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

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.

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

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).