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I asked this question on math stackexchange (see here) but didn't get any answer so I thought I would post it here too:

If $A$ and $B$ are groups we have the following short exact sequence: $$ 0 \to [A,B] \to A * B \to A \times B \to 0, $$ where the group $[A,B]$ is free (see e.g. Serre's Trees).

I am wondering if there is a "similar" sequence if we replace $A * B$ by $A *_C B$ (or by $A *_C$, an HNN-Extension). By "similar" I mean a short exact sequence where we have some particular knowledge about the left group (the "kernel") and the right group (the "quotient").

In this spirit (since I was thinking about the abelianization) I am searching for some references about the commutator subgroup of a free product with amalgamation (or HNN-Extension), since I am sure someone has already studied this subgroup.

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    $\begingroup$ There's no nontrivial such exact sequence in general. Indeed, amalgams of free groups that are simple (finitely generated) groups are known since the 50's. $\endgroup$
    – YCor
    Nov 29, 2015 at 0:20
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    $\begingroup$ It in principle easy to describe the abelianization of a amalgam, namely is it the cofibered product of the abelianizations in the category of $\mathbf{Z}$-modules. This does not precisely describe the kernel but is the first step to do (in particular it says how the derived subgroup intersects $A$ and $B$). $\endgroup$
    – YCor
    Nov 29, 2015 at 0:21
  • $\begingroup$ @YCor... thank you very much for your helpful comments. Concerning your second comment: Could you be a bit more precise regarding the conclusions for the kernel or do you have some references? You may write it as an answer to this question. $\endgroup$
    – M.U.
    Nov 29, 2015 at 11:47
  • $\begingroup$ If $C$ is central in both $A$ and $B$, then you can construct $A\times_C B$, which is a strong amalgam of $A$ and $B$ over $C$; it is the quotient of $A\times B$ modulo the central subgroup generated by elements $(c,c^{-1})$, with $c\in C$. In that situation, you also have a natural projection form $A*_CB$ to $A\times_CB$, whose kernel is also the cartesian $[A,B]$. $\endgroup$
    – Arturo Magidin
    Nov 30, 2015 at 3:13

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