# Calculating presentations for the normal subgroup of a semidirect product

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.

• 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? – S. Carnahan Apr 9 '10 at 4:38
• 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? – Ian Brown Apr 9 '10 at 4:39
• 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. – Ian Brown Apr 9 '10 at 4:42
• 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? – S. Carnahan Apr 9 '10 at 4:42
• 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? – S. Carnahan Apr 9 '10 at 4:45