Let $G$ be a group with a finite presentation $\langle S \cup S' | R \cup R'\rangle$. Assume that I happen to know that $G$ is the semidirect product of a normal subgroup $N$ and another subgroup $H$. Furthermore, assume that $H$ is the subgroup generated by $S'$ and that $H \cong \langle S' | R' \rangle$. Is there any algorithm to calculate a presentation for $N$? Assume that we know how to solve all the standard problems (the word problem, the conjugacy problem, the membership problem for $H$ and $N$, etc) in $G$ itself.

Here is an easy example to show what kinds of issues arise. Let $G$ be the free group on two letters $x$ and $y$. We then have a split short exact sequence

$$1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1,$$

where the group $H$ is the cyclic group generated by $y$ and $N$ is an infinite rank free group (it consists of all words in $x$ and $y$ the sum of whose $y$-exponents is $0$). The moral is that we cannot hope for a finite presentation for $N$, and our algorithm must return an infinite presentation.

EDIT : Here's two nice examples of what I am talking about. First, let $G = PSL_2(\mathbb{Z})$ and let $H = PSL_2(\mathbb{Z}/2\mathbb{Z})$. Then it turns out that $H$ is isomorphic to the symmetric group on $3$ letters and there is a splitting of the natural surjection $G \rightarrow H$. How can one compute the kernel of this map?

For another example, let $G = SP_4(\mathbb{Z})$ and let $H = SP_4(\mathbb{Z}/2\mathbb{Z})$. Then $H$ is isomorphic to the symmetric group on $6$ letters (this comes from the action on odd theta characteristics), and there is a splitting of the homomorphism $G \rightarrow H$. How can one compute the kernel of this map?

As you might guess from these examples, I have in mind applications to the theory of modular forms. I am looking for practical algorithms for computation, not theoretical results.

  • $\begingroup$ If you are allowing infinite presentations, can't you just take elements of G as generators, with relations defined by the multiplication table, together with additional relations sending anything to zero if its image in H is not the identity? $\endgroup$ – S. Carnahan Apr 9 '10 at 4:38
  • $\begingroup$ I'm not sure I follow. There is not necessarily a homomorphism from G to N -- remember, H does not need to be normal. What kinds of relations are you talking about? $\endgroup$ – Ian Brown Apr 9 '10 at 4:39
  • $\begingroup$ A related comment -- the category of groups has the funny property that a splitting H->G of a short exact sequence 1-->N-->G-->H-->1 does not imply that there is a retract homomorphism G-->N. $\endgroup$ – Ian Brown Apr 9 '10 at 4:42
  • $\begingroup$ My mistake. Instead, could you take all elements of G that map to the identity of H as generators, with multiplication in G defining relations? $\endgroup$ – S. Carnahan Apr 9 '10 at 4:42
  • $\begingroup$ I think the real question I should be asking is, what is your model of computation, so that we can distinguish between silly algorithms like mine (assuming I didn't mess up again) and something that would be useful to you? $\endgroup$ – S. Carnahan Apr 9 '10 at 4:45

If you have explicit finite presentations of G and H is small and finite, then you should be able to just ask GAP or magma for a presentation of the kernel of the projection. This is another disguise of the Schreier transversal. Let me know if you want code sample or algorithmic references. This will likely fail for things like Sp(6,13) which are simply too large, but I think it should work out for Sym(3) and Sym(6).


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