Simplicité de An pour n ≥ 5

Ann≥5
Ann≥5
n= 5
n > 5n= 5
H G
h∈H h H ∀g∈G, ghg−1∈H
h∈H g ∈G c =ghg−1h−1= (ghg−1)h−1H
h
G H
•n= 5
A560 15 2
20 3 3 24 5 5
3A52
τ= (ab)(cd)(e)τ0= (a0b0)(c0d0)(e0)σ∈ Anσ(a) = a0σ(b) = b0σ(e) = e0
τ0=στσ−1
HCA5H6={1}H3 2
5 5
5 5
H
25 = 24 + 1 21 = 20 + 1 16 = 15 + 1 60 H
|A5|= 60 H|H| ≥ 15+20+1 = 36 |H|= 60
H=A5
•n > 5
E={1, . . . , n}HCAn, H 6={1}σ∈H σ 6= 1 n= 5
σ H
5n−5
ρ=τστ−1σ−1τ∈ Anρ= (τστ−1)σ−1ρ H ρ
ρ=τ(στ−1σ−1)ρ τ
τ ρ
σ6= 1 a∈E b =σ(a)6=a c ∈E c 6=a, b, σ(b)τ
3τ= (acb)τ−1= (abc)ρ=τστ−1σ−1= (acb)(σa, σb, σc)b=σ(a)
F={a, b, c, σa, σb, σc}5ρ(F) = F ρ|E\F=Id|E\F
F|F|= 5 ρ
1ρ(b) = τσ(b)6=b σ(b)6=τ−1(b) = c
A(F)FA(F)A5A(F)
Anu7→ u
u|F=u;u|E\F=Id|E\F
H0={u∈ A(F)/ u ∈H}=H∩ A(F)H0A(F)
ρ|F∈H0ρ|F6=IdFA(F)' A5H0=A(F)u3
A(F)H0u3H
3AnHAn
H=An