Proposition 7

— Let $G$ be a commutative group annihilated by $n$ and $H$ a subgroup of $G$. Then the restriction homomorphism $\operatorname{Hom}(G, T) \to \operatorname{Hom}(H, T)$ is surjective.

For let $f : H \to T$ be a homomorphism; we shall prove that there exists a homomorphism of $G$ into $T$ extending $f$. Suppose first that $G$ is cyclic, generated by an element $g$ of order $r$ dividing $n$; denote by $t$ a generator of $T$. There exists a divisor s of r such that H is generated by $g^s$ (I, p. 50, Prop. 19), and for every $x \in \mathbf{Z}$ we have $f(g^{sx}) = t^{ax}$, where a is an integer such that n divides $ar/s$. Then $a/s = (ar/ns)(n/r)$ is an integer and the homomorphism $g^x \mapsto t^{(a/s)x}$, $x \in \mathbf{Z}$, of G into T extends f. In the general case consider the set of pairs $(H', f')$ where $H'$ is a subgroup of G containing H and $f'$ is a homomorphism of $H'$ into T extending $f$, and let us order this set by the relation $(H', f') \leq (H'', f'')$ if $H' \subset H''$ and the restriction of $f''$ to $H'$ is $f'$. By Set Theory, III, p. 154, Def. 3 and Th. 2, this set has a maximal element $(H_1, f_1)$ and it suffices to prove that $H_1 = G$; if this is not the case, there exists $g \in G, g \notin H_1$ and it is enough to prove that $f_1$ may be extended to a homomorphism T of the subgroup of G generated by $H_1$ and g. Now if C denotes the cyclic group generated by g, the restriction of $f_1$ to $C \cap H_1$ extends to a homomorphism $f_2$ of C into T and the homomorphism $xy \mapsto f_1(x) f_2(y), x \in H_1, y \in C,$ of $H_1C$ into T is the desired mapping.