Groups with trivial rational homology and their finite index subgroups

Question

For a short exact sequence $0 \to G \to H \to K \to 0$ of (discrete) groups with $K$ finite we have, as a consequence of the Hochschild-Serre spectral sequence, that $H^{\ast}(H;\mathbb Q) = H^{\ast}(G;\mathbb Q)^K$. This can be used to see that free and free abelian groups embedd with finite index into groups with trivial rational cohomology: $$H^{\ast}(F_n \rtimes_{\text{sign}} \mathbb Z/2;\mathbb Q) = H^{\ast}(\bigvee_n S^1;\mathbb Q)^{\mathbb Z/2} = H^{\ast}(\text{pt};\mathbb Q).$$ $$H^{\ast}(\mathbb Z^n \rtimes_{\text{sign}} (\mathbb Z/2)^n;\mathbb Q) = H^{\ast}(\prod_n S^1;\mathbb Q)^{(\mathbb Z/2)^n} = H^{\ast}(\text{pt};\mathbb Q).$$ Does this work for every group? Or writing down the opposite:

Is there a group $G$ such that every group $H$ which contains $G$ with finite index has nontrivial rational cohomology?

Answer 1

Thompson's group $T$ gives an example, i.e. if $T \le H$ has finite index, then $H^\ast(H;\mathbb{Q}) \neq 0$.

More specifically, there is always a non-trivial class in $H^4(H;\mathbb{Q})$.

Proof: Step 1: $T$ is normal in $H$

Since $T$ has finite index in $H$, $T_0 := \bigcap_{h \in H/T}hTh^{-1}\le T$ is a finite index, normal subgroup of $H$ (and $T$). But $T$ is infinite simple, so $T=T_0$ is normal in $H$.

Step 2: Since $T$ is normal, $H^\ast(H;\mathbb{Q})=H^\ast(T;\mathbb{Q})^H$.

By work of Ghys & Sergiescu, $H^2(T;\mathbb{Z})= \mathbb{Z}^2$ is generated by the Euler class $x'$ of $T$ and the Godbillion-Vey class $y'$. The corresponding rational classes $x=i(x')$ and $y=i(y')$ where $i: H^\ast(T;\mathbb{Z}) \to H^\ast(T;\mathbb{Q})$ generate the rational cohomology ring as $H^\ast(T;\mathbb{Q})=\mathbb{Q}[x,y]/(xy),\,\,|x|=|y|=2$.

I'll show that $x^2 + y^2$ is invariant under the action of $H$. Let $c: T \to T$ be conjugation by an element from $H$. Hence $c^\ast$ induces an isomorphism on $H^2(T;\mathbb{Z})$. Write $c^\ast(x')=ax'+by',\, c^\ast(y')=cx' + dy'$ with integers $a,b,c,d$ s.t. $ad-bc= \pm 1$. Since $c^\ast$ commutes with $i$, we find $c^\ast(x)c^\ast(y)=acx^2 + bdy^2$. But also $c^\ast(x)c^\ast(y)=c^\ast(xy)=0$ because $xy=0$. Hence $ac=0$ and $bd=0$. Thus $$c^\ast(x)=\pm x,\, c^\ast(y)=\pm y\quad \text{or}\quad c^\ast(x)=\pm y\,,c^\ast(y)=\pm x$$ In either case, $c^\ast(x^2+y^2)=x^2+ y^2$. qed

Answer 2

Not a complete answer. I have no idea how to approach this question if $G$ is not required to be normal. If it is, suppose that $G$ has both trivial center and trivial outer automorphism group. (Such groups are called complete.) Then every short exact sequence $1 \to G \to H \to K \to 1$ is trivial in the sense that $H \cong G \times K$, hence if $K$ is finite we have $H^{\bullet}(BH, \mathbb{Q}) \cong H^{\bullet}(BG, \mathbb{Q})$. So it suffices to find $G$ which

• has trivial center,
• has trivial outer automorphism group, and
• has nontrivial rational cohomology.

Such groups ought to exist. Explicitly, I think $\text{Aut}(F_2)$ is an example; apparently it's a result due to Dyer and Formanek that this group is complete, and most likely it has nontrivial rational cohomology...