Converse of Galois translation theorem

Question

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 :-)

Answer 1

As user8268 pointed out: If the Galois group $G$ of a Galois extension $F/k$ is the direct product $$G = H_1 \times H_2$$ of two normal subgroups, then the main theorem of Galois theory is the converse of the translation theorem I've looked for: Each fixed field $K_i = F^{H_i}$ has $Gal(K_i / F) = H_i$ for $i = 1,2$ and the compositum equals $F = K_1 K_2$.

My second question can be answered in the following way: If the group $Gal(L/k)$ acts on $Gal(M/L)$ trivially and $M/k$ is Galois, then we can write $$Gal(M/k) = Gal(M/L) \times Gal(L/k)$$ and the above statement gives a Galois extension $M'/k$ with Galois group $Gal(M'/k) = Gal(M/L)$. The trivial action could come from (different) prime group orders.

Thank you for your comments :-) That helped me a lot!