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Denote the three stated pair sums by $r_1,r_2,r_3$. They are pairwise distinct: for example,and the other differences factor similarly. Every Galois automorphism fixes each $r_i\in F$, and hence fixes each of the three corresponding partitions of four roots into two pairs. The kernel of the permutation action $S_4\to S_3$ on these partitions isThus $G$ embeds in the Klein four-group. By part (a), $4$ divides $|G|$, and the containing group has order four. ConsequentlyThis Split resolvent criterion for a Klein four Galois group uses the cubic resolvent of a quartic. The pair-sum differences remain nonzero in characteristic two, so this proof does not need the missing characteristic assumption from part (c).

Let $G$ be the Galois group of a polynomial of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given cubic resolvent of a quartic correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the Klein four-group $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the transitive subgroups of the symmetric group on four points, only $S_4$ has this image. Hence

Now put $f_p(t)=t^4+pt+p$. The Eisenstein criterion at $p$ proves that $f_p$ is an irreducible polynomial, and its cubic resolvent isIts discriminant of a depressed cubic isFor odd $p$, the $p$-adic valuation of ${\Delta }_p$ is three because $p\nmid 256$; for $p=2$, ${\Delta }_2=2^4\cdot 101$. Thus ${\Delta }_p$ is never a square in $\mathbb{Q}$.

By the rational root theorem, a reducible $g_p$ has an integer root among $\pm 1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in \{\pm 1,\pm p\}$, givesThe only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore

For a depressed quarticwith roots summing to zero, setThe cubic resolvent of a quartic is

The symmetric group $S_4$ acts on the three partitions of four roots into two unordered pairs. The resulting homomorphism $S_4\to S_3$ has kernel the Klein four-group $V_4$. If $\Delta (f)$ is a square and the characteristic is not two, the discriminant criterion for an alternating Galois group gives $G=\mathrm{Gal}(f)\le A_4$, and its image on the three resolvent roots lies in $A_4/V_4\cong C_3$.

For an irreducible quartic, $G$ is transitive on four roots, and its only possibilities inside $A_4$ are $V_4$ and $A_4$. The resolvent is irreducible exactly when the image acts transitively on its three roots, equivalently when that image is $C_3$. ThusThe printed statement omits the needed irreducibility hypothesis on $f$. Literally it is false: $f=x(x^3-3x-1)$ has square discriminant and Galois group $C_3$, and its resolvent is irreducible. Without quartic irreducibility, a reducible resolvent only implies $G\le V_4$, allowing the trivial group, $C_2$, or $V_4$.

For the specified polynomial $x^4+8x+12$,Its resolvent is $g(y)=y^3-48y-64$. Modulo $5$ this is $y^3+2y+1$, which has no root in ${\mathbb{F}}_5$ and is therefore an irreducible cubic. The quartic is given to be irreducible, so the preceding criterion yields

If an irreducible quartic over a field of characteristic zero has a cubic resolvent of a quartic whose Galois group of a polynomial is $S_3$, then the quartic has Galois group $S_4$. Irreducibility places the quartic group among the transitive subgroups of the symmetric group on four points, while the action on its three root pairings is surjective onto $S_3$. Of the possibilities $C_4,V_4,D_8,A_4,S_4$, only $S_4$ has this image.