# Fixed field of an Automorphism

$$\newcommand{\Aut}{\mathrm{Aut}}$$ Few Definitions:
Let $$E$$ be a field and $$\Aut(E)$$ denote the group of automorphisms of $$E$$.
Now, Let $$S\subset \Aut(E)$$
Define:$$\mathscr{F}(S):=\{a\in E\enspace|\sigma(a)=a \enspace\forall \enspace \sigma\in S \}$$ We call $$\mathscr{F}(S)$$ a fixed field of $$S$$.
Moreover, It is easy to see that $$\mathscr{F}(S)$$ is a subfield of $$E$$.

Question:

If $$S\subset \Aut(E)$$, then $$S\subset \Aut(E/\mathscr{F}(S))$$

I know that
$$\Aut(E/\mathscr{F}(S))=\{\sigma\in \Aut(E)|\sigma(x)=x\enspace \forall x\in \mathscr{F}(S)\}$$

I have to show that $$\forall\tau\in S\implies \tau\in \Aut(E/\mathscr{F}(S))$$

• @Enkidu Please read the question... I am not asking to prove that $Aut(E)\subset Aut(E/\mathscr{F}(S))$. In fact, I know that $Aut(E/\mathscr{F}(S))\subset Aut(E)$ – Kumar Sep 6 at 7:22

Suppose $$\tau \in S.$$ We need to show that $$\tau(x)=(x)$$ for every $$x\in \mathscr F (s)$$.
Let $$x\in\mathscr F (s)$$. Then $$\sigma(x)=x$$ for all $$\sigma \in S$$. In particular, $$\tau (x)=x$$.
If $$\tau\in S$$ then firstly, $$\tau\in\operatorname {Aut}(E)$$.
Secondly, by definition, for any $$x\in\mathscr F(S)$$, we have $$\tau (x)=x$$.
Thus $$\tau\in\operatorname {Aut}(E/\mathscr F(S))$$.