I have a question about amenable groups. The notion of amenability I am using is: The action of $G$ on $k$ ($k$ locally compact in a topological vector space) is amenable if there exists a point $x \in k$ that is $G$-fixed. $G$ is amenable if every action of a locally compact $G$-space $k$ is amenable.
Is the following statement true? $G/N$ is amenable if and only if $G$ is amenable.
Here is my thought. If $G/N$ is amenable, let $k$ be a $G$-space. Then I can define the action of $G/N$ as $$gN.x:=g.X$$ then it becomes a $G/N$ space and thus the action is amenable. For the other direction I would do the same, if $G$ is amenable, I take a $G/N$ space and define the $G$ action $$g.X:=gN.x$$ that is amenable.
Is the previous correct? My notes hint that this is wrong but I do not know why
Thank you!
If $G$ is amenable then so is $G/N$ by your second argument.
The problem with your first argument is that not every action of $G$ yields an action of $G/N$: in order for $gN.x := g.x$ to be well-defined, you need that $N$ acts trivially, or in other words: the action of $G$ is given by a homomorphism $\varphi\colon G \to \operatorname{Aut}(k)$, and if you want to obtain a $G/N$-action, you need $N \subseteq \ker{\varphi}$.
A trivial counterexample is obtained by taking a non-amenable group $G$ and $N = G$, then $G/N = \{1\}$ is amenable while $G$ is not. A less trivial counterexample is obtained by taking $G = F_2$, the free group on two generators, which isn't amenable, and $N = [G,G]$, its commutator subgroup. Then $G/N \cong \mathbb{Z}^2$ is amenable because it is abelian, but again $G$ is not amenable.
What is true, however, is that in a short exact sequence $N \rightarrowtail G \twoheadrightarrow Q$ of groups $G$ is amenable if and only if $N$ and $Q$ are amenable: The class of amenable groups is closed under passing to subgroups, quotients and extensions.