Suppose we let $\sigma: \mathbb{C} \to \mathbb{C}$ be the complex conjugate map, and we have a subfield $L \subseteq \mathbb{C}$ with $L/\mathbb{Q}$ a finite Galois extension.
I want to show that $\sigma(L)=L$
I was wondering whether it is necessary (or even possible) to use the Fundamental theorem of Galois Theory to answer this, saying as $L/\mathbb{Q}$ is Galois that as a result we have $\sigma(L)=L$ for any $\sigma \in \text{Aut}(L/\mathbb{Q})$.
The problem is I am not actually sure if $\sigma$ is in this group, as I feel I would need to know that the complex conjugate of any element of $L$ was also in $L$ to ensure that it is actually an automorphism, and if this is the case wouldn't it just follow directly anyway that $\sigma(L)=L$.
Sorry if this comes across incoherent I am just pretty confused, help appreciated as always :)