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.

  • $\begingroup$ You might mention that you have already asked this question a week ago on the group pub forum, without getting a satisfactory answer. $\endgroup$
    – Stefan Kohl
    Commented Sep 11, 2017 at 20:04
  • $\begingroup$ I'd indeed be interested for details on nilpotency of solvable quotients $\endgroup$
    – YCor
    Commented Sep 11, 2017 at 20:48
  • 2
    $\begingroup$ An idea of a counterexample, possibly wrong: If the quotient was always Abelian, would that imply that it is Abelian in the case of any infinite subdirect (i.e. subcartesian) product of simple groups? Now the free group $F_2$ is a subdirect product of $A_n, n\ge 2$. $\endgroup$
    – user6976
    Commented Sep 12, 2017 at 3:03
  • $\begingroup$ @MarkSapir That's interesting, although I don't immediately see how to use it to construct an example with finitely many $n$. $\endgroup$
    – Derek Holt
    Commented Sep 12, 2017 at 8:09
  • $\begingroup$ @MarkSapir it's definitely wrong: in case all factors are simple and distinct (as in your example $\prod_{n\ge 5}A_n$), a subdirect product has automatically all its projections (to $\prod_{5\le k\le n}A_k$, for all $n$) surjective, hence perfect. $\endgroup$
    – YCor
    Commented Sep 12, 2017 at 8:24

1 Answer 1


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 $60$.

  • $\begingroup$ I constructed what I think is your smallest example in Magma and verified that $G$ did indeed have a nilpotent quotient of class $2$. My example $G$ has order $2^{48}\cdot 3 \cdot 5$. $\endgroup$
    – Derek Holt
    Commented Apr 9, 2018 at 20:45
  • $\begingroup$ Can one get arbitrary large nilpotency class? $\endgroup$
    – YCor
    Commented Jul 13, 2018 at 20:50
  • $\begingroup$ Yes, that's the main point of the paper. $\endgroup$ Commented Jul 13, 2018 at 23:36

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