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I am not very familiar with group extensions, and I know that in general finding all possible extensions is an extremely hard problem, however I am interested in the following:

What are all possible extensions of a finite Möbius group $H$ by the cyclic group on two elements $C_2$. ie all groups $G$ such that there are homomorphisms such that

$$ 1 \longrightarrow C_2 \longrightarrow G \longrightarrow H \longrightarrow 1 $$

is a short exact sequence (since there is some confusion about what I mean).

This is of interest to me because it gives all possible automorphism groups of hyperelliptic surfaces, and I would like to be able to explain how we determine what these groups are. I have a list of the possible groups but I would like to know if they are equal to the set of extensions, or a proper subset of the possible extensions.

Edit: To be clear here, I am interested in the group theory behind finding these extensions, not in the automorphisms of certain curves, I already understand those.

A reference for this problem would also be a helpful answer. The only resource I have used so far is Milne's notes on group theory and they do not discuss how to solve this sort of problem.

EDIT It may be useful here to know that we can prove using properties of Riemann surfaces that $C_2$ is in the centre of the groups of interest. So I know the groups of interest will be central extensions. According to wikipedia the set of isomorphism classes of such groups are exactly the cohomolgy group $H^2(G,A)$ which is useful. I guess this answers my question if I restrict it accordingly.

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If $H$ has index 2 in $G$, then $H$ must be normal in $G$. So you need to examine all possible actions of $C_2$ on $H$ and construct the semidirect products: – alex.jordan Sep 13 '11 at 22:42
@alex.jordan : You have it backwards -- $H$ will be the quotient, not the subgroup. Also, not all extensions split, so you need to deal with things that are more general than semidirect products. – Adam Smith Sep 13 '11 at 22:52
@Adam: Oh yes, backwards, after looking up vocabulary. "An extension of $H$" makes me envision $H$ as being embedded in the extension. I guess I read it that way because if $K$ is a field "extension of $k$", then $k$ is embedded in $K$. Why the inconsistent vocabulary? – alex.jordan Sep 13 '11 at 23:09
Tradition I suppose. Also, field extensions are a little different than group extensions since you can't take a quotient of a field by a subfield (so you can't extend a field by another field). – Adam Smith Sep 14 '11 at 1:56
@Adam: I asked this question if you're interested in following it. Arturo already posted to it. – alex.jordan Sep 14 '11 at 22:47
up vote 2 down vote accepted

The answer to determining the possible automorphism groups of hyperelliptic Riemann surfaces is contained in the following paper:

R.Brandt, H.Stichtenoth, Die automorphismengruppen hyperelliptischer Kurven, Manuscripta Math 55 (1986), 83-92.

Of course, there is an extensive earlier literature as well.

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Unfortunately can't read German. Do you know if there is a translation? – AnonymousCoward Sep 14 '11 at 1:50
It's not a particularly famous paper, so I doubt that there is a translation. However, it is only 9 pages long. I'd recommend getting a dictionary and trying to read it yourself. It's probably not that bad. Certainly that's how I learned to read mathematical French and German back when I was a student! – Adam Smith Sep 14 '11 at 1:53
To be clear, I already know what the possible automorphism groups of hyperelliptic surfaces are. I'm interested in how one would find the list of groups in my question via group theory. – AnonymousCoward Sep 15 '11 at 1:58

Group cohomology. Specifically, $H^2(H,C_2)$ parametrizes the extensions (for a given action of $H$ on $C_2$).

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Up to equivalence, not isomorphism. – user641 Sep 14 '11 at 0:10
Plus, the action will always be trivial, so this will be central extensions. – user641 Sep 14 '11 at 0:11
Actually, although this doesnt answer my more general question, in the case of groups $G$ that will act on hyperelliptic surfaces, we further know that $C_2\in Z(G)$. So I can restrict my question to this. – AnonymousCoward Sep 15 '11 at 20:09

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