Let $G$ be a finite group, $K$ an abelian normal subgroup and $\Omega$ the set of all transversals of $K$ in $G$. Then $G$ acts by left multiplication on $\Omega$.
That is, $(S,x)\mapsto xS$, $x\in G, S\in\Omega$ is an action, this, according to the book I'm reading. The first thing that must be checked is that it is a function into $\Omega$. Let $S=\{s_1,...,s_n\}$. Multiplication on the right of the cosets of $K$ by elements of $G$ is an action. So if $x\in G$, $\{Ks_1,...,Ks_n\}=\{Ks_1x,...,Ks_nx\}$ and $\{s_1x,...,s_nx\}$ is a transversal. But I need $\{xs_1,...,xs_n\}$ to be a transversal. I can't conquer this difficulty. With the normality of $K$ all I get is that there is no distinction between left and right transversals. $K$ being abelian serves less.
For left multiplication to be an action I also need $xyS=yxS$. This is true iff $x^{-1}y^{-1}xyS=S$. That is I should have $S^{xy}=(S^x)^y$. But I can't prove this neither. Multiplication on the right would be easier. Any hint?
You have described the solution yourself. There is no distinction between left and right transversals. So why not prove that it acts on the set of left transversals instead? Then you just need to show that $xs_iK=xs_jK$ implies $i=j$, and you know that.
To be an action you only need that $x(yS)=(xy)S$, no? Certainly $xyS=yxS$ if and only if the quotient $G/S$ is abelian.
Generally, if you define a group action via multiplication by things in the group, it's automatically an action. That's not a mathematical statement of course, more of a guiding principle.
Edit: I don't see personally how $K$ being abelian makes much of a difference. Or $G$ finite. Or normality of $K$, for that matter, as long as you use left transversals for a left action.