Non-abelian finite groups with exactly $n$ normal subgroups. 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.

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.
