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(renamed) Torsion free group has finite commutator subgroup iff abelian

As Ycor points out this question is easy: A torsion free group (Wang or otherwise) is abelian if and only if its commutator subgroup is finite. For the first direction, abelian implies trivial ...
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S.E.S.'s in an Abelian Category

First of all, note that $(k \circ \ell) \circ f = k \circ h = 0$. Therefore, by $g$ being a cokernel of $f$, there exists a unique morphism $s : X \to Y$ such that $s \circ g = k \circ \ell$. ...
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S.E.S.'s in an Abelian Category

This looks like the "3x3 lemma": https://ncatlab.org/nlab/show/3x3+lemma or https://en.wikipedia.org/wiki/Nine_lemma. In particular, we have a commutative diagram $$ \begin{array}{ccccccccc} ...
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In a 3x3 diagram of modules, if the first two rows are split exact, is the third row also split?

Generalizing this answer, every exact sequence $X\xrightarrow{\alpha} Y\xrightarrow{\beta} Z\to 0$ appears as the bottom row of a diagram of the kind you describe, since you can take the diagram $$\...
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In a 3x3 diagram of modules, if the first two rows are split exact, is the third row also split?

I think for your first question, that square always commutes. Since $L$ is a coproduct in an abelian category, we can use its universal property. The map $L' \to L$ in your diagram and the section map ...

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