Let $\Gamma$ be a countable group and let $\Lambda\_1,\Lambda\_2<\Gamma$ be subgroups. We say that $\Lambda\_1$ is amenable relative to $\Lambda\_2$ if the action of $\Lambda\_1$ on $\Gamma/\Lambda\_2$ by left translation admits an invariant mean. 


Let now $\Sigma<\Lambda<\Gamma$ be subgroups and let $G<N\_\Gamma(\Sigma)$ be a subgroup of the normalizer of $\Sigma$ in $\Gamma$. Assume that $G$ is amenable relative to $\Lambda$. 


Question: Is $\Sigma G$ still amenable relative to $\Lambda$ ? 



Relative amenability of subgroups
Here is a counterexample. Take $\Delta$ a non-amenable group, and set $\Gamma = (\mathbb Z/2 \mathbb Z) \ltimes (\Delta \times \Delta)$, where the action is via the flip automorphism on $\Delta \times \Delta$. Let $\Sigma = \Lambda$ be the first copy of $\Delta$, and let $G$ be the second copy of $\Delta$.


Then $G$ is conjugate to $\Lambda$ and hence $\Gamma / \Lambda$ has a $G$-fixed point, and in particular, has a $G$-invariant mean. On the other hand, $\Gamma / \Lambda$ has two $\Sigma G$-orbits and neither can have an invariant mean since both witness the non-amenability of $\Delta$. 


