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Math-1806

Questions in Geometric Group Theory — Q 6.3

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Statement

Definitions and mathematical statement

Let $\mathbb QG$ denote the group algebra of a group $G$ over the rationals, and give $\mathbb Q$ its trivial $\mathbb QG$-module structure. Say that $G$ is of type $\mathrm{FP}$ over $\mathbb Q$ if there is an exact sequence \[ 0\longrightarrow P_d\longrightarrow\cdots\longrightarrow P_0 \longrightarrow\mathbb Q\longrightarrow0 \] for some finite integer $d$, with every $P_i$ a finitely generated projective $\mathbb QG$-module. Projective means a direct summand of a free module. Thus both the length of the resolution and the generation of its terms are finite.

Must there exist an integer $B=B(G)$ such that \[ |F|\leq B\qquad\text{for every finite subgroup }F\leq G? \] The bound may depend on the group and its resolution; no universal bound across all groups of type $\mathrm{FP}$ is asserted. The question concerns orders of subgroups, not their number or their number of conjugacy classes.

The coefficient field is essential: a finite group has a very different homological behavior over $\mathbb Q$ from its behavior over $\mathbb Z$. Also, $\mathrm{FP}$ here means a finite projective resolution, rather than merely $\mathrm{FP}_{\infty}$, which allows infinitely many nonzero terms. No integral finiteness hypothesis is added. Groups with many finite subgroups of one bounded order would still satisfy the desired conclusion; a negative example must contain finite subgroups of unbounded cardinality.

Short English statement

Does a finite projective resolution over the rational group algebra force a bound on all finite-subgroup orders?

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