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I have a group, $G$, with

$\alpha: G\rightarrow H$ surjective,

$\beta: H\rightarrow K$ surjective.

If I know the isomorphism class of $\operatorname{Ker}(\alpha)$ and of $\operatorname{Ker}(\beta)$ then can I calculate the isomorphism class of $\operatorname{ker}(\alpha\circ\beta)$?

Motivation: (added at the request of Theo Buehler, to try and entice an answer!) I am trying to compute $\operatorname{ker}(\alpha\circ\beta)$, which I would quite like to do via Reidemeister-Schrier (well, the variant which uses CW-complexes which I forget the name of, but I'm not fussy). However, $K$ is a Baumslag-Solitar group and it seems they have no "nice" normal form(s), so finding a transversal is, basically, impossible. On the other hand, $H$ is very nice so I can do R-S on it to get $\operatorname{ker}(\alpha)$, and I know what $\operatorname{ker}(\beta)$ is. I would therefore like to piece these together to get $\operatorname{ker}(\alpha\circ\beta)$.

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    Hmm...yes, maybe I hadn't thought that bit through (this is why I leave the specifics for my supervisor - it's less embarrassing!). However, I had found a paper which claimed that there were no normal forms where the normal element had least length (or something along those lines). I mean, yes, there is a normal form from the HNN-extension...but is it nice? I'll think about this tomorrow...thanks!2011-08-16

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