7
$\begingroup$

Apart from the direct products, what are some "interesting" or "naturally occurring" examples of extensions $$ 1 \to N \to G \to Q \to 1 $$ of finite groups such that neither $N$ nor $Q$ is solvable?

I feel a bit stupid for asking the question, but I don't think I know a single example that isn't split.

Just to set the bar pretty low: essentially the only example I can think of is to take the wreath product $\mathfrak{S}_n\wr\mathfrak{S}_m$ (group of permutations of $mn$ elements preserving a partition into $m$ sets of $n$), for $m,n\geq 5$, which is a semidirect product $(\mathfrak{S}_n)^m \rtimes \mathfrak{S}_m$. Anything which is not a trivial variation on this construction interests me.

Improve this question
$\endgroup$
6
  • $\begingroup$ To make the question precise you might require that $N,Q$ both have no abelian Jordan-Hölder factor. $\endgroup$
    – YCor
    Commented Jun 2, 2018 at 6:23
  • $\begingroup$ Joseph Ayoub once gave me an (somewhat involved) example of a non-split exact sequence of finite groups without abelian JH factors, relying on the existence of a nonabelain finite simple group $S$ for which $Aut(S)\to Out(S)$ does not split. $\endgroup$
    – YCor
    Commented Jun 2, 2018 at 9:14
  • $\begingroup$ PS Ayoub's example seem to exactly coincide with Derek's one. After embedding $Out(S)$ into a nonabelian simple group $K$, he uses an identification between non-abelian 1-cohomology pointed sets $H^1(Out(S),Out(S))=H^1(K,Out(S)^{K/Out(S)})$. But I've failed to understand details. (PS $S$ is not the same here and in Derek's post.) $\endgroup$
    – YCor
    Commented Jun 2, 2018 at 9:39
  • $\begingroup$ Of course is you allow $N$ and $Q$ to have abelian composition factors then it is much easier to find examples. For example you could take a central product of two (nonisomorphic) quasisimple groups with a common nontrivial centre. $\endgroup$
    – Derek Holt
    Commented Jun 2, 2018 at 9:47
  • $\begingroup$ @DerekHolt the central product might split. I think you take $N$ to be the central product and the left arrow in the exact sequence being given by the first factor. Say that the center $Z=\{1,s\}$ is cyclic of order 2, so $G$ is the quotient of $N\times N$ by $\{(1,1),(s,s)\}$. Then the diagonal yields a semidirect product decomposition (similar to a semidirect decomposition $SO(4)=SU(2)\rtimes SO(3)$). $\endgroup$
    – YCor
    Commented Jun 2, 2018 at 10:43

1 Answer 1

Reset to default
10
$\begingroup$

Here is a way you can construct nonsplit examples in which all composition factors are nonabelian.

Many finite nonabelian simple groups do not split over their automorphism groups. The smallest such example is the group often known as $M_{10}$, which is the point stabilizer in the Mathieu group $M_{11}$. It has order $720$ and is the extension of ${\rm PSL}(2,9)$ by a product of a field and a diagonal automorphism. So it has a normal subgroup $H$ of index $2$ with $H \cong A_6 \cong {\rm PSL}(2,9)$.

Now $M_{10}/H$ is of course solvable so we don't have an example yet, but we can use a wreath product type construction to get what we want.

Let $K$ be any finite nonabelian simple group, and let $S$ be any subgroup of order $2$ in $K$. Let $P$ denote the image of the permutation action of $K$ on the cosets of $S$. So, for example, if $K=A_5$ then $P$ is a subgroup of $S_{30}$.

Now the permutation wreath product $W$ of $H$ by $P$ is of course a split extension. $1 \to H^{|K:S|} \to W \to P \to 1$.

But there are in general other extensions of this type, and there is a theorem that says that the equivalence classes of such extensions are in one-one correspondence with the extensions of $H$ by $S$. So, if there is a nonsplit extension of $H$ by $S$, then there is also one of $H^{|K:S|}$ by $P$. (I will look for a reference for that result later.)

For example, if we take $H={\rm PSL}(2,9) \cong A_6$ as above, and $K=A_5$, then we get a nonsplit extension with normal subgroup $A_6^{30}$ and quotient $A_5$.

I found a reference for the result about wreath product type extensions in an old paper of my own. It is proved as Theorem 1 of D.F. Holt, Embeddings of group extensions into Wreath products, Quar. J. Math. (Oxford) 29 (1978), 463--468. But I suspect it was known earlier. The main result of that paper is a generalization of the Krasner Kaloujnine theorem about embedding group extensions into standard wreath products.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Is $A_6^{30} \cdot A_5$ the smallest non-split extension of one anti-solvable group by another? $\endgroup$
    – Daniel Sebald
    Commented Nov 17, 2022 at 1:43
  • 1
    $\begingroup$ @DanielSebald I think there should be an example $A_6^{15} \cdot A_5$ using the non-split extension ${\rm Aut}(A_6) \cong{\rm P\Gamma L}(2,9)$. I would guess that that is the smallest, but it might be tricky to prove it. $\endgroup$
    – Derek Holt
    Commented Nov 21, 2022 at 13:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .