It is known that for a presented group $G=F/N$ we have $$H_2(G;\mathbb{Z}) \cong \frac{[F,F]\cap N}{[F,N]}.$$ In general, the right side seems to be difficult to calculate. I am in the special situation where the group is finitely presented, so $G\cong \langle a_1,\dotsc,a_n\mid r_1,\dotsc,r_s\rangle$ and all relations are commutators, so $N\subseteq[F,F]$, and I am just interested in the rank of the free part of $H_2(G;\mathbb{Z})$. Is there any general approach to these kind of problems?
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4$\begingroup$ Deciding if $H_2(G)$=0 is not decidable in general. However, I don't know if this is known for the sort of presentations you are considering. books.google.com/… The issue here is that (I think) one may compute the rank of $H_2(K)$, where $K$ is a 2-complex presenting the group. But one needs to quotient by $\pi_2(K)$, which might not be algorithmic. If one knew more, like solvable word problem or small cancellation, then this might be doable. $\endgroup$– Ian AgolCommented Feb 6, 2019 at 4:59
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$\begingroup$ Thank you! In fact, I am doing all this in order to see that some $\pi_2(X)$ does not vanish because I believe that $X$ is no $K(G,1)$. ;-) $G$ is of the following form I have generators $a_1,\dotsc,a_5,b_1,\dotsc,b_5,c_1,\dotsc,c5$ and 74 relations which are of the form $[a_i,b_j]=0$ and of the form $b_ic_jb_i^{-1}=b_kc_jb_k^{-1}$ and of the form $a_ib_jc_k=c_kb_ja_i$. My first attempt was to look for relations which are already covered by previous ones, but I did not find one … $\endgroup$– FKranholdCommented Feb 6, 2019 at 10:26
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$\begingroup$ Are you throwing in all such relations? If so, I only count 70 (after throwing out obvious dependent relations). $\endgroup$– Ian AgolCommented Feb 6, 2019 at 19:33
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