Let $A$ be a quaternion algebra over a field $K$ of characteristic zero. By the famous Skolem-Noether theorem, every $K$-algebra automorphism $\varphi \colon A \to A$ is of the form $$ \varphi(x) = gxg^{-1}, $$ for some invertible element $g \in A^\times$.

Question: Does it follow that also every group automorphism of $A^1 := \{ x \in A: \text{nrd}(x) = 1\}$ or of $A^\times$ is given by conjugation with some element of $A^\times$?

For example, if you take $A = M_2(\mathbb{R})$, then this is true because it is known that every group automorphism of $\text{SL}_2(\mathbb{R})$ or $\text{GL}_2(\mathbb{R})$ is given by conjugation with some element of $\text{GL}_2(\mathbb{R})$. However, most proofs available of this fact use very different methods and not the Skolem-Noether theorem as far as I know. So is it just a coincidence in this case?

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    $\begingroup$ What you say for GL(2) is not true for "group automorphisms" (eg, math.stackexchange.com/q/2064343/11323). Maybe you mean automorphisms of algebraic groups? In this case, I think you can relate this to Skolem-Noether with Galois cohomology, but I haven't thought about the details. $\endgroup$ – Kimball Mar 12 at 13:56
  • $\begingroup$ Interesting, thanks for the correction. Now that I think of it, I was mostly wondering if one can use Skolem-Noether or some quaternion algebra methods to prove that every automorphism of the real Lie group $\text{SL}_2(\mathbb{R})$ is conjugation with some element of $\text{GL}_2(\mathbb{R})$ or if it is neccessary to use other methods here. $\endgroup$ – abenthy Mar 13 at 7:41
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    $\begingroup$ I'm not sure that the case over $\mathbb R$ is any easier---I guess you need a cohomology argument. About the only additional method you have for Lie groups is taking the derivative to get an automorphisms of Lie algebras, but these don't line up with automorphisms of central simple algebras. (Or rather you can connect them with Skolem-Noether but only in a convoluted way.) But I think this is a good question, and you could try asking on MathOverflow if you don't get an answer here. $\endgroup$ – Kimball Mar 13 at 13:19
  • $\begingroup$ You could try checking if the proof of Skolem-Noether could be generalized to composition algebras of split signature, since $SL_2(\Bbb R)$ is the group of unit-norm elements in the split quaternions (just like ${\rm Sp}(1)$ in the quaternions). $\endgroup$ – arctic tern Mar 14 at 7:39

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