Galois Basic Question (Cyclic Order 4 Extension of Q cannot contain i) Suppose $K$ is an extension of $\mathbb{Q}$ in $\mathbb{C}$, where $Gal(K/\mathbb{Q})$ is cyclic of order 4. Show that $i\notin K$. 
($i$ is the imaginary number $i^2=-1$.)
My Galois theory is quite weak, hope someone can check if my attempt is correct.

My attempt:
Suppose to the contrary $i\in K$. Let $\sigma\in Gal(K/\mathbb{Q})$.
Note that $\sigma (i)\sigma(i)=\sigma(i^2)=\sigma(-1)=-1$ since $\sigma$ fixes $\mathbb{Q}$.
This means that $\sigma(i)=i$ or $\sigma(i)=-i$. The first case is ruled out since $\sigma$ only fixes $\mathbb{Q}$.
So $\sigma(i)=-i$. This means that $\sigma(a+bi)=a-bi$, so $\sigma$ is effectively complex conjugation, which has order 2. 
Since $\sigma$ was arbitrary, this contradicts that $Gal(K/\mathbb{Q})$ has an element of order 4.

Is this ok?
Thanks.
Update: Now I see that my argument is clearly flawed. What would be the correct proof?

Here is a slick solution, I think. Consider generally a cyclic extension $F/k$ and try to embed it in an over-extension $K/F/k$ such that $K/k$ is cyclic. For simplification, suppose that $F/k$ has degree $p^n$, $K/F$ has degree $p$ and $k$ contains a primitive $p$-th root $\zeta$ of unity ($p$ a prime). Then, using Kummer theory, it can be shown that $K$ exists iff $\zeta$ is a norm in $F/k$ (see e.g. https://math.stackexchange.com/a/1691332/300700, where the situation is a bit more general). In our particular case here, $p=2$, $\zeta=-1$, $K/\mathbf Q$  is the given cyclic extension of degree 4. If $K$ contained $i$, take $F=\mathbf Q(i)$. Then $-1$ would be a norm in $F/\mathbf Q$, i.e. would be a sum of two squares in $\mathbf Q$ : impossible. Note that here, the previous embeddability criterion can be shown directly.
