Consider a finite group $G$ and a subgroup $H \subseteq G$. There is a transitive group action of $G$ on the set of left cosets $gH$ by left multiplication, and the stabilizer of $gH$ is $gHg^{-1}$. Now, restrict this action to $H$. It is clearly no longer transitive, as $1H$ is a fixed point.
What are some natural conditions on $G$ and $H$ such that $H$ acts transitively on the set of non-identity cosets? The only (necessary but not obviously sufficient) condition I can come up with is that either $[G:H] = 2$, or the normalizer of $H$ is just equal to $H$. This is necessary since if $g$ normalizes $H$, then $HgH = gHg^{-1} gH = gH$, so if there are more than one non-identity cosets, $g$ must be in $H$.
My motivation is that I'm working through Serre's Linear Representations of Finite Groups, and exercise 7.2 asks me to determine when the permutation representation $\rho$ of $G$ given by its action on the left cosets satisfies $\rho = 1 \oplus \tau$ for $\tau$ irreducible.
Letting $\chi$ be the character of $\rho$ and $\psi$ the character of $\tau$, we have $$(\psi, \psi) = (\rho, \rho) - 2(\rho, 1) + 1$$
Since $\rho$ is a transitive permutation representation, $(\rho, 1) =1$, so $\tau$ is irreducible iff $(\rho, \rho) = 2$.
Since $\rho = \text{Ind}_H^G(1)$, Fronenius reciprocity gives $$(\rho, \rho) = (\rho, \text{Ind}_H^G(1)) = (\rho|_H, 1)$$
Since the restriction of $\rho$ to $H$ is still a permutation representation, $(\rho|_H, 1)$ is the number of orbits of this action under $H$.