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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

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The extra-special group is a central power $H^n/Z$ of the Heisenberg group $H$ (of exponent $p$). Consider the subgroup $U$ of $H^n$ consisting of all constant functions (i.e. vectors where all coordinates are equal). Is the image of $U$ in $H^n/Z$ characteristic? – Mark Sapir Aug 18 '12 at 13:38
@Mark: Thanks Mark. But I have to say that I am not familiar with the Heisenberg group. I know the extra-special p-group in the way of abstract group theory. I don't know how to do it in your way. – Wei Zhou Aug 18 '12 at 15:41
It is easy to check that the original question has a positive answer iff the action of $Out(G)$ on the vector space $G/Z$ over $\mathbf{Z}/p$ is irreducible. – YCor Aug 18 '12 at 17:40
Wiki link for background and definition of extraspecial groups – YCor Aug 18 '12 at 18:07
up vote 8 down vote accepted

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.

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It would be helpful if you stated what you prove. – YCor Aug 18 '12 at 18:10
Did that and made a couple of other changes for clarity. – John Shareshian Aug 18 '12 at 18:28
@John: good to see you here! – Patricia Hersh Aug 18 '12 at 19:00
@John: Thanks for your answer. I am not familar with Symplectic group. Would you please give me a reference that $Sp(2n, p)$ is irreducible on $G/Z$? – Wei Zhou Aug 19 '12 at 1:11
In Section 20 of Aschbacher's "Finite Group Theory", you will find a proof of Witt's Lemma. A consequence of this lemma is that $Sp(2n,p)$ is transitive on $1$-dimensional subspaces of $G/Z$, from which irreducibility follows immediately. You can learn a lot about the symplectic group and other classical groups from Section 22 of that book. – John Shareshian Aug 19 '12 at 2:09

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