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The transitive subgroups of $S_{4}$ and their action on the three pairings

Statement

Up to conjugacy, the transitive subgroups of $S_{4}$ are $S_{4}, A_{4}, D_{4}, C_{4}$ , and $V_{4}$ , with the stated action on the three pairings. If

$V_{4} = {1, (12) (34), (13) (24), (14) (23)},$

then the action on the three partitions of ${1, 2, 3, 4}$ into two unordered pairs has kernel $V_{4}$ . The corresponding data are:

$H$	$∣ H \cap V_{4} ∣$	image on pairings	$H \leq A_{4}$
$S_{4}$	$4$	$S_{3}$	no
$A_{4}$	$4$	$A_{3}$	yes
$D_{4}$	$4$	$C_{2}$	no
$C_{4}$	$2$	$C_{2}$	no
$V_{4}$	$4$	trivial	yes

Facts & Assumptions

Given: The natural action of $S_{4}$ on four symbols; orbit-stabilizer cardinality (Orbit-stabiliser cardinality: $∣ G \cdot x ∣ = [G : G_{x}]$ whenever either side is finite, and $∣ G ∣ = ∣ G_{x} ∣ ∣ G \cdot x ∣$ for finite $G$ ); Lagrange's theorem (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ); conjugacy of Sylow subgroups (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class); and kernels as normal subgroups with quotient image (Normal subgroup: invariance under conjugation, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L1]

Thus $A_{n}$ consists exactly of the even permutations (The alternating group $A_{n} = \ker (sgn)$ of even permutations).

Proof

technique · direct

1.1

Acting on the explicitly listed pairings gives a homomorphism $θ : S_{4} \to S_{3}$ . A permutation fixes all three pairings exactly when it is the identity or one of the three double transpositions, so $\ker θ = V_{4}$ .

algebra

2.1

If $H \leq S_{4}$ is transitive on four symbols, orbit-stabilizer makes $4$ divide $∣ H ∣$ and Lagrange makes $∣ H ∣$ divide $24$ , so $∣ H ∣ \in {4, 8, 12, 24}$ . Order $24$ gives $S_{4}$ . An order- $12$ subgroup has index two and is normal; if it contained an odd permutation, the conjugates of that transposition or four-cycle would generate $S_{4}$ , so it is $A_{4}$ . An order- $8$ subgroup is Sylow and hence conjugate to the standard $D_{4}$ . For order $4$ the action is regular; an element of order four gives $C_{4}$ , and otherwise all nonidentity elements have order two and give $V_{4}$ .

step 1.1given

3.1

Intersecting representatives with the kernel in step 1.1 gives the second column of the table, and the quotient orders give the pairing images. By [L1], $S_{4}, D_{4}, C_{4}$ contain odd permutations, whereas $A_{4}$ and $V_{4}$ do not. These invariants give the displayed rows; only $D_{4}$ and $C_{4}$ share the same nontrivial intransitive pairing image, and their different kernel orders distinguish them.

step 1.1step 2.1L1algebra∎

Depends on

Used by

The five-case resolvent classification of an irreducible quartic Galois group Theorem

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Galois Groups of Cubics and Quartics, Table 3 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Quartic polynomials (standard reference, not scraped)

The transitive subgroups of S4S4​ and their action on the three pairings