## when is Aut(G) abelian

let $G$ be a group such that $Aut(G)$ is abelian. is then $G$ abelian?

this is a sort of generalization of the well-known exercise, that $G$ is abelian when $Aut(G)$ is cyclic. but I have no idea. at least, the finitely generated abelian groups $G$ such that $Aut(G)$ is abelian can be classified.

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From MathReviews:

MR0367059 (51 #3301) Jonah, D.; Konvisser, M. Some non-abelian $p$-groups with abelian automorphism groups. Arch. Math. (Basel) 26 (1975), 131--133.

This paper exhibits, for each prime $p$, $p+1$ nonisomorphic groups of order $p^8$ with elementary abelian automorphism group of order $p^{16}$. All of these groups have elementary abelian and isomorphic commutator subgroups and commutator quotient groups, and they are nilpotent of class two. All their automorphisms are central. With the methods of the reviewer and Liebeck one could also construct other such groups, but the orders would be much larger.

FYI, I found this via a google search.

The first to construct such a group (of order $64 = 2^6$) was G.A. Miller* in 1913. If you know something about this early American group theorist (he studied groups of order 2, then groups of order 3, then...and he was good at it, and wrote hundreds of papers!), this is not so surprising. I found a nice treatment of "Miller groups" in Section 8 of

http://arxiv.org/PS_cache/math/pdf/0602/0602282v3.pdf

(*): The wikipedia page seems a little harsh. As the present example shows, he was a very clever guy.

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 thanks :) . – Martin Brandenburg Dec 28 2009 at 11:23

The answer is No. (Pete beat me to it.)

The earliest example seems to be in a 1913 paper of GA Miller's A non-abelian group whose group of isomorphisms is abelian, Messenger of Math. 48 (1913) 124--125.

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Well, I'm voting for you, anyway. For such questions I admit to following the advice of SO immortal Jon Skeet: submit a brief version first and then edit to add the desired frills. – Pete L. Clark Dec 28 2009 at 11:29
That's very gracious of you -- thanks! – JosÃ© Figueroa-O'Farrill Dec 28 2009 at 14:38

1. Any group whose automorphism group is abelian must have nilpotency class at most two, because the inner automorphism group, being a subgroup of the automorphism group, is abelian.
2. For finite groups, being abelian and the automorphism group being abelian as well implies cyclic. In the infinite case, there are locally cyclic groups that are not cyclic, and these have abelian automorphism groups. For instance, the additive group of rational numbers has an abelian automorphism group (the multiplicative group of rational numbers).

Two other references (in addition to Miller and Jonah-Konvisser mentioned above) for examples of 2-groups with abelian automorphism groups:

1. Some nonabelian 2-groups with abelian automorphism groups by Rebecca Roth Struik, Archiv der Mathematik, ISSN 1420-8938 (Online), ISSN 0003-889X (Print), Volume 39,Number 4, Page 299 - 302(Year 1982); MathReviews Number: 0684397

2. Some new non-abelian 2-groups with abelian automorphism groups by Ali-Reza Jamali, Journal of Group Theory, ISSN 14435883 (print), ISSN 14434446 (online), Volume 5,Number 1, Page 53 - 57(Year 2002); MathReviews Number: 1879516

I have more notes on groups whose automorphism group is abelian here and here.

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A relate one: If $Aut(G)$ is cyclic then $G$ is cyclic.

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... which is not true. – Martin Brandenburg May 26 at 6:47
I find this: towson.edu/math/Zassenhaus_2011_Conference/… – Buschi Sergio May 26 at 15:06