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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.

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    $\begingroup$ I guess not really: even constructing such an explicit sequence having all finite simple groups as quotients will not be easier than constructing these finite simple groups. Of course there are trivial answers anyways such as "$G_n$ is the direct product of all nonisomorphic of order $n$", or, just a little better "$G_n$ is the free group on $n$ generators in the variety (in the sense of universal algebras) generated by finite groups of exponent $n$ — the latter $G_n$ being finite using Burnside's restricted problem. $\endgroup$
    – YCor
    Commented Feb 5, 2022 at 15:48
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    $\begingroup$ Thanks. I would like to have somethinig like: "all groups of order n are subgroups of the symmetric group S_n" but where we replace "subgroup" by "quotient" $\endgroup$
    – Vanja
    Commented Feb 5, 2022 at 16:54
  • $\begingroup$ I really guess there's nothing of this flavour. For instance, there's not any "easy" construction of any finite group having, say, the Monster sporadic group as quotient (or even as Jordan-Hölder factor). $\endgroup$
    – YCor
    Commented Feb 5, 2022 at 17:53
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    $\begingroup$ The obvious candidate would be the quotient $B(n,n)$ of the free group on $n$ generators modulo the relations generated by $g^n$ for all words $g$. Any group of order $n$ is a quotient of this, but alas these free Burnside groups have a tendency to be infinite themselves. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 6, 2022 at 2:54
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    $\begingroup$ The free group $F_m$ has a canonical finite characteristic quotient $Q_{m,n}:=F_m/C_n$, where $C_n$ is the intersection of all normal subgroups of index $n$. The group $Q_{m,n}$ surjects all $m$-generator finite groups of order $n$ by construction, so this gives a family with the desired property. Unfortunately I don't think anything is known about this family, so I don't think it satisfies the "nice" requirement. $\endgroup$
    – HJRW
    Commented Feb 6, 2022 at 9:20

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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 $\mathbb{Q}$.

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  • $\begingroup$ In the same vein I would have said "the class of all finite groups". Or "the class of all finite groups with nontrivial abelianization". $\endgroup$
    – YCor
    Commented Feb 6, 2022 at 10:07
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    $\begingroup$ That is a nice result (just to be clear, the Robinson is DJS Robinson, so I am not self-congratulating). $\endgroup$
    – Geoff Robinson
    Commented Feb 6, 2022 at 10:34
  • $\begingroup$ Thanks. My question arised exactly thinking about the inverse galois problem over $\mathbb{Q}$. $\endgroup$
    – Vanja
    Commented Feb 6, 2022 at 10:37

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