Aluffi IV.1.18 suggests this exercise: let $S$ be a set endowed with a transitive action of a finite group $G$, and assume $|S| \geq 2$. Prove that there exists a $g \in G$ without fixed points in $S$, that is, such that $gs \neq s$ for all $s \in S$.
There's also a hint to look at the isomorphism between the transitive action on a set and the left-multiplication on $G/H$ for $H$ a stabilizer of some element of the set and connect that with the fact that $G$ is not the union of conjugates of a proper subgroup.
Anyway, if I understand correctly, then the isomorphism between $G-\mathbf{Set}$s $S$ and $G/H$ translates the original claim $\exists g \in G. \forall s \in S. gs \neq s$ to $\exists g \in G. \forall g' \in G. gg'H \neq g'H$. But… $gg'H \neq g'H$ is equivalent to $g \notin g'H$, isn't it? If it is, then isn't the translated claim analogous to "$g$ doesn't belong to any left-coset of $H$"? And, since cosets form a partition, does it even make sense?
I'm surely terribly misunderstanding something here, but where am I wrong?
Best Answer
"But...$gg′H \neq g′H$ is equivalent to $g \notin g′H$, isn't it? "
No. It would be equivalent if $g'H$ would be a subgroup, but it is not.
Note that $gg'H=g'H$ is equivalent to $(g')^{-1}gg' \in H$ which is equivalent to $g \in g'H(g')^{-1}$.
So $gg′H \neq g′H$ is equivalent to $g \notin g′H(g')^{-1}$.