, then F_{k} \not\cong F_{l}.

Proof. There is a surjection F_{k} \longrightarrow F_{l} with nontrivial kernel.  If F_{k} \cong F_{l}, then we have contradicted the Hopf property. QED

Exercise 5. Give a different proof of the corollary above using homology and the Hurewicz Theorem.