Finitely generated image from commutative diagram Let the following Diagram commute with abelian groups

If $\text{im}(C \rightarrow F)$ and $\text{im}(B \rightarrow E)$ are finitely generated and $C \rightarrow D \rightarrow E$ is an exact sequence at D then the composition $\text{im}(A \rightarrow D \rightarrow G)$ is finitely generated.
I have some trouble proving this.
$D \rightarrow E$ can be represented by $\text{im}(A \rightarrow D) \rightarrow \text{im}(B \rightarrow  E)$ so we know there are finite many $d_1, ... , d_s \in \text{im}(A \rightarrow D)$ so, that any element of $\text{im}(A \rightarrow D)$ can be written as $d = c+ r_1d_1 + ... + r_sd_s$ for $r_i \in \mathbb{Z} , c \in \text{im}(C \rightarrow D)$.
Now I am stuck. Maybe I could show that $C$ is finitely generated and then argue with the exact sequence that D must be finitely generated? I am really unsure on how to continue this.
 A: I think it's clearer to use letters to name the maps in the diagram. So let me write $a$, $b$, $c$ and $d$ for the maps going down from $A$, $B$, $C$ and $D$ and  $\alpha$, $\gamma$, $\delta$ and $\phi$ for the maps going right from $A$, $C$, $D$ and $F$.
By assumption, there are finite sets $\{c_1, \ldots, c_t\} \subseteq C$ and $\{d_1, \ldots, d_s\} \subseteq D$, such that the $c(c_i)$ generate $\mathrm{im}(c)$ and the $\delta(d_j)$ generate $\mathrm{im}(\delta \circ a) = \mathrm{im}(b \circ \alpha) \subseteq \mathrm{im}(b)$ (since a subgroup of a finitely generated abelian group is finitely generated). (I think my $d_j$ are the same as the $d_j$ that you had in mind in your proof attempt.) If $x \in A$, then $a(x) = \gamma(y) + r_1d_1 + \ldots r_sd_s$ for some $y \in C$ and $r_j \in \Bbb{Z}$ (using $b(\alpha(x)) = \delta(a(x)) \in \mathrm{im}(\delta \circ \alpha)$ and the exactness of $C \mapsto D \to E$: we can choose $r_r$ such that $\delta(a(x)) = \delta(r_1d_1 + \ldots + r_sd_s)$ and then $a(x) - (r_1d_1 + \ldots + r_sd_s) \in \ker(\delta) = \mathrm{im}(\gamma)$). But then
$$\begin{align*}
d(a(x)) &= d(\gamma(y)) + d(r_1d_1) +  \ldots + d(r_sd_s) \\
 &= \phi(c(y)) + r_1d(d_1) +  \ldots + r_sd(d_s)
\end{align*}$$
But $c(y) = s_1c(c_1) + \ldots + s_tc(c_t)$ for some $s_i \in \Bbb{Z}$, so
$$\begin{align*}
d(a(x)) &= \phi(c(y)) + r_1d(d_1) +  \ldots + r_sd(d_s) \\
 &= s_1\phi(c(c_1)) + \ldots + s_t\phi(c(c_t)) + r_1d(d_1) +  \ldots + r_sd(d_s)
\end{align*}$$
So $\mathrm{im}(d \circ a)$ (i.e., $\mathrm{im}(A \to D \to G)$ in your notation) is generated by the set $\{\phi(c(c_1)), \ldots, \phi(c(c_t)), d(d_1), \ldots, d(d_s)\}$.
