TITLE: Exact factorization of finite groups
QUESTION [5 upvotes]: Suppose that $G=MN$ and $G=MP$ are two exact factorization of a finite group $G$. What is the relation between $M$ and $P$? Clearly if $G=MN$ then $G=M(mNm^{-1})$ is another factorization of $G$. Is this the only possibility to change $N$ into $P$?

REPLY [2 votes]: The answer of the problem is the following: 
A group G has two exact factorizations $G = M N = M P$  if and only if there exists a unitary bijective map (just a map, not a morphism of groups) $v : N \to P$ such that $n v(n)^{-1} \in M$, for all $n \in N$ and $v$ satisfy four natural compatibility conditions (not very transparent to write down here -- they are similar, mutatis mutandis, to the one given in Propositions 2.1 in http://front.math.ucdavis.edu/0903.5060 -- for example one of them is that v is a morphism of $M$-sets).