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?**



Groups with trivial rational homology and their finite index subgroups
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**


