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?

  • 2
    $\begingroup$ Take the image of the generators of $G$ and choose pullbacks of generators of $K$. $\endgroup$ Commented Jul 11, 2014 at 14:39
  • $\begingroup$ is that sufficient? $\endgroup$
    – jeevan
    Commented Jul 11, 2014 at 14:40
  • $\begingroup$ Yes. Say $g_1, g_2, \ldots, g_n \in H$ are the images of generators for $G$ and $k_1, k_2, \ldots, k_m \in H$ are pullbacks for generators for $K$. Then given $h \in H$, you can first find a word $k_{i_1} \cdots k_{i_\ell}$ so that $(k_{i_1} \cdots k_{i_\ell})^{-1} h$ is in the image of $G$ in $H$. Then you just use the generators of $G$ to knock it down to the identity. $\endgroup$ Commented Jul 11, 2014 at 14:45
  • $\begingroup$ I am assuming short exact sequence, which I just realized your question isn't assuming. I'm pretty sure in the most general case you're not going to get a finitely generated group. $\endgroup$ Commented Jul 11, 2014 at 14:47
  • $\begingroup$ how about long exact sequence $G\longrightarrow H \longrightarrow K \longrightarrow L \longrightarrow M \longrightarrow N$? Where I want to compute generators of L when I know the same for others? $\endgroup$
    – jeevan
    Commented Jul 11, 2014 at 15:01

1 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 $h - r_1c_1 - \ldots - r_mc_m = s_1a_1 + \ldots + s_na_n$.


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