Finding solvable groups with given Fitting subgroup

Let $K=\mathbb Z_2 \times \mathbb Z_2$. What finite solvable groups (up to isomorphism) have Fitting subgroup isomorphic to $K$?

So far I've found that $K, S_4, A_4$ satisfy that property. I don't know whether there are others... What follows is what I know. Let $G$ be a group we want to find. Then there is $n\in \mathbb N$ such that $G^{(n)}={1_G}$ (a derived subgroup is trivial). Its Fitting subgroup is the intersection of all the $p$-cores of $G$, or equivalently, the unique maximal normal nilpotent subgroup.

Since $p$-groups are nilpotent, if $G$ is a $p$-group, then $G=K$. Let's assume $G$ is not a $2$-group. Then there is $q\neq 2$ prime, dividing $|G|$. For every $p\neq 2$ dividing $|G|$, there are more than one Sylow $p$-subgroups, since its core is trivial.

Another important fact about finite solvable groups is that the centralizer of the Fitting subgroup is its center. In our case $C_G(K)=K$. Yet I can't see how this would help.

Is there a general strategy for any group $K$?

Since $C_G(K)=K$, the conjugation action of $G/K$ on $K$ maps $G/K$ into a subgroup of ${\rm Aut}(K) \cong S_3$. We can't have $|G/K| = 2$, so $|G/K| = 1,3,$ or $6$ giving the three examples that you have found. So there are no more examples.
The same strategy would work for any abelian $p$-group $K$. Otherwise it would get more complicated but could still be done because $G/Z(K)$ must be a subgroup of ${\rm Aut}(K)$, and hence $|G|$ is bounded.
• Thanks. Another question then arises: how can I recover $G$ from the quotient $G/K$, without computing all the groups with that order and checking the other properties? – Human Nov 2 '16 at 21:05
• You know the conjugation action of $G/K$ on $K$, so you don't have to consider all such groups. But in general there could be nonsplit extensions. – Derek Holt Nov 2 '16 at 23:22
• Sorry, I still can't see how to find $G$ from the conjugation action... – Human Nov 4 '16 at 20:04
• For a given conjugation action, one possibility id=s the semidirect product of $K$ with $G/K$. But there could also be nonsplit extensions. – Derek Holt Nov 4 '16 at 21:38
• I've just read the definition of split extension given by Wikipedia, but it doesn't help much... Let's go back to $K=\mathbb Z_2^2$. Let's say that the conjugation action of $G/K$ on $K$ maps $G/K$ into ${\rm Aut}(K)$. How do you know that $G \cong S_4$? – Human Nov 5 '16 at 8:45