5. The Hopf Property

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

latexcharcore

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