A $p$-group $G$ is called a ${\it UCS}$ $p$-group if $G$ has precisely three characteristic subgroups, namely $1$, $\Phi(G)$ and $G$.

Let $G$ be a finite UCS $p$-group of order $p^{2n}$ such that $\Phi(G)$ is elementary abelian $p$-group of order $p^n$. As an example of such a group we can give $G=\underbrace{\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p^{2}}\times\dots\times\mathbb{Z}_{p^{2}}}_{n\,\,times}$. Does there exist any nonabelian $p$-group with mentioned properties?


2 Answers 2


I think there are such examples for all odd primes $p$ and all $n \ge 3$.

There is a $p$-group $P$ of exponent $p$ of class $2$, with $\Phi(P)=Z(P)$ and $P/\Phi(P)$ and $\Phi(P)$ elementary abelian, with $|\phi(P)| = p^{n(n-1)/2}$, $|P/\Phi(P)|=p^n$, such that ${\rm Aut}(P)$ acts on $P/\Phi(P)$ as ${\rm GL}(n,p)$, where the induced action on $\Phi(P)$ is as the exterior square of the natural module for ${\rm GL}(n,p)$.

The group $P$ is defined by the presentation $\langle X \mid R \rangle$, where $$X=\{x_i:1 \le i \le n\} \cup \{y_{ij}^p: 1\le i<j\le n\} $$ and $$R=\{x_i^p:1 \le i \le n\}\cup \{y_{ij}^p: 1 \le i < j \le n\} \cup \{[x_i,x_j]y_{ij}^{-1}: 1 \le i < j \le n\} \cup C,$$ where $C$ consists of all commutators of all $y_{ij}$ with all other generators (to make the $y_{ij}$ central).

Now, by a well-known result of Zigmundy, there is a prime $q$ that divides $p^{n}-1$ but does not divide $p^r-1$ for any $r<n$, and ${\rm GL}(n,p)$ has a cyclic subgroup $Q$ of order $q$ that must act irreducibly on the natural module.

Now all nontrivial irreducible modules for $Q$ over ${\mathbb F}_p$ have dimension $n$, and in particular the exterior square $E$ of the natural module has a quotient module $E/K$ of dimension $n$. Let $R$ be subgroup of $\Phi(P)$ corresponding to $K$, and $G=P/R$. Then $|G|=p^{2n}$ with $|\Phi(G)|=p^n$, and since the cyclic group $Q$ acts irreducibly on both $\Phi(G)$ and $G/\Phi(G)$, the only characteristic subgroups of $G$ are $1$, $\Phi(G)$ and $G$.

  • $\begingroup$ How $P$ is constructed? $\endgroup$ Nov 26, 2018 at 11:16
  • $\begingroup$ I have added a presentation of $P$. $\endgroup$
    – Derek Holt
    Nov 26, 2018 at 12:20
  • $\begingroup$ @Farrokhi, Thank you so much for your valuable answers and comments. $\endgroup$ Nov 26, 2018 at 19:57
  • $\begingroup$ @Derek Holt, Thank you so much for your valuable answers and comments. $\endgroup$ Nov 26, 2018 at 19:57

Yes, there is an infinite series of these groups. Let $$A(n):=\left\{\begin{bmatrix}1&0&0\\a&1&0\\b&a^\theta&1\end{bmatrix}:a,b\in GF(2^n)\right\}.$$ be a Suzuki $2$-group of order $2^{2n}$ for $n\geq3$ with $\theta:x\mapsto x^2$ is a automorphism of $GF(2^n)$.

Then $G/\Phi(G)\cong\Phi(G)\cong C_2^n$ and $\{1,\Phi(G),G\}$ is the set of all characteristic subgroups of $G$.

Indeed, the same results hold for all primes $p$ and automorphisms $\theta:x\mapsto x^p$ of $GF(p^n)$.


  1. B. Huppert and N. Blackburn, Finite Groups II, Springer-Verlag, Berlin-New York, 1982. (pp. 294-299; also pp 299-316 for supplementary results)
  2. A. Mohammadian, A. Erfanian, M. Farrokhi D. G., and B. Wilkens, Triangle-free commuting conjugacy class graphs, J. Group Theory 19 (2016), no. 6, 1049–1061.

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