A question on UCS p-groups

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?

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 $$X=\{x_i:1 \le i \le n\} \cup \{y_{ij}^p: 1\le i and $$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 $$r, and $${\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 $$G$$.

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

See

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.