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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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    $\begingroup$ 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? $\endgroup$
    – user6976
    Aug 18, 2012 at 13:38
  • $\begingroup$ @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. $\endgroup$
    – Wei Zhou
    Aug 18, 2012 at 15:41
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Aug 18, 2012 at 17:40
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    $\begingroup$ Wiki link for background and definition of extraspecial groups $\endgroup$
    – YCor
    Aug 18, 2012 at 18:07

1 Answer 1

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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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  • $\begingroup$ It would be helpful if you stated what you prove. $\endgroup$
    – YCor
    Aug 18, 2012 at 18:10
  • $\begingroup$ Did that and made a couple of other changes for clarity. $\endgroup$ Aug 18, 2012 at 18:28
  • $\begingroup$ @John: good to see you here! $\endgroup$ Aug 18, 2012 at 19:00
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    $\begingroup$ 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. $\endgroup$ Aug 19, 2012 at 2:09
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    $\begingroup$ John: no nontrivial normal subgroup misses the center in a p- group. $\endgroup$
    – Steve D
    Aug 19, 2012 at 19:28

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