Question

Let $G$ be a extra-special $p$-group of order $p^{1+2r}$ with exponent $p$ (p odd). I want to know if $G$ has only $3$ characteristic subgroups?

Background: From [2], if $G$ is extra-special $5$-group of order $5^5$ with exponent $5^2$, then $G$ has more than $3$ characteristic subgroup. How about the exponent of the extraspecial $p$-group is $p$?

There are some references about this topic. [1]. D.R. Taunt, Finite groups having unique proper characteristic subgroups I, Proc. Cambridge Philos. Soc. 51 (1955) 25–36. [2]. S.P. Glasby, P.P. Pálfyb, Csaba Schneider p-groups having a unique proper non-trivial characteristic subgroup Journal of Algebra 348 (2011) 85–109

Answer 1

An extraspecial $p$-group of exponent $p$ contains exactly three characteristic subgroups, $1$, $G$ and the center of $G$.

Let $Z$ be the center of $G$ (so $Z=[G,G]=\Phi(G)$). The elementary abelian group $G/Z$ is a vector space of dimension $2r$ over the field of order $p$. The commutator map on $G$ induces a nondegenerate alternating bilinear form on $G/Z$. As shown in a paper of D. L. Winter in the Rocky Mountain Journal (1972), $Aut(G)$ has a subgroup $H$ of index $p-1$ such that $H/Inn(G)$ is isomorphic to the full stabilizer of the given form (this does not hold if $G$ does not have exponent $p$). Since this stabilizer is irreducible on $G/Z$, no characteristic subgroup of $G$ (other than $G$) strictly contains $Z$. Now assume for contradiction that $G$ has some nontrivial proper characteristic subgroup $X$ that does not contain $Z$. Then $XZ$ is characteristic in $G$ and strictly contains $Z$, which forces $XZ=G$. Now $X$ is maximal in $G$. However, this forces $Z=\Phi(G)\leq X$, a contradiction.