I'm interested in whether there exists a well known class of finite groups that generates all finite groups through quotienting.


I will be more formal: does there exist a class of ("nice") well known finite groups $\{G\_n\}$ such that for any finite groups $G$ we have that there exists $G\_n$ and a normal subgroup $N$ of $G\_n$ such that $G\_n/N=G$?


The answer is trivially positive if we remove the condition that $G\_n$ be finite: we can take the free groups of all finite ranks.



Families of finite groups of which every finite group is a quotient
It obviously depends a lot on your definition of "nice", but one class defined by a reasonably nice property, while also relevant in theory, is the class of all complete groups (where complete means that both the center and the outer automorphism group are trivial). That every finite group is a quotient of a complete group (even with some extra properties) is shown in Hartley, Robinson, "On finite complete groups" (1980).


I remember noticing a curious consequence of this fact some years ago: Complete groups have the property that, if they themselves are a normal subgroup of some finite group, then they must also be a quotient of that group. This implies that, if every finite group $G$ is a normal subgroup of some Galois group over, e.g., $\mathbb{Q}$, then (via embedding $G$ as quotient into a complete group and then using the above property), every finite group is indeed (a quotient of a Galois group, and hence itself) a Galois group over $\mathbb{Q}$.


