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.

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    $\begingroup$ 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, 2021 at 7:41
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    $\begingroup$ @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, 2021 at 12:30
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    $\begingroup$ So it just means that the automorphism group is elementary abelian. $\endgroup$
    – Derek Holt
    Feb 9, 2021 at 12:31
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    $\begingroup$ 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, 2021 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, 2021 at 12:34

2 Answers 2


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.

  • $\begingroup$ 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, 2021 at 2:15
  • $\begingroup$ @Nourddine Snanou: You are very welcome! $\endgroup$ Feb 11, 2021 at 9:12

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.

  • $\begingroup$ 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, 2021 at 1:05
  • $\begingroup$ The link seems to be missing a 4 at the end of the url. $\endgroup$
    – Cihan
    Mar 1, 2021 at 4:41
  • $\begingroup$ @Cihan Thanks, corrected! $\endgroup$
    – Derek Holt
    Mar 1, 2021 at 7:46

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