# What do we know about $\mathrm{Aut}(\mathbb{H})$?

I found the following exercise in my algebra syllabus:

Prove $\mathrm{Aut}(\mathbb{H})\cong S_4$.

Here $\mathrm{Aut}(\cdot)$ denotes the set of group automorphisms.

I assume this is a typo, since $\mathrm{Aut}(Q_8)\cong S_4$, and proofs for this are managable and can be found on this site. However, before I found out this is most likely a typo, I was thinking about how to find $\mathrm{Aut}(\mathbb{H})$, but I didn't find a way to solve this. Is there anything known about $\mathrm{Aut}(\mathbb{H})$? I couldn't find anything on the internet. I would be glad with either a reference or an answer.

• What type of automorphisms? The entire set of ring automorphisms? or $\mathbb R$ algebra automorphisms? Skolem-Noether applies in the latter case. – rschwieb Apr 24 '18 at 17:40
• But the quaternions only form a group under addition, and there the automorphism group is huge. – Tobias Kildetoft Apr 24 '18 at 17:43
• There is, but it seems unlikely to be what was meant. Basically, you can assume $n=1$, since that doesn't change the group up to isomorphism. Then the fact that the group is divisible implies that all group automorphisms are also automorphisms as $\mathbb{Q}$-vector spaces. And now we are left with this being the group of invertible $\mathbb{Q}$-linear maps between two vector spaces of uncountable dimension. – Tobias Kildetoft Apr 24 '18 at 17:51
• The automorphism group of $\mathbb{R}^n$ as an additive group is a horrible beast, even when $n=1$, because $\mathbb{R}^n$ is an uncountable dimensional vector space over $\mathbb{Q}$. So if you want to calculate the automorphism group of some mathematical object and you expect some nice answer, you might not want to ignore most of the natural structure on that object. The automorphism group of $\mathbb{R}^n$ as a vector space over $\mathbb{R}$, aka $GL(n,\mathbb{R})$, is much more interesting and useful than its automorphism group as an additive group. – Lee Mosher Apr 24 '18 at 17:51
• @VáclavMordvinov If you use the $\mathbb R$-algebra automorphisms instead, the group is isomorphic to $\mathbb H^\times/\mathbb R^\times$ – rschwieb Apr 24 '18 at 17:52