Minimal faithful permutation degree

How few points can a group be shuffled on? Every finite group G acts faithfully on |G| points by Cayley's theorem, but usually far fewer suffice. The minimal faithful permutation degree is

μ(G) = min { m : G embeds in Sₘ }.

The action realizing it need not be transitive; the transitive minimum is the larger μ_t(G).

Cosets and cores

A permutation representation is a disjoint union of transitive ones, and each transitive piece is the action of G on the cosets of a subgroup H ≤ G, of degree [G : H]. Its kernel is the normal core core(H) = ⋂_g∈G gHg⁻¹, the largest normal subgroup contained in H. A union of pieces H₁, …, H_r is faithful exactly when the cores meet trivially, so

μ(G) = min { [G : H₁] + ⋯ + [G : H_r]  :  core(H₁) ∩ ⋯ ∩ core(H_r) = 1 }.

Equivalently: the union is faithful iff every minimal normal subgroup N of G escapes at least one member — N ≰ Hᵢ for some i. This is the practical test, since N ≤ H ⟺ N ≤ core(H) for minimal normal N.

Transitive or not

Always μ(G) ≤ μ_t(G), taking the one-member family. Equality holds whenever G is monolithic (a unique minimal normal subgroup N): some member must escape N, and having escaped the only minimal normal subgroup its core is trivial, so that member alone is already faithful.

The gap is real otherwise. In C₃ × C₃ every subgroup is normal, so the only core-free subgroup is {e} and the smallest faithful transitive action is the regular one on 9 points — but two independent C₃-orbits are faithful already, so μ(C₃ × C₃) = 3 + 3 = 6.

Abelian groups

Theorem (Johnson). For abelian G with primary cyclic decomposition G ≅ ⊕_j C_{p_j^a_j},

μ(G) = Σ_j p_j^a_j.

One orbit per primary factor, each carrying a full cycle. So μ(C₆) = μ(C₂ × C₃) = 2 + 3 = 5 — the permutation (1,2)(3,4,5) has order 6 — while μ(C₄) = 4 and μ(C₂ × C₂) = 4.

Note what the formula does not say: it sums the primary factors, not the invariant factors. C₆ × C₆ ≅ C₂² × C₃² gives 2 + 2 + 3 + 3 = 10.

When Cayley cannot be beaten

μ(G) = |G| for exactly three families (Johnson, 1971): the cyclic p-groups C_p^k, the generalized quaternion groups Q₂ⁿ, and the Klein four-group C₂ × C₂. In the first two a unique subgroup of order p sits inside every nontrivial subgroup, forcing the regular action; the Klein four-group is a small coincidence (2 + 2 = 2·2).

Direct products

μ(G × H) ≤ μ(G) + μ(H) always — put the two faithful actions on disjoint point sets. Equality is not general: Saunders exhibits direct products with μ(G × H) < μ(G) + μ(H). It does hold in two common cases:

Coprime orders. gcd(|G|, |H|) = 1 forces every subgroup of G × H to split as a product, and the bound follows.
Both nilpotent (Wright): μ(P × Q) = μ(P) + μ(Q).

Values

group	`\|G\|`	`μ`	`μ_t`
`C₄`	`4`	`4`	`4`
`C₆`	`6`	`5`	`6`
`C₃ × C₃`	`9`	`6`	`9`
`D₄`	`8`	`4`	`4`
`Q₈`	`8`	`8`	`8`
`D₆ ≅ C₂ × S₃`	`12`	`5`	`6`
`Sₙ` (`n ≥ 3`)	`n!`	`n`	`n`
`Aₙ` (`n ≥ 4`)	`n!/2`	`n`	`n`
`PSL(2,𝔽₇)`	`168`	`7`	`7`

D₄ vs Q₈ (both of order 8). D₄ acts faithfully on the 4 vertices of a square: a reflection ⟨s⟩ ≅ C₂ is core-free of index 4. Q₈ cannot improve on the regular action — its only subgroup of order 2 is the center {±1}, which lies inside every nontrivial subgroup.
D₆ ≅ C₂ × S₃ splits its 5 points as 2 + 3: the central reflection swaps a pair, the S₃ permutes a triangle. No faithful action on 4 points exists, and a transitive one needs 6.
PSL(2,𝔽₇). As GL(3,𝔽₂) it acts on the 7 points of the Fano plane, below its 8-point projective-line action.