TITLE: Subdirect product of perfect groups
QUESTION [25 upvotes]: Let $G$ be a subdirect product of finitely many perfect groups $S_1,\ldots,S_n$. That is, $G \le S_1 \times \cdots \times S_n$, and $G$ projects onto each of the direct factors $S_i$.
It is not difficult to construct examples in which $G$ is not perfect. For example, there is a subdirect product of ${\rm SL}_2(5)^2 = (2.A_5)^2$ with the structure $2^2.A_5$ that is not perfect.
Let $G/N$ be a solvable quotient of $G$. I would really like to know whether $G/N$ is necessarily abelian. It can be proved that $G/N$ is nilpotent of class at most $n-1$ (I can give more details of that on request), so we can assume that $n \ge 3$.
If it helps to assume that the $S_i$ are finite, then please do so.
This problem arose in a study of the complexity of certain algorithms for finite permutation and matrix groups. The group $G$ in the applications is the kernel of the action of a transitive but imprimitive permutation (or matrix) group on a block system. So in that situation the $S_i$ are all isomorphic, and ${\rm Aut}(G)$ induces a transitive action on the direct factors $S_i$, but I am doubtful whether assuming those properties would help much in trying to answer the question.
Here is a proof that $G/N$ is nilpotent of class at most $n-1$. This proof is due to Keith Kearnes.
Call a normal subgroup $N$ of $G$ co-${\mathcal P}$, if $G/N$ has property ${\mathcal P}$, which could be solvable, nilpotent, or perfect.
We claim that if $G$ has co-perfect normal subgroups $K_1,\ldots,K_n$, with $\cap_{i=1}^n K_i = \{ 1 \}$, and $N$ is a co-solvable normal subroup of $G$, then $G/N$ is nilpotent of class at most $n-1$.
We can then apply this claim to the original problem with $K_i$ equal to the subgroups of $G$ that projects trivially onto $S_i$ to deduce the required result.
To prove the claim, observe first that, since each $NK_i$ is both c-perfect and co-solvable, we have $NK_i = G$ for all $i$.
In general, for normal subgroups $A,B$ of a group $X$, we have $[AB,X] = [A,X][B,X]$. We write $[A,B,C]$ for $[[A,B].C]$, etc.
Using this gives
$$[G,G,\ldots,G] = [NK_1,NK_2,\ldots,NK_n] = [K_1,K_2,\ldots,K_n]\prod_i [A_{i1},A_2,\ldots,A_{in}],$$
where, in each of the terms $[A_{i1},A_2,\ldots,A_{in}]$, at least one of the $A_{ij}$ is equal to $N$.
Now the first term $[K_1,K_2,\ldots,K_n]$ lies in $\cap_{i=1}^n K_i$ and so is trivial, while each of the remaining terms in the product lies in $N$. So $[G,G,\ldots,G] \le N$, and hence $G/N$ is nilpotent of class at most $n-1$, as claimed.

REPLY [12 votes]: In this problem one has
a finite sequence of perfect groups
$P_1,\ldots,P_k$, a subdirect subgroup
$G\leq (P_1\times \cdots\times P_k)$,
and a normal subgroup $N\lhd G$.
The question is:
Under the above assumptions,
if $G/N$ is solvable, must it be abelian?
The answer is: No.
This is explained in 
a manuscript Peter Mayr,
Nik Ruskuc and I recently 
submitted to the arxiv.
We show that if the quotient
$G/N$ is solvable, then it must be nilpotent,
but there is no restriction on the 
possible nilpotence class.
The smallest example in the paper where
$G/N$ is solvable but not abelian
involves a perfect group $P$ satisfying 
$|P|=2^{22}\cdot 3\cdot 5$,
a subdirect subgroup $G\leq P^4$, and
a normal $N\lhd G$ with $G/N$ 
nilpotent of class $2$.
The group $P$, which is constructed
as a subgroup of $\text{SL}_6(\mathbb F_4)$, 
has a nilpotent
normal subgroup of order $2^{20}$
and a complementary subgroup isomorphic to
the simple group of order $60$.