Let $\mathfrak{N}$ be the class of all non-abelian finite groups and define $\nu: \mathfrak{N} \rightarrow \mathbb{N}_{\gt 1}$ by $\nu(G)=|\{{1} \leq N \leq G: N$ normal in $G\}|$. Is the map $\nu$ surjective? In other words, for any positive integer $n \gt 1$, does there always exist a non-abelian group with exactly $n$ normal subgroups? Example: if $n$ is a (non-trivial) power of $2$, then it is guaranteed: take a direct product of $n$ copies of a non-abelian simple group.

  • $\begingroup$ It doesn't matter much for the problem statement, but $\mathfrak N$ is of course a proper class, not a set. $\endgroup$ Feb 20, 2014 at 9:39
  • $\begingroup$ Ah, right will edit. $\endgroup$ Feb 20, 2014 at 9:41
  • $\begingroup$ You can't get $1$ because the trivial group is abelian. And if you specify proper subgroups then your example needs to change. $\endgroup$ Feb 20, 2014 at 10:03
  • $\begingroup$ Mark, thanks, that is correct, but I do not want proper normal subgroups. By the way, this is only a translation of -2 of the map $\nu$. $\endgroup$ Feb 20, 2014 at 10:10

2 Answers 2


The answer is yes. Let $p$ be an odd prime, and consider the dihedral group of order $2p^n$:

$$G = D_{2p^n} = \langle x, y \mid x^2 = y^{p^n} = 1, x^{-1}yx = y^{-1} \rangle$$

Then the proper normal subgroups of $G$ are precisely $\langle y^d \rangle$, where $d$ is a divisor of $p^n$. Thus $G$ has exactly $n+2$ normal subgroups.

  • 1
    $\begingroup$ This is neat, thanks! $\endgroup$ Feb 20, 2014 at 10:53
  • $\begingroup$ I agree with Nicky. Very nice! $\endgroup$
    – user1729
    Feb 20, 2014 at 10:54

Yes. For $n \ge 1$, we can construct solvable examples $G_n$ having exactly $n+1$ normal subgroups that form a chain. We can start with $G_1 = C_2$, which is abelian, but they will be nonabelian for $n>1$, and to solve your problem for $n=2$ you need to take a nonabelian simple group.

Assume that we have constructed $G_n$ with this property. So $G_n$ has a unique minimal normal subgroup $N$. Choose a prime $p$ and an irreducible ${\mathbb F}_pG$-module $V$, such that $N$ does not act trivially on $V$. We can certainly do that if we choose $p$ not dividing $|G|$. Then, since $N$ is the unique minimnal normal subgroup of $G_n$, $V$ must be a faithful module. Now we can define $G_{n+1} = V \rtimes G_n$, and it has the required properties, with the elementary abelian subgroup $V$ being its unique minimal normal subgroup.

I think you can probably do this also using just two alternating primes. So, with primes $2,3$, you could choose $G_1=C_2$, $G_2=S_3$, $G_3=S_4$, $G_4=3^3:S_4 = {\tt SmallGroup(648,704)}$, $G_5=2^6:3^3:S_4$, etc.


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