# p-groups with unique normal minimal subgroup

Have $p$-groups with a unique normal minimal subgroup been classified? Is there any article on the subject?

• In a p-group each central element of order p generates a normal minimal subgroup. So the p-groups with a unique normal minimal subgroup are exactly the p-groups with cyclic center. There is a classification of p-groups with cyclic center and cyclic commutator subgroup. But I don't know if there is a classification of p-groups with cyclic center. – Todd Leason Feb 9 '16 at 21:43
• The question is about normal minimal subgroup not minimal subgroup. – Mohsen Feb 9 '16 at 23:02
• @YCor: I though the same thing; and $p$-groups with a unique subgroup of order $p$ (hence unique normal subgroup) must be either cyclic or generalized quaterniong (e.g., Theorem 5.46 in Rotman's "Introduction to the theory of groups" 4th edition). But there are $p$-groups with unique minimal normal subgroups that do not have unique minimal subgroups, e.g., the group of order $p^3$ and exponent $p$. – Arturo Magidin Feb 9 '16 at 23:04
• OK actually I read correctly the question mut misread the other question I linked. Certainly the class is big; it contains for instance the upper unipotent groups over fields of prime order. I would not expect any kind of reasonable classification. – YCor Feb 9 '16 at 23:10

As Todd Leason said, a $p$-group has a unique minimal normal subgroup if and only if it has cyclic center. Here is some evidence that there is no reasonable classification of such groups:
1. The class of these groups is big: It contains the upper unipotent groups over prime fields, as YCor said, it contains the Sylow $p$-subgroups of the symmetric group $S_{p^n}$, the $p$-groups of maximal nilpotency class, the extraspecial groups and lots of other groups.
2. Every $p$-group embeds into such a group (follows from each of the first two examples in 1.).
3. Every $p$-group $P$ has factor groups of this type, and at least one factor group with the same nilpotency class as $P$: Namely, if $D\colon P \to \operatorname{GL}(d,\mathbb{C})$ is an irreducible representation, then $P/\operatorname{Ker}(D)$ has cyclic center, and there must be an irrep such that its kernel does not contain the last non-trivial term, $P^c$, of the decending central series of $P$.