Suppose, $G$ is an arbitrary group.
Define
$$F_G = \{ B \subset G | \exists \text{ finite } C \subset G \text{ such that } CB = \{cb|c \in C, b \in B\} = G\}$$
and
$$I(A,B)=\min\{ |C||A \subset CB\},$$
where $A \subset G$, $B \in F_G$ (note, that $I(A, B)$ is always non-zero), and
$$M(A) = \inf\left\{ \frac{I(A, B)}{I(G, B)} | B \in F_G\right\}.$$
Suppose, $A$ and $B$ are subsets of $G$.
Does $A \cap B = \emptyset$ always imply that $M(A \cup B) = M(A) + M(B)$?
In case if $G$ is finite, it definitely does: That follows from the trivial fact, that in finite groups $M(A)|G| = |A|$.
When $G$ is infinite, the things become more difficult: The only facts, that I managed to deduce, is following inequalities: $$M(A \cup B) \geq \max(M(A), M(B))$$ (that is due to the fact, that $A \subset B$ implies $M(A) \leq M(B)$) and $$M(A \cup B) \leq M(A) + M(B)$$ (as $I(A \cup B, C) \leq I(A, C) + I(B, C)$). From them easily follows, that if $M(A) = 0$, then $$M(A \cup B) = M(B) = M(A) + M(B).$$
However, I do not know the answer for the question in case $M(A) \neq 0$, $M(B) \neq 0$ and $G$ is infinite.
Any help will be appreciated.