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I'm not so sure the answer is obviously "yes" if the surface has boundary components (as opposed to punctures). For instance, the mapping class group of a pair of pants is $\mathbb{Z}^3$, which does not embed in $SL_2(\mathbb{Z})$, the outer automorphism group of its fundamental group. ADDED: I guess it's OK, since you can always glue some other stuff onto the boundary components to make a punctured surface. But there is a little something to do!
The title asks about presentations of groups, but the question seems to be about presentations of the theory of groups. These are very different things -- perhaps the title should be altered to something less misleading?
It is easy to see from Gersten's presentation that $\lambda_{ij}^2=1$. Indeed, by (vii) (in Proposition 2.5 of Kielak's paper), setting $\epsilon_i=1$ forces $\rho_{ij}=\lambda_{ji}^{-1}$, and also $\rho_{ij}^2=1$.
@LSpice: I'm not sure why the OP is so reluctant to mention the paper they have been looking at. A plausible guess is Kielak's arxiv.org/abs/1103.1624 -- see Proposition 2.8 therein. The question can now be restated as: "1. How does the stated fact follow from Mennicke's proof of the congruence subgroup property (CSP)? 2. Is there a proof that avoids the CSP?" Both seem like reasonable questions. It should be possible to deduce it from Gersten's presentation (Proposition 2.5 in Kielak's paper). At the very least, Mennicke's main result doesn't apply when $n=2$.
I think one detail in some parts of the above discussion is slightly wrong. The “icosahedral group” (ie $A_5$) is not a subgroup of $SU(2)$; it’s a subgroup of $SO(3)\cong SU(2)/\{\pm I\}$. Its preimage in $SU(2)$ is the binary icosahedral group, which has order 120 and has a non-trivial abelian normal subgroup.
I wonder why Nemethi’s computation doesn’t mention the genus, which is an important part of the Seifert invariant. Compare for instance, with the explanation on Wikipedia: en.m.wikipedia.org/wiki/Seifert_fiber_space . It provides a quick sanity check, since if $g>0$ it can’t be a homology sphere.
Where did you get your conditions? (You don’t give a source.) I haven’t completely parsed them, but they look more like conditions to be an integer homology sphere. Indeed, the examples you give are rational homology spheres but not integral ones.
@BenWieland: Sorry for the slow response. This sounds interesting, but it's quite quick in a comment! I would be happy to hear more details, explained at greater length.
Well, since it's so important I am glad it is now on my radar! Nevertheless, I would be amazed and very pleasantly surprised if any arguments as "soft" as the one you attempted could be used to construct finite acyclic complexes with interesting group actions.
@KoljaKnauer: Well, "regular embeddings" sound like they are precisely the finite subgroups of the respective homeomorphism groups. So they seem much closer to lots of mainstream work in geometric topology, and should be well understood.
@KoljaKnauer: Thanks! That's pretty complicated. It's interesting that the planar Cayley graphs can all be embedded in such a way that the action extends to the sphere; whereas that can't be true for this example.
Thanks for the explanation. After a couple of attempts, I found this nice account of Elmendorff’s theorem, and managed to figure out the statement. golem.ph.utexas.edu/category/2018/06/elmendorfs_theorem.html . It may be useful to organise one’s thinking about topological groups, but for discrete groups, this looks like a tautology that isn’t going to help much. Note that, as in the $A_5$ example, it requires an actual construction to build one of these spaces!
