For any group $H$ and any $n \in \mathbb{N}$, it is always possible to find a group $G$ which contains $H$ as an $n$-index subgroup. For example, we can simply add a single generator $x$ to $H$ which commutes with the generators of $H$ and which has order $n$. In this simple case, one transversal takes the form $\{1, x, ... , x^{n-1}\}$ which forms a multiplicative set.
Given a group $H$ and $n\in\mathbb{N}$, can we always define relations on the formal elements $\{g_1, ... , g_n\}$ so that $G = g_1H \sqcup ... \sqcup g_nH$ forms a group and that $\{g_1, ... , g_n\}$ isn't multiplicative? How about if we force $H = J*K$?
And finally, what about $H\subseteq G$ which admit no multiplicative transversal?
See here for some context: Examples of non-free group actions on trees with finite edge-stabilizers
Any advice/insight would be greatly appreciated.