I would like to a get a result similar to the third isomorphism theorem, but I want to deal with sets of cosets, rather than a quotient subgroup.
Let me formalize it.
Say $G$ is a group, with subgroups $A\subset B\subset G$. We may assume $A$ is normal in $B$, but no other normality may be assumed.
Let $O$ be a set of sets of cosets $O=\{C_i\}_{i\in I}$ where $C_i=\{Ac_{ij}\}_{j\in J_i}$ for $c_{ij}\in G$, that is $O\subset P(G/A)$ where $P(X)$ is the set of subsets, and $G/A$ is the space of $A$-cosets. We may assume that the elements of $O$ are pairwise disjoint, although I doubt it will be necessary.
Define $O^\prime=\{D_i\}_{i\in I}$ where $D_i=\{Bc_{ij}\}_{j\in J_i}$. That is, we look at the same cosets, but this time with respect to $B$.
I would like to give a lower bound on the cardinality of $O^\prime$.
I would think that the desired $|O^\prime|\geq \frac{1}{[B:A]} |O|$ should hold, where $[B:A]$ is the index.
An attempt of a proof would be the mimic the proof of the third isomorphism theorem, which pretty much gives this result if everything was a normal subgroup.
I would define $f:O\to O^\prime$ by $f(C_i)=D_i$ and would try to prove that $|f^{-1}(D_i)|\leq [B:A]$. If $C_i$ are singletons, this is easy. If these are larger sets, it looks like it doesn't quite work, or at least I cant seem to be able to do it. Any help would be appreciated.
P.S: $[B:A]$ may be assumed to be finite. If necessary, $O$ may be assumed to be finite.