Finite-index normal subgroup acts amenably on compact Hausdorff space $X$

Question

Let $\Gamma$ be a discrete group acting on a compact Hausdorff space $X$. The action is called (topologically) amenable if there exists a net of continuous maps $m_i: X\rightarrow \text{Prob}(\Gamma)$ such that for each $s \in \Gamma$, $$\lim_{i\rightarrow\infty}\left(\sup_{x\in X}\left\Vert s.m_{i}^{x}-m_{i}^{s.x}\right\Vert _{1}\right)=0,$$ where $s.m_i^x(g)=m_i^x(s^{-1}g)$. Here $\text{Prob}(\Gamma)$ denotes the set of propability measures on $G$ and continuous means that for every convergent net $x_j \rightarrow x\in X$ we have $m_i^{x_j}(g) \rightarrow m_i^x (g)$ for all $g \in \Gamma$

Question: Let $\Gamma$ be a discrete group acting on $X$. Is it true that if $N\vartriangleleft\Gamma$ is a finite-index normal subgroup that acts amenably on the space $X$, then the action action $\Gamma \curvearrowright X$ is amenable as well?

Answer 1

This is a corollary of the following fact, found as Proposition 5.1.11 in Brown & Ozawa's "$C^*$-Algebras and Finite-Dimensional Approximations".

Let $\Gamma$ be a discrete group, $N$ be a normal subgroup and $\overline \Gamma=\Gamma/N$. If $Y$ is a compact amenable $\overline \Gamma$-space and $X$ is a $\Gamma$-space which is amenable as a $N$-space, then $X\times Y$ (with the diagonal $\Gamma$-action) is an amenable $\Gamma$-space.

Given this, the result is straightforward: Since $\Gamma/N$ is a finite group, it's trivial action on a one-point space $\{p\}$ is amenable. As the action $N\curvearrowright X$ is also amenable, this implies that the diagonal action $\Gamma\curvearrowright X\times\{p\}=X$ is amenable.