Finiteness of non-abelian cohomology

Question

Let $H$ be a finite group acting on a group $G$. Are there general conditions one can place on $G$ so that the set $H^1(H, G)$ is guaranteed to be finite? For instance, does finiteness hold if $G$ is finitely presented, or if $K(G, 1)$ is a finite CW complex?

Added: According to Reid Barton's answer here, $H^1(H, G)$ is identified with $\pi_0(X^{hH})$, where $X$ is the space $K(G, 1)$ and the $hH$ denotes homotopy invariants. Thus it would be enough to know the following: if a finite group $H$ acts on a finite CW complex $X$, is $\pi_0(X^{hH})$ finite?

Answer 1

There is an interesting paper of Adem:

A. Adem: Automorphisms and Cohomology of Discrete Groups, J. of Algebra 182(1996), 721-737,

where he considers non-abelian cohomology with coefficients in a discrete group $\Gamma$ of finite cohomological dimension (in particular $\Gamma$ is torsion-free).

Among others he cites (and gives a proof) of the following result due to Serre:

1) Let $G$ be a finite subgroup of $\operatorname{Aut}(\Gamma)$, let $\bar{\Gamma} = \Gamma \rtimes G$ be the semi-direct produt and let $\kappa: \bar{\Gamma} \to G$ be the natural projection. Then for each $H \le G$ there is a bijection between the conjugacy classes $\lbrace \bar{H} \le \bar{\Gamma} | \kappa(\bar{H}) = H \rbrace / \Gamma$ and $H^1(H;\Gamma)$.

In particular $H^1(G;\Gamma)$ is finite iff there are (up to $\Gamma$-conjugacy) only finitely many $\bar{G} \le \bar{\Gamma}$ such that $\bar{G}\Gamma/\Gamma \cong G$.

Adem uses this result to prove:

2) If $P$ is a $p$-subgroup of $\operatorname{Aut}(\Gamma)$ then $|H^1(P;\Gamma)| \le \dim_{\mathbb{F}_p} H^{\ast}(\Gamma;\mathbb{F}_p)$. In particular $H^1(P;\Gamma)$ is finite.

Perhaps you can adjust some of the methods used there to the situation you have in mind.

Answer 2

Take $H=\langle a|a^2=1\rangle$ and $G=\langle b,c|b^2=1\rangle$ with $H$ acting trivially on $G$. Then $H^1(H;G)=\text{Hom}(H,G)=\{g\in G| g^2=1\}$, and this contains the infinite set $\{c^nbc^{-n}|n\in\mathbb{Z}\}$. Thus, it is not enough to assume that $G$ is finitely presented.

Now suppose that there is a finite-dimensional CW complex of homotopy type $K(G,1)$. Let $E$ denote the universal cover. We then see that for every subgroup $G'\leq G$, the quotient $E/G'$ is a model for $K(G',1)$, so for any coefficient ring $R$ we have $H^k(K(G',1);R)=0$ for $k\gg 0$. This means that $G'$ cannot be a finite cyclic group, so we see that $G$ is torsion-free. Thus, if $H$ acts trivially on $G$ we have $H^1(H;G)=\text{Hom}(H,G)=1$. If the action is nontrivial then Ralph's answer becomes relevant.