Proof. We have to show that, if $G$ is a divisible Group, $\varphi: U \to G$ is any homomorphism, and $U$ is a subgroup of a Group $H$ , there is a homomorphism $\psi: H \to G$ such that the restriction $\psi|_U = \varphi$ . In other words, we want to extend $\varphi$ to a homomorphism $H \to G$ .

Let $\mathcal{D}$ be the set of pairs $(K, \psi)$ such that $K$ is a subgroup of $G$ containing $U$ and $\psi: K \to G$ is a homomorphism with $\psi|_U = \varphi$ . Then $\mathcal{D}$ ist non-empty since it contains $(U, \varphi)$ , and it is partially ordered by $$ (K, \psi) \leq (K', \psi') :\Longleftrightarrow K \subseteq K' \text{ and } \psi'|_K = \psi. $$

For any ascending chain $$ (K_1, \psi_1) \leq (K_2, \psi_2) \leq \dots, $$ in $\mathcal{D}$ , the pair $(\bigcup_{i \in \mathbb{N}} K_i, \bigcup_{i \in \mathbb{N}} \psi_i)$ is in $\mathcal{D}$ , and it is an upper bound for this chain. Therefore, by Zorn's Lemma, $\mathcal{D}$ contains a maximal element $(M, \chi)$ .

It remains to show that $M = H$ . Suppose the opposite, and let $h \in H \setminus M$ . Let $\langle h \rangle$ denote the subgroup of $H$ generated by $h$ . If $\langle h \rangle \cap M = \{0\}$ , the sum $M + \langle h \rangle$ is in fact a direct sum, and we can extend $\chi$ to $M + \langle h \rangle$ by choosing an arbitrary image of $h$ in $G$ and extending linearly. This contradicts the maximality of $(M, \chi)$ .

Let us therefore suppose $\langle h \rangle \cap M$ contains an element $nh$ , with $n \in \mathbb{N}$ minimal. Since $nh \in M$ , and $\chi$ is defined on $M$ , $\chi(nh)$ exists, and furthermore, since $G$ is divisible, there is a $g \in G$ such that $ng = \chi(nh)$ . It is now easy to see that we can extend $\chi$ to $M + \langle h \rangle$ by defining $\chi(h) := g$ , in contradiction to the maximality of $(M, \chi)$ .

Therefore, $M = H$ . This proves the statement.