[proofplan] The proof identifies all transitive subgroups of $S_4$ (since $f$ is irreducible, $G$ acts transitively on the four roots) and then distinguishes them using two computable invariants: the discriminant $\Delta_f$ and the factorisation of the resolvent cubic $g$. The discriminant detects containment in $A_4$ via the [Discriminant and the Alternating Group](/theorems/1325). The resolvent cubic encodes the action of $G$ on the three partitions of the root set into pairs, and its splitting behaviour over $\mathbb{Q}$ reveals the size of the image of $G$ in $S_3$ under this action. Combining these two tests classifies $G$ completely. [/proofplan]

[step:Enumerate all transitive subgroups of $S_4$]Let $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ be the roots of $f$ in $K$. Since $f$ is irreducible over $\mathbb{Q}$, the Galois group $G = \operatorname{Gal}(K/\mathbb{Q})$ acts transitively on $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$: for any two roots $\alpha_i, \alpha_j$, there exists $\sigma \in G$ with $\sigma(\alpha_i) = \alpha_j$ (because $\alpha_i$ and $\alpha_j$ have the same minimal polynomial $f$ over $\mathbb{Q}$, so the [Extension of Isomorphisms to Splitting Fields](/theorems/3313) provides such an automorphism). By the [Galois Group as a Subgroup of $S_n$](/theorems/3326), $G$ embeds as a transitive subgroup of $S_4$. By the [Orbit-Stabiliser Theorem](/theorems/796), transitivity gives $4 \mid |G|$, since the orbit of any root has size $4$ and $|G| = |\operatorname{Orb}_G(\alpha_1)| \cdot |\operatorname{Stab}_G(\alpha_1)|$. Moreover, $|G| \mid |S_4| = 24$ by [Lagrange's Theorem](/theorems/782). The transitive subgroups of $S_4$ (up to conjugacy) are exactly: 1. $S_4$ of order $24$, 2. $A_4$ of order $12$, 3. $D_8$ of order $8$ (the dihedral group of the square, e.g., $\langle (1234), (13) \rangle$), 4. $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ of order $4$ (the Klein four-group), 5. $\mathbb{Z}/4\mathbb{Z} = \langle (1234) \rangle$ of order $4$. This is verified by checking all subgroups of $S_4$ of order divisible by $4$ and selecting those acting transitively on $\{1,2,3,4\}$.[/step]

[guided]Why must $G$ be transitive? If $f$ is irreducible over $\mathbb{Q}$, then all four roots are conjugate over $\mathbb{Q}$: they share the same minimal polynomial. Given any two roots $\alpha_i$ and $\alpha_j$, the $\mathbb{Q}$-isomorphism $\mathbb{Q}(\alpha_i) \cong \mathbb{Q}(\alpha_j)$ sending $\alpha_i \mapsto \alpha_j$ extends to a $\mathbb{Q}$-automorphism of the splitting field $K$ by the [Extension of Isomorphisms to Splitting Fields](/theorems/3313). This automorphism maps $\alpha_i$ to $\alpha_j$, so $G$ acts transitively on the roots. Transitivity forces $4 \mid |G|$ by the [Orbit-Stabiliser Theorem](/theorems/796). Combined with $|G| \mid 24$ from [Lagrange's Theorem](/theorems/782) applied to $G \leq S_4$, the possible orders are $4, 8, 12, 24$. A case-by-case enumeration of subgroups of $S_4$ at these orders, checking transitivity, yields exactly the five groups listed. There are other subgroups of orders $4$ and $8$ (e.g., $\langle (12) \rangle \times \langle (34) \rangle \cong V_4$ acting intransitively), but they fail transitivity.[/guided]

[step:Use the discriminant to detect containment in $A_4$] By the [Discriminant and the Alternating Group](/theorems/1325), $G \subseteq A_4$ if and only if $\Delta_f$ is a perfect square in $\mathbb{Q}$. Among our five candidates: - $S_4 \not\subseteq A_4$ and $D_8 \not\subseteq A_4$ (both contain odd permutations), - $\mathbb{Z}/4\mathbb{Z} = \langle (1234) \rangle \not\subseteq A_4$ (the $4$-cycle $(1234)$ is an odd permutation), - $A_4 \subseteq A_4$ and $V_4 \subseteq A_4$. Hence $\Delta_f$ is a perfect square in $\mathbb{Q}$ if and only if $G \cong A_4$ or $G \cong V_4$. [/step]

[step:Introduce the resolvent cubic and its roots]The three elements \begin{align*} \theta_1 &= \alpha_1\alpha_2 + \alpha_3\alpha_4, \\ \theta_2 &= \alpha_1\alpha_3 + \alpha_2\alpha_4, \\ \theta_3 &= \alpha_1\alpha_4 + \alpha_2\alpha_3 \end{align*} correspond to the three ways of partitioning $\{1,2,3,4\}$ into two pairs. The resolvent cubic $g \in \mathbb{Q}[x]$ is the polynomial with roots $\theta_1, \theta_2, \theta_3$; its coefficients lie in $\mathbb{Q}$ because $\theta_1, \theta_2, \theta_3$ are the elementary symmetric functions of the $\theta_i$ expressed via the coefficients of $f$ (see [Resolvent Cubic of a Quartic](/theorems/1307)). The Galois group $G$ acts on $\{\theta_1, \theta_2, \theta_3\}$ by permuting the root partitions. This defines a group homomorphism \begin{align*} \psi: G \to S_3, \end{align*} where $S_3$ is the symmetric group on $\{\theta_1, \theta_2, \theta_3\}$.[/step]

[guided]The resolvent cubic is the key tool that "compresses" the action of $G$ on four roots to an action on three objects. Each $\theta_i$ corresponds to a partition of $\{1,2,3,4\}$ into two pairs: - $\theta_1 = \alpha_1\alpha_2 + \alpha_3\alpha_4$ corresponds to $\{\{1,2\},\{3,4\}\}$, - $\theta_2 = \alpha_1\alpha_3 + \alpha_2\alpha_4$ corresponds to $\{\{1,3\},\{2,4\}\}$, - $\theta_3 = \alpha_1\alpha_4 + \alpha_2\alpha_3$ corresponds to $\{\{1,4\},\{2,3\}\}$. A permutation $\sigma \in S_4$ permutes the roots, which permutes the partitions, which permutes the $\theta_i$. This gives the homomorphism $\psi: G \to S_3$. Why is this useful? Because $\theta_i \in \mathbb{Q}$ if and only if $\sigma(\theta_i) = \theta_i$ for all $\sigma \in G$. In other words, the splitting behaviour of $g$ over $\mathbb{Q}$ is controlled by the image $\psi(G) \leq S_3$, which we can compute for each candidate group.[/guided]

[step:Compute the image $\psi(G)$ for each candidate group] We compute $\psi(G) \leq S_3$ for each of the five transitive subgroups, using the explicit action on partitions. **$G = S_4$:** Every permutation of the three partitions is realised, so $\psi(S_4) = S_3$. The resolvent cubic $g$ is irreducible over $\mathbb{Q}$ (it has no rational root, since $G$ acts transitively on $\{\theta_1, \theta_2, \theta_3\}$). **$G = A_4$:** The even permutations of four elements still act transitively on the three partitions (e.g., $(123)$ sends $\theta_1 \mapsto \theta_2 \mapsto \theta_3 \mapsto \theta_1$). We have $\psi(A_4) = S_3$ (the kernel of $\psi|_{A_4}$ is $V_4 = A_4 \cap \ker\psi$, so $|{\psi(A_4)}| = |A_4|/|V_4| = 12/4 = 3$... Let us verify: the kernel of $\psi: S_4 \to S_3$ is $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$, since these are exactly the permutations that fix each partition. So $\ker(\psi|_{A_4}) = V_4 \cap A_4 = V_4$ (since $V_4 \subset A_4$). Hence $|\psi(A_4)| = 12/4 = 3$, so $\psi(A_4) \cong A_3$, a cyclic group of order $3$. This means $A_4$ acts transitively on $\{\theta_1, \theta_2, \theta_3\}$ but only via even permutations. The resolvent cubic $g$ is irreducible over $\mathbb{Q}$. **$G = D_8 = \langle (1234), (13) \rangle$:** The $4$-cycle $(1234)$ sends $\theta_1 = \alpha_1\alpha_2 + \alpha_3\alpha_4 \mapsto \alpha_2\alpha_3 + \alpha_4\alpha_1 = \theta_3$, and $\theta_2 = \alpha_1\alpha_3 + \alpha_2\alpha_4 \mapsto \alpha_2\alpha_4 + \alpha_3\alpha_1 = \theta_2$, and $\theta_3 \mapsto \theta_1$. So $(1234)$ acts as the transposition $(\theta_1\, \theta_3)$ fixing $\theta_2$. Thus $\theta_2 \in \mathbb{Q}$ (it is fixed by all of $G$, since $(13)$ also fixes $\theta_2$: $(13)$ sends $\alpha_1\alpha_3 + \alpha_2\alpha_4 \mapsto \alpha_3\alpha_1 + \alpha_2\alpha_4 = \theta_2$). So $g$ has exactly one rational root $\theta_2$, and factors as a linear times an irreducible quadratic over $\mathbb{Q}$. **$G = V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$:** Each element of $V_4$ fixes all three $\theta_i$. Indeed, $(12)(34)$ sends $\theta_1 = \alpha_1\alpha_2 + \alpha_3\alpha_4 \mapsto \alpha_2\alpha_1 + \alpha_4\alpha_3 = \theta_1$, and similarly for $\theta_2$ and $\theta_3$. Hence $\psi(V_4) = \{e\}$, and all three $\theta_i$ lie in $\mathbb{Q}$. The resolvent cubic $g$ splits completely over $\mathbb{Q}$. **$G = \mathbb{Z}/4\mathbb{Z} = \langle (1234) \rangle$:** As computed above, $(1234)$ sends $\theta_1 \mapsto \theta_3$, $\theta_2 \mapsto \theta_2$, $\theta_3 \mapsto \theta_1$. So $\theta_2 \in \mathbb{Q}$ but $\theta_1, \theta_3 \notin \mathbb{Q}$ (they are swapped by a generator of $G$). The resolvent cubic has exactly one rational root. [/step]

[step:Distinguish the five cases using $\Delta_f$ and $g$]We now combine the discriminant test and the resolvent cubic factorisation to classify $G$. **Case 1: $\Delta_f$ is not a perfect square and $g$ is irreducible.** From the discriminant test, $G \not\subseteq A_4$, ruling out $A_4$ and $V_4$. The irreducibility of $g$ means no $\theta_i$ is rational, ruling out $D_8$ and $\mathbb{Z}/4\mathbb{Z}$ (both of which fix $\theta_2$). Hence $G \cong S_4$. **Case 2: $\Delta_f$ is a perfect square and $g$ is irreducible.** From the discriminant test, $G \subseteq A_4$, so $G \cong A_4$ or $V_4$. Since $g$ is irreducible, no $\theta_i$ is rational, ruling out $V_4$ (which fixes all three). Hence $G \cong A_4$. **Case 3: $\Delta_f$ is a perfect square and $g$ splits completely.** The discriminant forces $G \subseteq A_4$. The complete splitting means all $\theta_i \in \mathbb{Q}$, i.e., $\psi(G) = \{e\}$. Among our candidates with $G \subseteq A_4$, only $V_4$ has $\psi(G) = \{e\}$ (since $\psi(A_4) \cong A_3 \neq \{e\}$). Hence $G \cong V_4$. **Case 4: $\Delta_f$ is not a perfect square and $g$ has exactly one rational root.** The discriminant forces $G \not\subseteq A_4$, so $G \cong S_4$, $D_8$, or $\mathbb{Z}/4\mathbb{Z}$. The existence of exactly one rational root among $\theta_1, \theta_2, \theta_3$ rules out $S_4$ (which has no rational root — $g$ would be irreducible). Both $D_8$ and $\mathbb{Z}/4\mathbb{Z}$ have exactly one rational root. To distinguish $D_8$ from $\mathbb{Z}/4\mathbb{Z}$: they have orders $8$ and $4$ respectively. By the [Order of the Galois Group](/theorems/3325), $|G| = [K : \mathbb{Q}]$. Since $f$ is irreducible of degree $4$, we have $[\mathbb{Q}(\alpha_1) : \mathbb{Q}] = 4$, and $[K : \mathbb{Q}(\alpha_1)]$ equals the number of ways the remaining roots can be expressed over $\mathbb{Q}(\alpha_1)$. For $G \cong D_8$, $[K:\mathbb{Q}] = 8$; for $G \cong \mathbb{Z}/4\mathbb{Z}$, $[K:\mathbb{Q}] = 4$, meaning $K = \mathbb{Q}(\alpha_1)$ (the splitting field equals the stem field). In practice, $G \cong \mathbb{Z}/4\mathbb{Z}$ if and only if $g$ has exactly one rational root and $\Delta_f$ is not a perfect square and the quadratic factor of $g$ over $\mathbb{Q}$ splits over $\mathbb{Q}(\alpha_1)$ (equivalently, $[K:\mathbb{Q}] = 4$). Otherwise $G \cong D_8$.[/step]

[guided]The classification works as a decision tree with two tests: 1. **Is $\Delta_f$ a perfect square?** This splits the five candidates into two groups: $\{A_4, V_4\}$ (yes) and $\{S_4, D_8, \mathbb{Z}/4\mathbb{Z}\}$ (no). 2. **How does $g$ factor over $\mathbb{Q}$?** This refines within each group: - Among $\{A_4, V_4\}$: $g$ irreducible $\Rightarrow A_4$; $g$ splits completely $\Rightarrow V_4$. - Among $\{S_4, D_8, \mathbb{Z}/4\mathbb{Z}\}$: $g$ irreducible $\Rightarrow S_4$; $g$ has one rational root $\Rightarrow D_8$ or $\mathbb{Z}/4\mathbb{Z}$. The final distinction between $D_8$ and $\mathbb{Z}/4\mathbb{Z}$ requires a finer test. Both have $|G|$ divisible by $4$ but $D_8$ has order $8$ while $\mathbb{Z}/4\mathbb{Z}$ has order $4$. The group $G \cong \mathbb{Z}/4\mathbb{Z}$ occurs precisely when $[K:\mathbb{Q}] = 4$, i.e., when adjoining a single root already gives the entire splitting field. This happens when the remaining roots of $f$ can be expressed as rational functions of $\alpha_1$ over $\mathbb{Q}$. Note that the case "$\Delta_f$ is a perfect square and $g$ has exactly one rational root" does not arise among our five candidates: the only groups with $G \subseteq A_4$ and exactly one fixed $\theta_i$ would need an element of $A_4$ that transposes two of the $\theta_i$ while fixing the third — but any such transposition on partitions lifts to an odd permutation, contradicting $G \subseteq A_4$. Similarly, "$\Delta_f$ not a square and $g$ splits completely" cannot occur: $g$ splitting means $V_4 \leq \ker\psi \leq G$, but if all three $\theta_i$ are rational and $G \not\subseteq A_4$, then $G$ contains an element not in $A_4$ that fixes all $\theta_i$; the only such elements in $S_4$ lie in $V_4 \subset A_4$, a contradiction. So the four cases in the theorem statement, together with the $D_8$ vs $\mathbb{Z}/4\mathbb{Z}$ refinement, are exhaustive.[/guided]