Why is it that: $\forall K \supseteq \mathbb{Q}(\mathbb{i}), G=Gal(K / \mathbb{Q}) = \langle \sigma \rangle \implies \sigma (\mathbb{i}) = - \mathbb{i}$? (Note: I am guessing that $\sigma ≠ \mathbb{1}$ here.) Moreover, why must there not be a field extension $K \supseteq \mathbb{Q}(\mathbb{i}): K/ \mathbb{Q}$ is Galois with Galois group $G$ is cyclic with order $|G|=4$? I see the connection but why could there not be another one? Perhaps because an automorphism $\tau$ that results in complex conjugation is a power of $\sigma$? That is all that I can come up with but even that does not immediately relate as far as I can tell.
I am going over notes in class and the teacher said that these things were obvious, but not to me...
Note: this formulation may not be perfectly correct because (1) it was sort of on the fly during class (rather than carefully written out), and (2) I was taking notes, so I may have written something down incorrectly.
Let us assume $\;\Bbb Q(i)\subset \Bbb K\;$ is such that $\;\Bbb K/\Bbb Q\;$ is a Galois cyclic extension, and thus $\;Gal(\Bbb K/\Bbb Q)=\langle\sigma\rangle\;$.
Now, since $\;\sigma (q)=q\;\;\;\forall\,q\in\Bbb Q\;$ ,we have that
$$\sigma(i)=i\iff \sigma=\text{Id.}_{\Bbb Q(i)}$$
But this can't be by the Fundamental Theorem of Galois Extensions, since the fixed field of $\;\sigma\;$ is $\;\Bbb Q\;$ ...