Welcome to MathOverflow

Question

9

3

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 Autmorphism 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.

edited Oct 9 2011 at 10:31

asked Sep 10 2011 at 13:58

Chandrasekhar
2,020●1●11●28

$GL_2(\mathbb{Z})$ 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) – Alain Valette Sep 10 2011 at 14:18

4

warm-up exercise: show that there is no group having $\mathbb Z$ as automorphism group. – a-fortiori Sep 10 2011 at 15:32

@a-fortiori: Couldn't think of anything for proving the warm up exercise. An hint would be helpful. – Chandrasekhar Sep 10 2011 at 18:51

1

Don't forget that sometimes that automorphism is trivial... – Andrew Dudzik Sep 14 2011 at 4:26

1

But if that's trivial then $G$ is a vector space over $\mathbb{Z}/2$. – Adrian Keet Oct 9 2011 at 11:16

show 3 more comments

Benjamin Steinberg · Answer 1 · 2011-09-10 19:19:22Z UTC

9

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.

answered Sep 10 2011 at 19:19

Benjamin Steinberg
10.7k●1●14●49

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. – Pasha Zusmanovich Oct 21 2011 at 15:15

Alain Valette · Answer 2 · 2011-09-11 13:19:04Z UTC

Another not quite answer: In the beautiful paper

Automorphism groups of polycyclic by finite groups and arithmetic groups, (with F. Grunewald), Publ. Math. IHES 104 (2006), no. 1

the authors show that in any cases the automorphism groups of polyciclic-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).

Example of a Group which has $\text{SL}_{n}(\mathbb{Z})$ as the automorphism group

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.