Complex no's and Quaternions Automorphism I read in research paper that 
There is only one norm preserving automorphism(identity map) for 
$\mathbb{R}$. There are two norm preserving automorphism(identity map and dual map) for $\mathbb{C}$.For $\mathbb{H}$ the norm preserving automorphism is SO(3).
For $\mathbb{C}(\mathbb{F}_{q})$, the automorphism could be obtained by combining the Frobenius automorphism and dual automorphism.
The automorphism group for $\mathbb{H}(\mathbb{F}_{q})$ could be obtained by combining the Frobenius automorphism and the special orthogonal group SO(3,$\mathbb{F}_{q}$).
I understand the first paragraph but problem is in understanding second paragraph
My questions are


*

*What is dual automorphism? I know about about automorphism and Frobenius automorphism !

*What is meant by combining in above statements. Is it meant multiplication, addition or any other mathmatical operation

 A: I assume $\mathbb{C}(\mathbb{F}_q)$ and $\mathbb{H}(\mathbb{F}_q)$ are the $\mathbb{F}_q$-algebras respectively spanned by $\{1,i\}$ and $\{1,i,j,k\}$  satisfying the usual multiplication rules. Note $\mathbb{C}(\mathbb{F}_q)$ is $\cong\mathbb{F}_{q^2}$ if $\mathbb{F}_q$ has no $\sqrt{-1}$ (equivalently, if $q\not\equiv1\bmod4$) and is $\cong\mathbb{F}_q\oplus\mathbb{F}_q$ as a ring otherwise, and we can similarly split the isomorphism type of $\mathbb{H}(\mathbb{F}_q)$ into cases but it will never be a skew field (no noncommutative finite skew fields exist).
(1) By "dual automorphism" they mean complex conjugation, essentially. Writing elements of $\mathbb{C}(\mathbb{F}_q)$ as $a+bi$ with $a,b\in\mathbb{F}_q$, this automorphism is $\overline{a+bi}=a-bi$. In the event this is $\mathbb{F}_{q^2}$, it also happens to be a Frobenius automorphism $x\mapsto x^q$, which acts as the identity on $\mathbb{F}_q$.
(2) In general, saying a elements of a group $G$ are "combinations" of elements from subsets $X$ and $Y$ means $G$ is generated by the subsets, written $G=\langle X,Y\rangle$. A tighter condition is that one subset is a subgroup and the other a transversal. A tighter condition still is that they are both subgroups and $G$ is a knit product. A tighter condition than that is that they are both subgroups and $G$ is a semidirect product of them. (And the tightest you can go is that $G$ is that they are both normal subgroups and $G$ is a direct product.) In this case, we mean semidirect product at least, and it turns out Frobenius automorphisms of $\mathbb{H}(\mathbb{F}_q)$ will commute with those coming from $\mathrm{SO}(3,\mathbb{F}_q)$ so it will be a direct product.
