Lemma 3

Let A and B be commutative groups such that A is divisible. Let C be a subgroup of B and $\varphi$ a morphism of C into A. There exists a morphism of B into A which extends $\varphi$.

Let $\mathscr{O}$ be the set of pairs $(X, f)$, where X is a subgroup of B containing C and $f$ a morphism of X into A extending $\varphi$. Let us order $\mathscr{O}$ by the relation “ $X\subset X'$ and $f'$ extends $f$ ”. One verifies that $\mathscr{O}$ is inductive. Let $(X, f)$ be a maximal element of $\mathscr{O}$ (E, III, p. 20 , Theorem 2). If $X\not= B$, let us take an element $b$ of B-X and let $X'$ be the subgroup generated by X and $b$. The commutativity of B shows that $X'$ is the set of elements $b^nx$ for $n\in \mathbf{Z}$ and $x\in X$. Suppose first that $b^n\notin X$ for every integer $n\not= 0$ and define $f'$ of $X'$ into A by taking an arbitrary element $y\in A$ and putting $f'(b^nx) =y^nf(x)$ for every $n\in \mathbf{Z}$ and every $x\in X$. Since A is commutative, $f'$ is a morphism, and it extends $f$. Suppose now that there exists $n\not= 0$ such that $b^n\in X$ and let $m >0$ be such that $m\mathbf{Z}=\{n\in \mathbf{Z}|b^n\in X\}$. Since A is divisible, there exists an element $y\in A$ such that $y^m=f(b^m)$. One then extends $f$ to a morphism of $X'$ into A by $f'(b^nx) =y^nf(x)$ for $n\in \{0,1, . . . , m-1\}$ and $x\in X$. In both cases, $(X, f)$ would not be maximal. Hence X = B and the lemma is proved.