generators of groups from exact sequence

Question

Suppose I have a middle term exact sequence of finitely generated abelian groups $G \longrightarrow H \longrightarrow K$. How do I get the generators of $H$ if I know the same for other two groups?

Answer 1

The exact sequence $G \xrightarrow{f} H \xrightarrow{g} K$ gives a short exact sequence $\DeclareMathOperator{\im}{im}$ $$0 \to \im(f) \to H \to H/\im(f) \cong \im g \to 0$$ Then $G,K$ finitely generated (f.g.) abelian groups $\implies H$ f.g. (since $\im g$ is a subgroup of $K$, hence is f.g., using only that $\mathbb{Z}$ is a Noetherian ring, and also $\im(f)$ is f.g., being a quotient of $G$. Thus $H$ is an extension of two f.g. abelian groups, hence is f.g.) Explicitly, if $\im(f) = \ker g = \langle a_1, \ldots, a_n \rangle$, $\im g = \langle b_1, \ldots, b_m \rangle$, then for any choice of lifts $c_i$ with $g(c_i) = b_i$, one has $H = \langle a_1, \ldots, a_n, c_1, \ldots, c_m \rangle$.

Indeed, for any $h \in H$, one has $g(h) = r_1g(c_1) + \ldots + r_mg(c_m)$ for some $r_i \in \mathbb{Z}$, so $g(h - r_1c_1 - \ldots - r_mc_m) = 0$, and thus $h - r_1c_1 - \ldots - r_mc_m = s_1a_1 + \ldots + s_na_n$.