I have been puzzled by this lately. And first, thank you to whoever takes the times to read through these questions.
Consider a finite group $G$ which is solvable. Let $F(G)$ define the Fitting subgroup of $G$ and $soc(G)$ the socle of $G$.
Consider for the following cases that $G$ is a primitive group (There is a maximal subgroups $M<G$ such that $Core_G(M) = \{id_G\}$).
Then, from “Finite Soluble Groups” of Doerk and Hawkes, we have that (Theorem 15.6, pp54)
Let $G$ be a primitive soluble group with stabilizer $M$. Then $G$ has a unique minimal normal subgroup $N$, the stabilizer $M$ complements $N$ in $G$, and $N=C_G(N) = F(G)$.
(I believe here that "the stabilizer $M$ complements $N$ in $G$" means that $G = F(G)M$.
A few pages ahead we read (pp. 56 - Concluding Remarks)
If $G$ is a primitive soluble group, we can write $G=NM$, with $N=soc(G)$ and $M$ a stabilizer.
First question: If $G$ is a primitive solvable group, then $soc(G)=F(G)$?
EDIT : Removed the second part of the question
Now, consider that $G = N\rtimes A$ is a solvable (but not necessarily primitive) group, with $N$ being an abelian normal subgroup of $G$, $N\cap A = \{id_G\}$ and $Core_G(A)=\{id_G\}$. Let $k = |N|$. Computationally I can see that, for small cases, when $gcd(k,2) = 1$, then $N = F(G)$.
Second question: Is this a known fact? If so, is there a reference about it?
Third question: Based on this, given a $G$ solvable group, if $gcd(|F(G)|,2)=1$, then there is a $A\leq G$ such that, $F(G)\cap A =\{id_G\}$ and $G = F(G) \rtimes A$? Does this need $A$ to be core-free (or else it would be in the $p$-cores of $F(G)$)?
Final question: Is there anywhere where I could find more information on the structure of these kind of solvable groups (with more or less restrictions)?
Thank you!
Let $N=F(G)$ be the minimal normal subgroup of $G$ in Theorem 15.6 of Doerk and Hawkes.
If $L$ was another minimal normal subgroup of $G$ then we would have $L \cap N = \{1\}$ and so $[L,N] = 1$ contradicting $C_G(N)=N$. So $N$ is the unique minimal normal subgroup and hence $N = {\rm soc}(G)$.