The translation theorem of Galois theory (as I learned it) roughly states that two Galois extensions $L,K$ of a ground field $k$ that are "far enough" (i.e. $L \cap K = k$) inside the same algebraic closure $\bar k$ of $k$ give rise to an extension $M = LK$ (the field compositum) of $k$ having both, $L$ and $K$ as subextensions with Galois group
$$Gal(M/k) = Gal(L/k) \times Gal(K/k)$$
(I know that there are some generalisations that not both extensions have to be Galois and that the Galois group $Gal(L/k)$ is isomorphic to the Galois group $Gal(M/L)$ but that should not play a role for my question)
Is there a converse statement to the translation theorem of Galois theory?
I'm thinking of a statement that whenever one can express the Galois group $H$ of an extension $M$ over a field $k$ as a direct product of groups, then those groups have to be Galois groups of certain extensions of $k$ and one can realize them? Or is this just what normal Galois theory gives?
EDIT And if we start from a Galois extension $M$ over $L \supseteq k$, under which conditions do we get a Galois extension $M'$ over $k$ which has the same Galois group $Gal(M'/k) \cong Gal(M/L)$
Thank you :-)