I was solving the following problem:
Let $char(F)\ne 2,3$ and $f(x) = x^4 +px^2 +qx+r$ be an irreducible separable polynomial in $F [x]$ with roots being $α_1, α_2, α_3, α_4$. Let $L = F(α_1,α_2,α_3,α_4) $ and $G = Gal(L/F) ≤ S_4$. Set $β_1 = α_1α_2 + α_3α_4, β_2 =α_1α_3+α_2α_4$ and $β_3 =α_1α_4+α_2α_3$. Show that there exists $i$ such that $β_i \in F$ if and only if $G≤D_8.$
Attempt: I first showed that $Gal(F(\beta_1,\beta_2,\beta_3)/F)\cong G/(G\cap V)$, where $V =\{id,(12)(34),(13)(24),(14)(23)\}$. Assume $\beta _1\in F$, then $[F(\beta_2,\beta_3):F]=\dfrac{|G|}{|G\cap V|}\ge \dfrac{|G|}{4}$. For contradiction, I supposed $G\gneq D_8\implies |G|\gneq 8, [F(\beta_2,\beta_3):F]\gneq2$, but this actually turned out to make sense to me, and I couldn't get a contradiction, which is imo weird. Is there anything I have missed? What did I do wrong?
From your remark $Gal(F(\beta_1,\beta_2,\beta_3)/F)\cong G/(G\cap V)$ we see this group must be a subgroup of $S_3$.
If it contains an element of order $3$ then the $\beta_i$ are all conjugate and so none can lie in $F$.
If it does not contain an element of order $3$ then it is either trivial or cyclic of order $2$; in each case one $\beta_i$ lies in an orbit of size $1$ and so is in $F$.
The only subgroups of $S_4$ not containing an element of order $3$ are subgroups of $D_8$.