If group G has a set of generators and set of relations, and given H, K two normal subgroups. Suppose one can write down the elements in H and K explicitly (also in terms of generators and relations). Now how to obtain an element which is in the intersection H\cap K but not in the commutator subgroup [H,K]?

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    $\begingroup$ Your question is very vaguely stated. An element of the intersection is precisely an element of the kernel of the product map $G\to G/H\times G/K$. Does that help? $\endgroup$ – HJRW Apr 3 '12 at 20:26
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    $\begingroup$ adm.lnpu.edu.ua/downloads/issues/2004/N4/adm-n4-3.pdf seems relevant. $\endgroup$ – Mustafa Gokhan Benli Apr 3 '12 at 20:32
  • $\begingroup$ Why do you think there is such element? A priori $H \cap K=[H,K]$, e.g. $H=K=G$ and $G$ is perfect. So there cannot be a general procedure without more conditions. $\endgroup$ – Yiftach Barnea Apr 18 '12 at 7:34

I would just like to mention that this question has topological relevance, as shown in

R. Brown ``Coproducts of crossed $P$-modules: applications to second homotopy groups and to the homology of groups'', Topology 23 (1984) 337-345.

Denote the classifying space of a group $G$ by $BG$. Given normal subgroups $M,N$ of $G$ one can form the space $X$ as the homotopy pushout (i.e. double mapping cylinder) of the two maps $BG \to B(G/M), BG \to B(G/N)$. Then the second homotopy group of $X$ is isomorphic to

$$(M \cap N)/ [M,N] . $$

Actually the result of the paper says more, namely that the homotopy 2-type of $X$ is described by the crossed module $M \circ N \to G$ where $\circ$ is the coproduct of the title of the paper. It is feasible that the question asked could be helped by an analysis of this coproduct. (Note that normal subgroups of a group are special cases of crossed modules.)

  • $\begingroup$ Thank you! Indeed I was considering the condition that $(M\cap N)=[M,N]$. I will check your paper. $\endgroup$ – Changlong Apr 4 '12 at 16:18
  • $\begingroup$ I think the following papers are also relevant: Gilbert, N. D. Identities between sets of relations. J. Pure Appl. Algebra 84 (1993), no. 3, 263–276. Bogley, W. A.; Gilbert, N. D. The homology of Peiffer products of groups. New York J. Math. 6 (2000), 55–71. $\endgroup$ – Ronnie Brown Jun 5 '12 at 11:52

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