Unique normal subgroups of every possible order

Question

Can one characterize groups $G$ for which there is a unique normal subgroup of order $d$ for every divisor of the order of $G$? For example must they be solvable?

Answer 1

Note: I'm interpreting the assumption on $G$ as: for every $d$ dividing $|G|$, the set of normal subgroups of $G$ of order $d$ has size exactly $1$.

Such a group must be cyclic. If $G$ is such a group, then for each prime $p$ dividing $|G|$, there is a normal Sylow $p$-subgroup of $G$, which is then unique, so $G$ is nilpotent and a product of its Sylow subgroups.

Now let $P$ be a Sylow $p$-subgroup of $G$. Since $P$ is a direct factor of $G$, a subgroup $H \le P$ is normal in $P$ iff it is normal in $G$. Now any maximal subgroup of $P$ is normal in $P$, hence normal in $G$, hence unique (in $G$ or $P$), so $P$ has a unique maximal subgroup, which implies $P$ is cyclic.

Thus all Sylow subgroups of $G$ are cyclic and normal, so $G$ is a product of cyclic groups (of relatively prime order), hence is cyclic. Conversely, cyclic groups do satisfy the hypothesis.

Update: Since it was asked in the comments, and the proof is so simple, let me say why $P$ having a unique maximal subgroup implies $P$ cyclic: if $P$ is not cyclic, but has a unique maximal subgroup $H$, then for every $a \in P$, $\langle a \rangle \subsetneq P$, hence $a \in H$, but then $P \subseteq H$, contradiction. This proof only needs that every proper subgroup is contained in a maximal subgroup, which holds e.g. if the group is finite.