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Let $G$ be a fixed group. By definition, a $G$-group is a group $X$ with a $G$-action that respects the group operation of $X$. A free $G$-group means a group freely generated by a free $G$-set. A "projective $G$-group" means a projective object in the category of $G$-groups.

Can there be projective $G$-groups which are not free $G$-groups? If yes, for which groups $G$ does it happen?

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  • $\begingroup$ What is a $G$-group? $\endgroup$
    – abx
    Commented Oct 27, 2018 at 15:59
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    $\begingroup$ Related: mathoverflow.net/questions/313860/… (a $G$-subgroup of a free $G$-group can fail to be a free $G$-group). $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 16:09
  • $\begingroup$ @abx $G$-group is a group with a $G$-action, which is compatible with the group operation $\endgroup$
    – user49822
    Commented Oct 27, 2018 at 16:39
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    $\begingroup$ As discussed in the comments to the question YCor links to, there are 2 possible definitions of free $G$-group. Unlike that question, I suspect the answer here depends on which definition you choose. $\endgroup$
    – Mark Grant
    Commented Oct 27, 2018 at 17:13

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No (there do not exist projective $G$-groups which are not free $G$-groups).

By standard arguments, a $G$-group is projective if and only if it is a retract of a free $G$-group. So we only have to show that a retract of a free $G$-group is a free $G$-group.

This is not immediate, because a $G$-subgroup of a free $G$-group need not be a free $G$-group, as noted in the comments. However it follows from considerations of equivariant group cohomology.

Let $H$ be a $G$-group, and let $M$ be a $\mathbb{Z}[H\rtimes G]$-module. Cegarra, García-Calcines and Ortega define the equivariant group cohomology of $H$ with coefficients in $M$ to be $H^*(H\rtimes G,G;M)$, the Takasu relative cohomology of the pair (which agrees with the singular cohomology of the pair $(B(H\rtimes G),BG)$ for suitably chosen classifying spaces). This has the usual functoriality properties for homomorphisms of $G$-groups. Define the equivariant cohomological dimension of the $G$-group $H$ to be the maximal $n$ such that $H^n(H\rtimes G,G;M)\neq 0$ for some coefficient module $M$.

In a recent preprint with Kevin Li, Ehud Meir and Irakli Patchkoria, we prove the following:

Theorem: A $G$-group has equivariant cohomological dimension $\le1$ if and only if it's a free $G$-group.

Now if the identity on $H$ factors as $H\to F\to H$ where $F$ is a free $G$-group, then the equivariant cohomological dimension of $H$ is less than or equal to $1$. Hence projective $G$-groups are free $G$-groups.

(Actually this is just one of three possible definitions of equivariant cohomological dimension, which are compared in the preprint.)

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  • $\begingroup$ Since my comment is no longer relevant, I have deleted it (and will delete this one soon). $\endgroup$
    – LSpice
    Commented Jan 16, 2023 at 17:06
  • $\begingroup$ Mark, maybe I’m missing something but if you have two non trivial groups $G, K$, a free $G$ group is a retract of a free $G\times K$ group so it should be projective over the product. $\endgroup$
    – Fernando Muro
    Commented Feb 25, 2023 at 8:20
  • $\begingroup$ @FernandoMuro I'm not sure I understand. In order to view a free $G$-group as a retract of a free $G\times K$-group, I have to first view it as a $G\times K$-group. I could do this by letting the $K$ act trivially...but then it won't generally give a free $G\times K$-group (thinking of the case where $G$ and $K$ are finite, you could have a free $G$ group of rank $|G|$, which cannot have a free basis of cardinality a multiple of $|G\times K|$). In short, how do you move back and forth between the category of $G$-groups and the category of $G\times K$-groups? $\endgroup$
    – Mark Grant
    Commented Feb 26, 2023 at 8:23
  • $\begingroup$ @MarkGrant I somehow imagined that the retraction $\langle G\times K\rangle\twoheadrightarrow \langle G\rangle\hookrightarrow \langle G\times K\rangle$ was $G\times K$-equivariant, which obviously isn't (the section fails). $\endgroup$
    – Fernando Muro
    Commented Mar 1, 2023 at 22:42

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