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$\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 $$\forall\tau\in S\implies \tau\in \Aut(E/\mathscr{F}(S))$$

I am unable to realize it. Please help.

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  • $\begingroup$ @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 $Aut(E/\mathscr{F}(S))\subset Aut(E)$ $\endgroup$ – Kumar Sep 6 at 7:22
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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 $.

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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 $\tau\in\operatorname {Aut}(E/\mathscr F(S))$.

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