I'm studying the Galois group $G$ of a separable quartic polynomial $f$ over a field $K$ whose splitting field is $F$. Proposition V.4.11 in Algebra by Hungerford gives a classification of the Galois group $G$ of $f$:
Proposition 4.11. Let $K$ be a field and and $f\in K[x]$ be an (irreducible) separable quartic with Galois group $G$ (considered as a subgroup of $S_4$). Let $u_i$ ($i=1,\ 2,\ 3,\ 4$) be the roots of $f$ and write $\alpha = u_1 u_2 + u_3 u_4$, $\beta = u_1 u_3 + u_2 u_4$, $\gamma = u_1 u_4 + u_2 u_3$. Let $m = [K(\alpha,\ \beta,\ \gamma):K]$. Then:
(i) $m=6$ iff $G=S_4$;
(ii) $m=3$ iff $G=A_4$;
(iii) $m=1$ iff $G=V$, the Klein vierergruppe embedded in $S_4$;
(iv) $m=2$ iff $G\simeq D_4$ or $G\simeq \mathbb Z/4$, where $D_4$ is the group of isometries of a square. In this case, $G\simeq D_4$ if $f$ is irreducible over $K(\alpha,\ \beta,\ \gamma)$. Otherwise, $G\simeq \mathbb Z/4$.
In the proof, the author says:
(...) In view of the discussion preceeding the theorem, it suffices to prove only the implications $\Leftarrow$ in each case. (...)
So I repeat the discussion here:
We are now in a position to compute the Galois group of any (irreducible) separable quartic $f \in K[x]$. Since its Galois group $G$ is a transitive subgroup of $S_4$ whose order is divisible by $4$ (Theorem 4.2 in the book), $G$ must have order $24$, $12$, $8$ or $4$. Verify that the only transitive subgroups of orders $24$, $12$, and $4$ are $S_4$, $A_4$, $V(\simeq \mathbb Z/2 \oplus \mathbb Z/2)$ and the various cyclic subgroups of order $4$ generated by $4$-cycles. One transitive subgroup of $S_4$ of order $8$ is the dihedral group generated by $(1234)$ and $(24)$. Since $D_4$ is not normal in $S_4$, and since every subgroup of order $8$ is a Sylow $2$-subgroup, it follows from the Sylow theorems that $S_4$ has precisely three subgroups of order $8$, each isomorphic to $D_4$.
But I don't see why this paragraph gives the proof of $\Rightarrow$ part. For instance if $m=6$, then it means that there is a subgroup of $G$ of index $6$; so the only choices of $|G|$ are $12$ and $24$. Now what should I do?