Recall • ϵ:S_n→Z\2Z σ↦{■8(0 ̅&σ is a product of even number of transposition@1 ̅&σ is a product of odd number of transposition)┤ • ϵ^′:S_n→Z\2Z σ↦{■8(0 ̅&σ(Δ)=Δ@1 ̅&σ(Δ)=−Δ)┤ • Δ≔∏_(1≤i j≤n)▒(x_i−x_j ) , σ(Δ)≔∏_(1≤i j≤n)▒(x_σ(i) −x_σ(j) ) Proposition 40 • Statement ○ Let n∈Z( 0) ○ If τ∈S_n is transposition, then ϵ^′ (τ)=1 ̅ • Example ○ When n=4, τ=(1 2) ○ Δ=(x_1−x_2 )(x_1−x_3 )(x_1−x_4 )(x_2−x_3 )(x_2−x_4 )(x_3−x_4 ) ○ τ(Δ)=(x_2−x_1 )(x_2−x_3 )(x_2−x_4 )(x_1−x_3 )(x_1−x_4 )(x_3−x_4 ) ○ τ(Δ)=−Δ⇒ϵ^′ (τ)=1 ̅ • Proof: Suppose τ=(1 2) ○ Say (x_i−x_j ) is a factor of Δ ○ Then τ(i) τ(j)⟺i=1,j=2 ○ Thus τ(Δ)=−Δ ○ So ϵ^′ (τ)=1 ̅ • Proof: Suppose τ=(i j), 1≤i j≤n ○ Let λ∈S_n denote the following permutation § λ(1)=i § λ(2)=j § λ(i)=1 § λ(j)=2 § λ(k)=k,k∉{1,2,i,j} ○ (i j)=λ(1 2)λ § [λ(1 2)λ](i)=[λ(1 2)](1)=λ(2)=j § [λ(1 2)λ](j)=[λ(1 2)](2)=λ(1)=i § Without loss of generality, assume i,j∉{1,2} § [λ(1 2)λ](1)=[λ(1 2)](i)=λ(i)=1 § [λ(1 2)λ](2)=[λ(1 2)](j)=λ(j)=2 § For k∉{1,2,i,j} § [λ(1 2)λ](k)=[λ(1 2)](k)=λ(k)=k ○ We know ϵ′ is a homomorphism, so § ϵ^′ (i j)=ϵ^′ (λ(1 2)λ) § =ϵ^′ (λ)+ϵ^′ (1 2)+ϵ^′ (λ) § =2ϵ^′ (λ)+1 ̅ § =0 ̅+1 ̅=1 ̅ Corollary 41 • Statement ○ ϵ is well-defined, and ϵ=ϵ^′ • Proof ○ Let σ∈S_n ○ Say σ=τ_1⋯τ_n where τ_i is a transposition, then ○ ϵ^′ (σ)=ϵ^′ (τ_1 )+…+ϵ^′ (τ_n )=⏟(1 ̅+⋯+1 ̅ )┬(n copies)=n ̅ ○ Thus if n is odd, σ cannot be written as a product of an even number of transpositions, and vice versa ○ This shows ϵ is well-defined, and ϵ=ϵ^′ Corollary 42 • Statement ○ If n≥2, ϵ is surjective • Proof ○ ϵ(1)=0 ̅, and ϵ(1 2)=1 ̅ ○ Since Z\2Z has only 2 elements, ϵ is surjective Alternating Group • Definition ○ The alternative group, denoted as A_n is the kernel of ϵ ○ That is, A_n contains of all even permutations in S_n • Order of A_n ○ By the First Isomorphism Theorem ○ We have an isomorphism S_n \A_n≅Z\2Z ○ By Lagrange s Theorem, |A_n |[S_n:A_n ]=|S_n | ○ ⇒|A_n |=|S_n |/[S_n:A_n ] =n!/2 • Note ○ We showed earlier that, if (a_1…a_t )∈S_n, ○ (a_1…a_t )=⏟((a_1 a_t )(a_1 a_(t−1) )⋯(a_1 a_2 ) )┬(t−1 terms) ○ t\cycle is even when t is odd, and vise versa ○ Thus, (a_1…a_t )∈A_n⟺t is odd • Examples ○ A_2= trivial group ○ A_3={(1), (1 2 3),(1 3 2)}=⟨(1 2 3)⟩ ○ A_4= {(1), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} • Subgroups of A_4 Order Subgroup 1 {(1)} 2 3 cyclic subgroups 3 4 cyclic subgroups 4 Homework 6 None 12 A_4 Converse of Lagrange s Theorem • A_4 has no subgroup of order 6 • This shows that the converse of Lagrange s Theorem is false ○ If ├ d┤|├ |G|┤, there is not necessarily a subgroup of G with order d • But the converse does hold for finite cyclic groups • Cauchy s Theorem ○ If p is a prime, and ├ p┤|├ |G|┤, then G contains a subgroup of order p • Sylow s Theorem ○ If |G|=p^α m, where p is prime and p∤m ○ Then G contains a subgroup of order p^α
