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Presentation of a semi-direct products of $N$ by $H$ can be written from presentations of $N$ and $H$. But for other extensions of $N$ by $H$ (cyclic, central etc.), which are not semi direct products, can we write presentation of the extension from presentation of $N$ and $H$?

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In general, you need to know two things: what each relator in H is equal to in N, and what generators of H conjugate generators of N to. For example, suppose I have the extension C_2 -> G -> C_2, with presentations <x | x^2> and <y | y^2>. Suppose I know x^y = x and y^2=x, then my presentation for G is <x,y | x^2, y^2=x, x^y=x>. –  Steve D Jan 22 '11 at 7:15
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The question is: how is the extension given to you in the first place? –  Alex B. Jan 22 '11 at 8:25
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This is a duplicate of this (very poorly worded, subsequently closed) question: mathoverflow.net/questions/44631/… , which received a very good answer from Derek Holt. –  HJRW Jan 22 '11 at 16:08
    
As per Holder's program, it is possible construct all groups if we know simple groups. "Semi-direct product" is a nice tool to construct many groups from two known groups; but when we try to find all "p-groups" then semi-direct products are not sufficient. We move towards general extension of groups. There may be many extensions of a group $N$ by $H$, which are isomorphic. But if we have presentations of these extensions, we can determine isomorphism classes easily. Therefore, I would like to know, whether presentation of extension can be written from presentation of $N$ and $H$. –  RDK Jan 23 '11 at 4:37
    
What do you mean by 'it is possible to construct all groups'? How do you want to describe the groups? As presentations? It is clearly possible to list all presentations. On the other hand, the isomorphism problem for groups is unsolvable, so you can't possibly hope to classify groups up to isomorphism. –  HJRW Jan 23 '11 at 18:50

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To start with, following Alex's comment above, I think you need to look at some books on cohomology of groups to specify how the extension is going to be given. That means looking at the Schreier theory.

Many years ago the Schreier theory of extensions was adapted to give exactly this by Turing.(Little known paper of his.) Ronnie Brown and I gave a more modern treatment of it in a paper in the Proceedings Royal Irish Academy.(On the Schreier theory of non-abelian extensions: generalisations and computations, (Proc.Royal Irish Acad., 96A, (1996) 213-227.)) That is quite general, but a simple version of the question can found discussed in many books on combinatorial group theory, such as D. L. Johnson, 1980, Topics in the theory of group presentations , number 42 in London Math. Soc Lecture NotesMS Lecture Notes, Cambridge University Press. and D. L. Johnson, 1997, Presentations of groups , volume 15 of London Mathematical Society Student Texts , Cambridge University Press, Cambridge. The advantage of these is that they do not require an imense expenditure of time to get to the heart of the problem.

(An interesting follow on is to examine ways in which to start with an extension of groups, plus resolutions of the two ends and give a resolution of the middle term.)

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