We are given that $E/\Bbb{Q}$ is a splitting field extension for some quartic polynomial $f(x)\in \Bbb{Q}[x]$ such that $G = \operatorname{Gal}(E/\Bbb{Q}) = S_4$. We are told $f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)\in E[x]$, and we must determine the degree of $[B:\Bbb{Q}]$ where $B = \Bbb{Q}(\alpha_1+\alpha_2)$, and we must determine the degree of the subfields of $B$.
Progress Thus Far
I believe I have determined the degree $[B:\Bbb{Q}]$. We need all the permutations that will fix $\alpha_1+\alpha_2$, which are the following: $$ H = \{(1), (12), (34),(12)(34)\}.$$ This means that $[B:\Bbb{Q}] = |G:H| = 24/4 = 6.$
Now to determine the subfields, I want to find the subgroups of $S_4$ that contain $H$. I believe the only possibilities are $$H_1 = \{(1),(12),(34),(12)(34), (1324),(1423),(14)(23),(13)(24)\} \simeq D_4,$$ and $S_4$ itself. Clearly, the fixed field of $S_4$ is $\Bbb{Q}$ and $[\Bbb{Q}:\Bbb{Q}] = 1$. I know that the degree over $\Bbb{Q}$ of the fixed field of $H_1$ would be $24/8 = 3$, but I am having a hard time figuring out the fixed field of $H_1$.
It seems like the field $B_1 = \Bbb{Q}((\alpha_1+\alpha_2)(\alpha_3+\alpha_4))$ would be fixed by $H_1$, but how would I know that this is THE fixed field? Also, how do I know that this is in fact a subfield of $B$, because it is not clear to me that it is? Is there some other way that we could write $B$ that would make it more clear that $B_1$ is a subfield of $B$?
I have some insight into why $B_1 = \Bbb{Q}((\alpha_1+\alpha_2)(\alpha_3+\alpha_4))$ is a subfield of $B$. First, if we expand out $f$ we can find that $\alpha_1+\alpha_2+\alpha_3+\alpha_4$ is one of the coefficients. This means $\alpha_1+\alpha_2+\alpha_3+\alpha_4 \in \Bbb{Q}$, which also implies that $(\alpha_1+\alpha_2+\alpha_3+\alpha_4) - (\alpha_1+\alpha_2) = \alpha_3 + \alpha_4$ must be in $B$ as well. So in fact $(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)\in B$ since $B$ is a field.