Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
1  
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
1  
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. –  Yves Cornulier Aug 18 '12 at 17:40
1  
Wiki link for background and definition of extraspecial groups –  Yves Cornulier Aug 18 '12 at 18:07
add comment

1 Answer

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.

share|improve this answer
    
It would be helpful if you stated what you prove. –  Yves Cornulier 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
1  
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
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.