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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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No, without further information you can't. You only have a short exact sequence $0 \to \ker{\alpha} \to \ker{\beta\alpha} \to \ker{\beta} \to 1$ (by the snake lemma for example). –  t.b. Aug 16 '11 at 11:32
    
Hmm...well, what other information might I need? I am trying to bypass Reidemeister-Schreier, as I cannot find a decent transversal for $K$ (it is a Baumslag-Solitar group), but I know what $ker(\beta)$ is, and $H$ is nice so I can do Reidemeister-Schreier to find $ker(\alpha)$. –  user1729 Aug 16 '11 at 11:39
    
Well, I don't know enough combinatorial group theory in order to tell you anything you don't know already. I would suggest that you include more details about the situation you're interested in into your question because then some of the group theorists here might be able to help you. In the generality you asked your question I doubt that there is much more to say than what I said in my previous comment. –  t.b. Aug 16 '11 at 11:48
    
Your right, I could ask my question outright. However, I prefer asking more general questions, and save the specific bits for my supervisor! –  user1729 Aug 16 '11 at 12:03
    
But a Baumslag-Solitar group is an HNN extension. –  user641 Aug 16 '11 at 14:33
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