Let $G$ be a group and $K$ be a subgroup. Suppose $g \in G$ commutes with every element of $K$. Is it true that conjugation by $g$ will act trivially on $H_*(G,K)$?
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4$\begingroup$ What do you mean by $H_*(G,K)?$ $\endgroup$– Gregory AroneCommented Nov 13, 2020 at 5:04
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$\begingroup$ By $H_*(G,K)$, I mean the homology of the mapping cone $BK \to BG$. $\endgroup$– qqqqqqwCommented Nov 13, 2020 at 17:18
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3$\begingroup$ I don't think this is true. If I did not miscalculate then the integral Heisenberg group, and its central subgroup, gives a counterexample. $\endgroup$– Oscar Randal-WilliamsCommented Nov 13, 2020 at 20:03
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1 Answer
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According to the paper "Comparison of relative group (co)homologies" there are distinct versions (which agree for certain pairs of groups). But I expect the answer to be yes in any sensible version, by mimicking the proof in Brown's bible (Corollary III.8.2).
For example, if K is a normal subgroup of G, then Corollary 4.29 of that "comparison" paper says $H_\ast(G,K)\cong H_\ast(G/K)$. The action should respect this isomorphism, hence trivial.
answered Nov 13, 2020 at 13:24