Suppose we have the subgroups $H \subseteq K \subseteq G$ (not necessarily normal), all finite groups. Denote $\mathcal{R}^G_H$ be a complete set of representatives of the cosets of $G$ in $H$, and similarly for the other combinations.
Is it then true that: $$ \mathcal{R}^G_H = \mathcal{R}^G_K \cdot \mathcal{R}^K_H, $$ At least that the product $\mathcal{R}^G_K \cdot \mathcal{R}^K_H$ (i.e. the pairwise product of every pair of elements from those sets) is a complete set of representatives of the cosets of $G$ in $H$?
I didn't manage to prove it and I'm not sure if it's true in the first place.
Let $H \le K \le G$ and let $R_H^K$ and $R_K^G$ be right transversals of $H$ in $K$ and of $K$ in $G$, respectively (i.e. $K = \cup_{k \in R_H^K} Hk$ and $G = \cup_{g \in R_K^G} Kg$). Then $R_H^KR_K^G = \{ k \in R_H^K,g \in R_K^G \}$ is a right transversal of $H$ in $G$.
It is immediately clear that $G = \cup _{ k \in R_H^K,g \in R_K^G } Hkg$.
If $Hk_1g_1=Hk_2g_2$ then, $g_1 = g_2$, and hence $Hk_1=Hk_2$, so $k_1=k_2$.