Automorphism group of quaternions Let $\mathbb{H}$ be the algebra of quaternions. How to prove that the automorphism group $Aut(\mathbb{H})$ is isomorphic to the group $SO_3$?
I found that all derivations on algebra $\mathbb{H}$ are inner.  Moreover, they are isomorphic to the Lie algebra $so_3$. It remains to understand whether it is simply connected or not. But I don’t know how to show that the group $Aut(\mathbb{H})$ is not simply connected.
 A: An automorphism $\alpha$ must preserve $1$, so by $\mathbb{R}$-linearity it preserves the real line.
Also $\alpha$ must preserve the set of square roots of $-1$, which is the unit sphere $S^2$, and all of their real multiples, which forms the pure imaginary subspace $\mathbb{R}^3$ of 3D vectors.
If $\mathbf{u}$ and $\mathbf{v}$ are vectors, the real part of $\mathbf{uv}$ is minus the dot product, i.e. $-\mathbf{u}\cdot\mathbf{v}$. Since $\alpha$ preserves real parts, this means it preserves dot products, so it acts as a linear isometry of $\mathbb{R}^3$. So $\alpha\in\mathrm{O}(3)$.
Similarly the imaginary part of $\mathbf{uv}$ is the cross product $\mathbf{u}\times\mathbf{v}$. See if you can show $\alpha$ preserves cross products, and this will further mean $\alpha\in\mathrm{SO}(3)$.
Conversely, any element of $\mathrm{SO}(3)$ preserves dot products and cross products, so preserves the quaternion multiplication (which is built out of them). Because of $S^3\to\mathrm{SO}(3)$, this means all automorphisms of $\mathbb{H}$ are inner.
