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According to Wikipedia the automorphism group of the octonions is the exceptional group $G_2$. Are there analogous groups for the real numbers, the complex numbers and the quaternions?

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  • $\begingroup$ My abstract algebra is really rusty. Does it matter if we're looking at these as groups versus looking at them as fields? $\endgroup$
    – Teepeemm
    Apr 30, 2016 at 20:20
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    $\begingroup$ "automorphism" with respect to what structure though? $\endgroup$ Jul 22, 2019 at 18:30

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The automorphism group of $\mathbb{R}$ is trivial.

The automorphism group of $\mathbb{C}$ is extremely complicated (at least if you accept the axiom of choice) : see http://www.jstor.org/stable/2689301?seq=1#page_scan_tab_contents for instance.

The automorphism group of $\mathbb{H}$ is $\mathbb{H}^*/\mathbb{R}^*$ : according to the Skolem-Noether theorem, every $\mathbb{R}$-algebra automorphism of $\mathbb{H}$ is inner, so it just remains to quotient out the center. But every ring automorphism must induce an automorphism of the center, and since the center $\mathbb{R}$ does not have not-trivial automorphisms, every ring automorphism is actually a $\mathbb{R}$-algebra automorphism.

Note that this group is actually isomorphic to $SO(3)$ : the conjugation action is by isometry on the subspace of pure quaternions.

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    $\begingroup$ Indeed, if one only looks at continuous automorphisms of $\mathbb{C}$ though, then there are only two automorphisms (the identity and conjugation). $\endgroup$ Apr 30, 2016 at 12:18
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Regarding $R,$ let $h:F \to G$ be an isomorphism between subfields $F,G$ of $R .$

If $\forall x\in F\; (x>0\implies \sqrt x\in F),$ then $h=id_F. $ Because, for $x,y\in F,$ we have $$x>y\implies 0\ne h(x)-h(y)=(h(\sqrt {x-y}))^2 \implies h(x)>h(y).$$ And $h|Q=id_Q.$ So $\{q\in Q:q<x\}=\{q\in Q:q<h(x)\}$ for all $x\in F.$ In particular if $F=R$ then $h=id_R$ and $G=R.$

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