Suppose we are given group $G=\langle a_1,\ldots,a_n \mid R_1=1,\ldots R_m=1 \rangle$. Is there an algorithm which computes a finite presentation for the Schur multiplier, i.e. second homology group $K=H^2(G,\mathbb{Z})$? Can one at least solve a decision problem, whether $K$ is trivial or not?
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$\begingroup$ This is implemented in GAP and thus probably accessible via sage. See the first answer to mathoverflow.net/questions/53730/… for some code. $\endgroup$– Franz LemmermeyerCommented Aug 18, 2011 at 14:08
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4$\begingroup$ @Franz: "implemented" does not necessary mean that there is an algorithm :) $\endgroup$– Igor RivinCommented Aug 18, 2011 at 14:18
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$\begingroup$ @Franz: The question mentioned by you is about finite groups, and I am mostly interested in infinite groups. As I understand, the function AbelianInvariantsMultiplier from GAP works only for finite groups. Also it should be mentioned that question is about "theoretical" algorithm (with any space and time constraints), but it should work for any group presentation. $\endgroup$– Al TalCommented Aug 18, 2011 at 14:36
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$\begingroup$ There are some partial results in intlpress.com/hha/v12/n1/a3. If G is finitely presented, and you're okay with field coefficients, there is an algorithm that can give an upper bound to H_2. $\endgroup$– JoshCommented Aug 18, 2011 at 15:44
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1 Answer
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It is proved in "Some embedding theorems and undecidability questions for groups" by C. Gordon, that there is no algorithm that computes a presentation for the Schur multiplier from a presentation of $G$, and moreover there is no algorithm for the deciding whether the Schur multiplier is trivial or not. Some other similar related results are proved, as well.
answered Aug 18, 2011 at 14:30