Smallest normal subgroup making quotient abelian, nilpotent, solvable 
Given a finite group $G$ that is not abelian, nilpotent, or solvable, what is the smallest normal subgroup $H$ in each case such that $G/H$ is abelian, nilpotent, or solvable (respectively)?

In the abelian case it seems clear that the correct subgroup is the commutator subgroup. But what of the other two? Perhaps I'm just shaky with the concepts (of nilpotency and solvability) but I'm having trouble finding the right way to tackle this problem.
Edit: Well I've tried the most obvious thing and it worked out.
For the nilpotent case I took the intersection of all the subgroups in the lower central series, and for the solvable case I took the intersection of all subgroups in the derived/commutator series. 
 A: The things you are asking about are quite well studied:
Suppose $G$ is a group and $\mathfrak{U}$ is a variety of groups (a class of groups, that is closed under taking subgroups, quotients and direct products; examples of varieties include abelian groups, nilpotent groups of class less than $c$ for any constant $c$ and soluble groups of class less than $c$ for any constant $c$). Then the smallest normal subgroup $H$ of $G$, such that $\frac{G}{H} \in \mathfrak{U}$ is called verbal. Basic properties of verbal subgroups can be seen under this link
What's about the classes of all nilpotent (soluble) groups, then this does not work (because they are not varieties). However, we can look at the intersection of all verbal subgroups corresponding to the varieties of nilpotent (soluble) groups of class less than $c$ for all possible constants $c$. This subgroup will be normal and there are two cases possible:
1) Quotient by it is nilpotent (soluble). Then it is the subgroup you are looking for.
2) Quotient by it is non-nilpotent (non-soluble). Then the subgroup you were looking for does not exist.
