I have recently encountered Frobenius maps in the context of orthogonal and symplectic groups. However, some aspects still strike me as odd. Maybe I'm making some stupid mistakes here. Let's just assume that $G$ is an orthogonal group of dimension $n$ over the algebraic closure of a finite field with $q$ elements. Let's view $G$ as a matrix group and say that $F$ is the standard Frobenius morphism (of the general linear group) raising each entry of a given matrix to the $q$-th power. Why is $F$ then a Frobenius morphism for $G$? I understand that the basic concept involved is $F$-stability, but... Is it obvious?
Frobenius maps and orthogonal groups
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abstract-algebra
group-theory
finite-groups
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1I am finding this question a bit hard to pin down: what is your definition of "a Frobenius morphism for $G$"? What do you mean by $F$-stability? Also, it sounds like you are asking questions based on some specific text(s) you are reading. Telling us what these are could only help us help you... – 2011-12-06
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0Among other texts, I've been working through the first chapter of Carter's _Finite Groups of Lie Type_. I don't have the book at hand at the moment, but I believe that Frobenius morphisms are introduced in I.17. I'm still quite confused with the concept, hence the confusion in my question. What I'm trying to ask/understand: If $F$ is the Frobenius map raising the entries of a matrix to the $q$-th power, why is $F(G) \subseteq G$? I don't see why the map should respect quadratic forms. It's probably quite easy to see, I'm just not sure how to deal with the concept.. Thanks in advance! – 2011-12-06