Alternating Groups

The alternating group on a finite type α is the subgroup of the permutation group Perm α consisting of the even permutations.

Main definitions

• alternatingGroup α is the alternating group on α, defined as a Subgroup (Perm α).

Main results

• two_mul_card_alternatingGroup shows that the alternating group is half as large as the permutation group it is a subgroup of.

• closure_three_cycles_eq_alternating shows that the alternating group is generated by 3-cycles.

• alternatingGroup.isSimpleGroup_five shows that the alternating group on Fin 5 is simple. The proof shows that the normal closure of any non-identity element of this group contains a 3-cycle.

Tags

alternating group permutation

TODO

• Show that alternatingGroup α is simple if and only if Fintype.card α ≠ 4.

def alternatingGroup (α : Type u_1) [Fintype α] [DecidableEq α] : Subgroup (Equiv.Perm α)

The alternating group on a finite type, realized as a subgroup of Equiv.Perm. For $A_n$, use alternatingGroup (Fin n).

Equations

• alternatingGroup α = Equiv.Perm.sign.ker

instance alternatingGroup.instFintype (α : Type u_1) [Fintype α] [DecidableEq α] : Fintype { x : Equiv.Perm α // x ∈ alternatingGroup α }

Equations

• alternatingGroup.instFintype α = Subtype.fintype (Membership.mem (alternatingGroup α))

instance instUniqueSubtypePermMemSubgroupAlternatingGroupOfSubsingleton (α : Type u_1) [Fintype α] [DecidableEq α] [Subsingleton α] : Unique { x : Equiv.Perm α // x ∈ alternatingGroup α }

Equations

• instUniqueSubtypePermMemSubgroupAlternatingGroupOfSubsingleton α = { default := 1, uniq := ⋯ }

theorem alternatingGroup_eq_sign_ker {α : Type u_1} [Fintype α] [DecidableEq α] : alternatingGroup α = Equiv.Perm.sign.ker

@[simp]

theorem Equiv.Perm.mem_alternatingGroup {α : Type u_1} [Fintype α] [DecidableEq α] {f : Equiv.Perm α} : f ∈ alternatingGroup α ↔ Equiv.Perm.sign f = 1

theorem Equiv.Perm.prod_list_swap_mem_alternatingGroup_iff_even_length {α : Type u_1} [Fintype α] [DecidableEq α] {l : List (Equiv.Perm α)} (hl : ∀ g ∈ l, g.IsSwap) : l.prod ∈ alternatingGroup α ↔ Even l.length

theorem Equiv.Perm.IsThreeCycle.mem_alternatingGroup {α : Type u_1} [Fintype α] [DecidableEq α] {f : Equiv.Perm α} (h : f.IsThreeCycle) : f ∈ alternatingGroup α

theorem Equiv.Perm.finRotate_bit1_mem_alternatingGroup {n : ℕ} : finRotate (2 * n + 1) ∈ alternatingGroup (Fin (2 * n + 1))

theorem two_mul_card_alternatingGroup {α : Type u_1} [Fintype α] [DecidableEq α] [Nontrivial α] : 2 * Fintype.card { x : Equiv.Perm α // x ∈ alternatingGroup α } = Fintype.card (Equiv.Perm α)

instance alternatingGroup.normal {α : Type u_1} [Fintype α] [DecidableEq α] : (alternatingGroup α).Normal

Equations

• ⋯ = ⋯

theorem alternatingGroup.isConj_of {α : Type u_1} [Fintype α] [DecidableEq α] {σ : { x : Equiv.Perm α // x ∈ alternatingGroup α }} {τ : { x : Equiv.Perm α // x ∈ alternatingGroup α }} (hc : IsConj ↑σ ↑τ) (hσ : (↑σ).support.card + 2 ≤ Fintype.card α) : IsConj σ τ

theorem alternatingGroup.isThreeCycle_isConj {α : Type u_1} [Fintype α] [DecidableEq α] (h5 : 5 ≤ Fintype.card α) {σ : { x : Equiv.Perm α // x ∈ alternatingGroup α }} {τ : { x : Equiv.Perm α // x ∈ alternatingGroup α }} (hσ : (↑σ).IsThreeCycle) (hτ : (↑τ).IsThreeCycle) : IsConj σ τ

@[simp]

theorem Equiv.Perm.closure_three_cycles_eq_alternating {α : Type u_1} [Fintype α] [DecidableEq α] : Subgroup.closure {σ : Equiv.Perm α | σ.IsThreeCycle} = alternatingGroup α

theorem Equiv.Perm.IsThreeCycle.alternating_normalClosure {α : Type u_1} [Fintype α] [DecidableEq α] (h5 : 5 ≤ Fintype.card α) {f : Equiv.Perm α} (hf : f.IsThreeCycle) : Subgroup.normalClosure {⟨f, ⋯⟩} = ⊤

A key lemma to prove $A_5$ is simple. Shows that any normal subgroup of an alternating group on at least 5 elements is the entire alternating group if it contains a 3-cycle.

theorem Equiv.Perm.isThreeCycle_sq_of_three_mem_cycleType_five {g : Equiv.Perm (Fin 5)} (h : 3 ∈ g.cycleType) : (g * g).IsThreeCycle

Part of proving $A_5$ is simple. Shows that the square of any element of $A_5$ with a 3-cycle in its cycle decomposition is a 3-cycle, so the normal closure of the original element must be $A_5$.

theorem alternatingGroup.nontrivial_of_three_le_card {α : Type u_1} [Fintype α] [DecidableEq α] (h3 : 3 ≤ Fintype.card α) : Nontrivial { x : Equiv.Perm α // x ∈ alternatingGroup α }

instance alternatingGroup.instNontrivialSubtypePermFinHAddNatOfNatMemSubgroup {n : ℕ} : Nontrivial { x : Equiv.Perm (Fin (n + 3)) // x ∈ alternatingGroup (Fin (n + 3)) }

Equations

• ⋯ = ⋯

theorem alternatingGroup.normalClosure_finRotate_five : Subgroup.normalClosure {⟨finRotate 5, ⋯⟩} = ⊤

The normal closure of the 5-cycle finRotate 5 within $A_5$ is the whole group. This will be used to show that the normal closure of any 5-cycle within $A_5$ is the whole group.

theorem alternatingGroup.normalClosure_swap_mul_swap_five : Subgroup.normalClosure {⟨Equiv.swap 0 4 * Equiv.swap 1 3, ⋯⟩} = ⊤

The normal closure of $(04)(13)$ within $A_5$ is the whole group. This will be used to show that the normal closure of any permutation of cycle type $(2,2)$ is the whole group.

theorem alternatingGroup.isConj_swap_mul_swap_of_cycleType_two {g : Equiv.Perm (Fin 5)} (ha : g ∈ alternatingGroup (Fin 5)) (h1 : g ≠ 1) (h2 : ∀ n ∈ g.cycleType, n = 2) : IsConj (Equiv.swap 0 4 * Equiv.swap 1 3) g

Shows that any non-identity element of $A_5$ whose cycle decomposition consists only of swaps is conjugate to $(04)(13)$. This is used to show that the normal closure of such a permutation in $A_5$ is $A_5$.

instance alternatingGroup.isSimpleGroup_five : IsSimpleGroup { x : Equiv.Perm (Fin 5) // x ∈ alternatingGroup (Fin 5) }

Shows that $A_5$ is simple by taking an arbitrary non-identity element and showing by casework on its cycle type that its normal closure is all of $A_5$.

Equations

• alternatingGroup.isSimpleGroup_five = ⋯