I am looking for smallest example of a group $G$ such that:

  • $G$ is a finite, non-abelian group
  • $G$ is not simple
  • $G$ has non-trivial, proper, normal subgroups: $H_1, H_2, \dots $
  • $H_1, H_2, \dots $ are not (all) abelian
  • There is more than one normal subgroup (hence $H_1, H_2, \dots $)

In particular I would like to generate the multiplication table for such a group, if possible with permutations (all finite groups are subgroups of $S_n$, right!).

I do have access to gap, but I am just beginning with that

Since smallest possibility for non-abelian subgroup is order $6$, and I look for at least $2$ normal subgroups, then I believe $\text{Order}(G)$ must be at least $18$ (as if index of $H_1$ was $2$, then $H_1$ would be the unique normal subgroup). But none of the order $18$ finite groups fit all the above criteria.


  • 1
    $\begingroup$ Try $\operatorname{SL}(2,3)$, of order $24$. $\endgroup$
    – M. Vinay
    Commented May 1, 2016 at 2:48
  • $\begingroup$ Should there be (at least) two non-Abelian normal subgroups? $\endgroup$
    – M. Vinay
    Commented May 1, 2016 at 3:04
  • 1
    $\begingroup$ There are a few groups of order 16 that have both the dihedral group and the quaternion group of order 8 as subgroups, for example the semidihedral group of order 16. (see groupprops.subwiki.org/wiki/Semidihedral_group:SD16) $\endgroup$
    – verret
    Commented May 1, 2016 at 10:20
  • $\begingroup$ I believe the order 16 cases do not fit as dihedral and quaternion subgroups are not both normal in the parent? My understanding is that, in general, if there is an index 2 normal subgroup then it is uniquely normal of that order? $\endgroup$ Commented May 1, 2016 at 21:31
  • $\begingroup$ @KevinMaguire Here's why that can't be right: Abelian groups are obvious counterexamples. All their subgroups are normal, and of course they can have subgroups of index $2$ as well as other indexes (even apart from the trivial ones). $C_6$, for example. Okay, let's say you meant only non-Abelian groups. The quaternion group of order $8$ has three different subgroups of index $2$, and a subgroup of index $4$, all of which are normal! One last example: Any dihedral group $D_{2n}$ with $n$ even has an index $2$ subgroup, necessarily normal, and a non-trivial center of order $2$ (index $n$). $\endgroup$
    – M. Vinay
    Commented May 2, 2016 at 2:17

1 Answer 1


$D_{12} \cong C_2 \times S_3$ is nonabelian of order $12$, has at least $2$ normal subgroups ($C_2 \times 1$ and $1 \times S_3$), and they are not all abelian.

  • $\begingroup$ $C_2\times1\cong C_2$ is abelian $\endgroup$
    – janmarqz
    Commented May 1, 2016 at 3:03
  • $\begingroup$ @janmarqz That's what I was thinking, but after seeing the answer, I notice that OP hasn't explicitly asked for two non-Abelian normal subgroups. $\endgroup$
    – M. Vinay
    Commented May 1, 2016 at 3:06
  • 1
    $\begingroup$ Yes, as OP I wanted that at least 1 of H1, H2, ... were not abelian, not necessarily all of them. So thanks. $\endgroup$ Commented May 1, 2016 at 21:34
  • 1
    $\begingroup$ I guess SL(2,3) is first case where H1 and H2 are both abelian. I wish I knew how to generate a (the?) permutation group representation? Aside from doing by hand - tedious, but possible. $\endgroup$ Commented May 1, 2016 at 21:44
  • $\begingroup$ In fact, $C_2 \times S_3$ contains two subgroups isomorphic to $S_3$, both of which are normal (since they have index 2). The less obvious one consists of $(0, \sigma)$ for $\sigma \in A_3$, together with $(1, \sigma)$ for $\sigma \notin A_3$. Viewing $D_{12}$ as the symmetries of a hexagon, one of these is the symmetries of a triangle whose vertices are three vertices of the hexagon, and the other is the symmetries of a triangle whose vertices are midpoints of edges. $\endgroup$ Commented May 6, 2016 at 0:34

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