My question is a general one about computing quotient groups of (a subgroup of) a direct sum of $\mathbb{Z}$'s and $\mathbb{Z}/n\mathbb{Z}$'s, but it arises from computing the Baer sum of two extensions. To find all 6 inequivalent short exact sequences $$ 0 \rightarrow \mathbb{Z} \rightarrow E \rightarrow \mathbb{Z}/6\mathbb{Z} \rightarrow 0,$$ I start with the sequence

$$ 0 \rightarrow \mathbb{Z} \xrightarrow{\cdot 6} \mathbb{Z} \rightarrow \mathbb{Z}/6\mathbb{Z} \rightarrow 0 \tag{1} $$ and compute successive Baer sums of $(1)$ with itself.

The way that I learnt of computing Baer sum is as follows:

The Baer sum of two extensions (of $R$-modules) $0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$ and $0 \rightarrow A \xrightarrow{f'} B' \xrightarrow{g'} C \rightarrow 0$ is the extension $0 \rightarrow A \rightarrow X \rightarrow C \rightarrow 0$, where $X$ is the homology group of the complex $$ A \xrightarrow{(f, -f')} B \oplus B' \xrightarrow{g-g'} C, $$ the map $A \rightarrow X$ is obtained from $(f, 0)$, and the map $X \rightarrow C$ is obtained from $g$.

Using this, the middle term in the Baer sum $(1)+(1)$ is $$X = \frac{\{(m,n) \in \mathbb{Z} \oplus \mathbb{Z}: 6|(m-n)\}} {\langle(6, -6)\rangle},$$

which turns out to be isomorphic to $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

Similarly, the middle term in the Baer sum $(1)+(1)+(1)$ is $$Y = \frac{\{(m,n, [0]) \in \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}: 6|(m-n)\} \cup \{(m,n, [1]) \in \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}: 6|(m-n+3)\} } {\{(6n, -3n, [n])\in \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}: n \in \mathbb{Z}\}},$$

which turns out to be isomorphic to $\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$.

I have tried plotting some points on the integer lattice plane and do see a pattern, but how can I find an explicit isomorphism between $X$ and $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, and between $Y$ and $\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$? I'd prefer a general method over an ad hoc method. Thank you!



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