I'm struggling with this question:
Let $G$ be a group that acts freely on a set $X$, and $K < G$. Prove that the normaliser of $K$, $N(K)$, acts freely on $X/K$.
The question was posed to me like this, but it doesn't even seem to be fully specified. How can we prove that $N(K)$ acts freely on $X/K$ when we don't even know how it acts on $X/K$? Do we assume that it acts in the natural way, i.e. that $n *(K \cdot x) = (nK)\cdot x$? I've made this assumption but keep getting stuck when I try to prove it.
It is $N(K)/K$ which acts freely on $X/K$, not $N(K)$ since $K$ acts trivially on $X/K$.
We denote by $p:X\rightarrow X/K$ the quotient map. Let $g\in N(K)$ and $x\in X$, we define $g.p(x)=p(g.x)$. This action is well defined, if $p(x)=p(y)$, $y=kx, k\in K$, $gy=gkx=gkg^{-1}g(x)$ since $g\in N(K)$, $gkg^{-1}\in K$ and $p(g.x)=p(g.y)$. Since $K$ acts trivially on $X/K$ this induces an action of $N(K)/K$ on $X/K$.
Let $\bar g\in N(K)/K$ Suppose that $\bar g.p(x)=p(x)$. Let $g\in N(K)$ whose is image by the quotient map $N(K)\rightarrow N(K)/K$ is $\bar g$, we deduce that $gx=kx, k\in K$, we deduce that $g^{-1}kx=x$ since $G$ acts freely, $g=k$