$\DeclareMathOperator\cd{cd}$Are there any known examples of nonfree groups with a property that $\cd(G)+1 = \cd(G \times G)$, or, less restrictive, $G, H$ with $\cd \neq 1, \infty$ such that $\cd(H)+1 = \cd(G \times H)$?
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1$\begingroup$ This question mathoverflow.net/questions/272715/… is relevant $\endgroup$– Benjamin SteinbergCommented Mar 11, 2022 at 13:31

1$\begingroup$ Although question is relevant, (currently absent) answer to it would not provide an answer to my question as far as I can see and vise versa. $\endgroup$– Denis TCommented Mar 11, 2022 at 14:32

$\begingroup$ That questions shows that the "typical" behavior is $cd(G\times G) =2cd(G)$ for $FP_{\infty}$groups for example and also if you ask $cd(G)=cd(G\times G)$ then $G$ has infinite cohomological dimension. I didn't claim it answers your question. It's for context. $\endgroup$– Benjamin SteinbergCommented Mar 11, 2022 at 15:04

$\begingroup$ For the less restrictive you want neither one free right? $\endgroup$– Benjamin SteinbergCommented Mar 11, 2022 at 16:03

$\begingroup$ ...I think I need both have cd greater than than one for my possible application (namely, some clarifications/upgrades in KanThurston construction; such groups can provide a source for suspensionlike functor on crossed modules with extra structure, in order to get full functor from hoTop to a certain localisation of category of that contraptions), but I guess any example can be pretty illuminating. $\endgroup$– Denis TCommented Mar 11, 2022 at 16:30
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1 Answer
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Let $G=(\mathbb{Q},+)$. Then ${\rm cd}(G)=2$ and ${\rm cd}(G\times G)=3$.


2$\begingroup$ $K(Q,1)$ can be built as the mapping telescope, or homotopy colimit, of the sequence of self maps of $S^1$ of degrees 2,3,4,5... Geometrically this can be realized by gluing together a sequence of cylinders $S^1 \times [0,1]$ by the appropriate maps at the ends. Hence $cd(Q) \leq 2$, and it is not 1 by Swan. $K(Q \oplus Q, 1)$ can be built similarly from cylinders on 2tori, hence $cd \leq 3$. To see that $cd(Q\oplus Q)$ is not 2, note that the uncountable group $Ext^1_Z(H_2(K(Q \oplus Q,1),Z),Z) = Ext^1_Z(Q\oplus Q, Z)$ injects into $H^3(K(Q\oplus Q,1), Z)$ by universal coefficients. $\endgroup$ Commented Mar 12, 2022 at 15:00

$\begingroup$ According to Kropholler's answer here math.stackexchange.com/questions/2320997/… this works more generally for any solvable group of cohomological dimension 2 $\endgroup$ Commented Mar 12, 2022 at 15:34