# On $p$-groups with abelian automorphism group

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.

• 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? Feb 9, 2021 at 7:41
• @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)
– YCor
Feb 9, 2021 at 12:30
• So it just means that the automorphism group is elementary abelian. Feb 9, 2021 at 12:31
• 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?
– YCor
Feb 9, 2021 at 12:33
• @DerekHolt Thank you for your remark. Of course it is just elementary abelian p-group. I will edit it. Sorry. Feb 9, 2021 at 12:34

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.

• 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). Feb 11, 2021 at 2:15
• @Nourddine Snanou: You are very welcome! 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.

• 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. Feb 11, 2021 at 1:05
• The link seems to be missing a 4 at the end of the url. Mar 1, 2021 at 4:41
• @Cihan Thanks, corrected! Mar 1, 2021 at 7:46