For the past one week, I have been trying to learn more about automorphism groups of different groups. Very recently one of my friend asked this question to me:

  • What is the automorphism group of $(\mathbb{Q}^{\ast},\times)$. In short, what is $\text{Aut}(\mathbb{Q}^{\ast})$?

I emailed couple of friends and got the answer as:

  • $\text{Aut}(\mathbb{Q}^{\ast})$ is isomorphic to the automorphism group of a free abelian group of countable rank. In particular, it will contain $\text{GL}(n,\mathbb{Z})$ for all $n$.

My question would be :

  • Can we realize $\text{SL}_{n}(\mathbb{Z})$ to be the automorphism group of some group?

  • Are there groups which are which are "very difficult" to be realized as the automorphism group of a certain group.

So suppose someone comes and asks me: Is $S_{3}$ or $\text{GL}_{2}(\mathbb{Z})$ the automorphism group of some group, then how can I answer the question? I am particularly interested in seeing how to think for a solution.

  • 2
    $\begingroup$ $GL_2(\mathbb{Z})$ is the automorphism group of $\mathbb{Z}^2$. And $S_3$ is the automorphism group of $C_2\times C_2$ (where $C_2$ is the 2-element group) $\endgroup$ Sep 10, 2011 at 14:18
  • 6
    $\begingroup$ warm-up exercise: show that there is no group having $\mathbb Z$ as automorphism group. $\endgroup$
    – user2035
    Sep 10, 2011 at 15:32
  • $\begingroup$ @a-fortiori: Couldn't think of anything for proving the warm up exercise. An hint would be helpful. $\endgroup$
    – C.S.
    Sep 10, 2011 at 18:51
  • 2
    $\begingroup$ Don't forget that sometimes that automorphism is trivial... $\endgroup$ Sep 14, 2011 at 4:26
  • 1
    $\begingroup$ But if that's trivial then $G$ is a vector space over $\mathbb{Z}/2$. $\endgroup$
    – arkeet
    Oct 9, 2011 at 11:16

2 Answers 2


This doesn't quite answer your question, but Bumagin and Wise proved every countable group is the outer automorphism group of a finitely generated group.

  • 1
    $\begingroup$ And G\"obel and Paras proved in [Automorphisms of metabelian groups with trivial center, Illinois J. Math. 42 (1998), N2, 333-346] that every group is an outer automorphism group of a metabelian group. I believe there are quite a few results of such sort. $\endgroup$ Oct 21, 2011 at 15:15

Another not quite answer: In the beautiful paper

Automorphism groups of polycyclic by finite groups and arithmetic groups, by Oliver Baues and Fritz Grunewald, Publ. Math. IHES 104 (2006), no. 1 (arXiv version, MathReviews, journal version)

the authors show that in any cases the automorphism groups of polycyclic-by-finite groups are "arithmetic" (basically subgroups of $SL(n, \mathbb{Z})$ (results of this sort have actually been know for quite a while for narrower classes of groups -- check the references of the paper).


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