8
$\begingroup$

Let $G$ be a $p$-group of order $p^{n}\geq p^{7}$ and its automorphism group is elementary abelian $p$-group. Then, it is clear that $G$ is nilpotent of class $2$. However, the converse is not true in general (take for example the unitriangular group of $3\times 3$ upper triangular matrices over $\mathbb{F}_{p}$). In fact, Marta Morigi proves that there is a $p$-group of order $p^{7}$ and class $2$ and its automorphism group is elementary abelian $p$-subgroup of order $p^{12}$. Furthermore, I think the converse can be true for 2-groups $G$ with abelian direct factor but I don't know how to do this.

Are there any assumption to add to p-groups of class $2$ to get the converse for arbitrarily integer $n$?

Let $p$ be an odd prime and $G$ be a purely non-abelian p-group of class $2$ and order $p^{n}\geq p^{7}$. Does the automorphism group of $G$ is abelian?

Are there some other class of $p$-groups with elementary abelian automorphism $p$-groups?.

Any response or reference may be very helpful. Thank you in advance.

$\endgroup$
7
  • 1
    $\begingroup$ I am sorry but I do not understand what you mean by "its automorphism group is elementary abelian $p$-subgroup" in the first sentence. Subgroup of what? $\endgroup$
    – Derek Holt
    Feb 9 at 7:41
  • 1
    $\begingroup$ @DerekHolt the group of automorphisms of a group $G$ is a subgroup of the group of permutations of $G$ (I agree the phrasing is unusual) $\endgroup$
    – YCor
    Feb 9 at 12:30
  • 1
    $\begingroup$ So it just means that the automorphism group is elementary abelian. $\endgroup$
    – Derek Holt
    Feb 9 at 12:31
  • 1
    $\begingroup$ I'd wonder, with $p$ odd start from the free 2-nilpotent $p$-group of exponent $p$ and $n$ generators. This has cardinal $p^{n+n(n-1)/2}$, with derived subgroup = center of elementary abelian of cardinal $p^{n(n-1)/2}$. Fix a dimension $k$ (with both $k$ and $n(n-1)/2-k$ not too small), and mod out by a "random" $k$-dimensional subspace of the derived subgroup. Does, generically, the quotient have the required property? $\endgroup$
    – YCor
    Feb 9 at 12:33
  • $\begingroup$ @DerekHolt Thank you for your remark. Of course it is just elementary abelian p-group. I will edit it. Sorry. $\endgroup$ Feb 9 at 12:34
11
$\begingroup$

I suggest you to have a look at the following paper and references therein:

V.K. Jain, P.K. Rai, M.K. Yadav: On finite $p$-groups with abelian automorphism group. Internat. J. Algebra Comput. 23 (2013), no. 5, 1063--1077.

$\endgroup$
2
  • $\begingroup$ Thank you very much for the above interesting reference. In fact, I have seen almost all interesting references you suggest and which study a lot of beautiful examples of p-groups with elementary abelian automorphism group as the one given in the above. It seems that all have class 2. So this allows us to ask whether a p-group of class 2 have abelian (or elementary abelian) automorphism p-group (see the first tow questions). $\endgroup$ Feb 11 at 2:15
  • $\begingroup$ @Nourddine Snanou: You are very welcome! $\endgroup$ Feb 11 at 9:12
6
$\begingroup$

This is just an amplification of my comment above.

Theorem 3.3 of this this survey article, A Survey on Automorphism Groups of Finite p-Groups, by Geir T. Helleloid (2006) describes a result published in

U. M. Webb, The number of stem covers of an elementary abelian p-group, Math. Z. 182 (1983), no. 3, 327–337.

A stem cover of a group $Q$ is a maximal stem extension of $Q$; that is, a maximal (under group extensions) group $G$ with normal subgroup $N$ such that $G/N \cong Q$ and $N \le Z(G) \cap [G,G]$. (So, in a stem cover, $N$ is isomorphic to the Schur Multiplier of $Q$.)

This result says that, for $p$ an odd prime, the proportion of stem covers of an elementary abelian group of order $p^n$ for which the automorphism group is an elementary abelian $p$-group approaches 1 as $n \to \infty$. Elementary abelian gropups have lots of distinct stem covers, so this is producing plenty of examples of the type you are looking for.

$\endgroup$
3
  • $\begingroup$ thank you very much for the above interesting reference. I understand that, as $n \to \infty$, almost all stem covers of an elementary abelian group have elementary abelian automorphism p-groups and clearly have class 2. In fact, I' m interested to the converse of the statement, so for more clarification, I have asked a second question concerning p-groups of class 2. Thank you again for your time. $\endgroup$ Feb 11 at 1:05
  • $\begingroup$ The link seems to be missing a 4 at the end of the url. $\endgroup$
    – Cihan
    Mar 1 at 4:41
  • $\begingroup$ @Cihan Thanks, corrected! $\endgroup$
    – Derek Holt
    Mar 1 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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