Consider the polynomial $f(t) = t^5 -4t+2$ over $\mathbb{Q}$. By Eisenstein's criterion, this is irreducible. By graphing this function, we see that it has 3 real roots, hence 2 (distinct, because they are conjugates) non-real roots. Therefore as 5 is prime, we can use a well known lemma to see that the Galois group of group of $f$ over $\mathbb{Q}$ is $S_5$. Now I want to look at the Galois group of $f$ over $\mathbb{Q}(i)$. Now I have managed to show that the group is either $A_5$ or $S_5$. Let the roots of $f$ be $\alpha_1,...,\alpha_5$ with $$\alpha_1 = x +iy$$ $$\alpha_2 = x-iy$$ $$y\ne0$$ Complex conjugates. Now let $\phi$ be the element of $Gal(f)$ over $\mathbb{Q}$ defined by complex conjugation. I.e. it fixes $\alpha_3,\alpha_4,\alpha_5$ and swaps $\alpha_1$ with $ \alpha_2$. This clearly does $not$ fix $i$, so $Gal(f)$ over $\mathbb{Q}(i)$ (call this group $G$) is not all of $S_5$, hence is $A_5$.
Although the line of reasoning here seems valid, it leads to some weird consequences. Given that $G$ = $A_5$, when we represent automorphisms by cycle types, any double transposition must be in the Galois group, i.e. must fix $i$. In particular $$(\alpha_1,\alpha_2)(\alpha_3,\alpha_4)$$ Must fix $i$. But when we apply this to $\alpha_1$, say, it would seem to involve conjugating it, and hence moving $i$. A contradiction. Yet if my line of reasoning is correct, this double transposition must leave $i$ fixed. How is this possible? Have I made an error somewhere, or just missing something obvious?
Many thanks.