(Definition)设α∈Sn,且α=∏i=1tβi(βi∈Sn)是分解为不相交轮换的完全轮换分解:定义sgn(α)=(−1)n−t.(Definition)设α∈Sn,且α=∏i=1tβi(βi∈Sn)是分解为不相交轮换的完全轮换分解:定义sgn(α)=(−1)n−t.
(Theorem)∀α,β∈Sn,sgn(αβ)=sgn(α)sgn(β).(Theorem)∀α,β∈Sn,sgn(αβ)=sgn(α)sgn(β).
(Theorem)α是偶置换,iff.sgn(α)=1,否则α是奇置换.(Theorem)α是偶置换,iff.sgn(α)=1,否则α是奇置换.
(Proposition 2.2)α∈Sn,则sgn(α−1)=sgn(α)(Proposition 2.2)α∈Sn,则sgn(α−1)=sgn(α)
证明:由完全轮换分解定理α−1=(∏i=1tβi)−1=∏i=t1β−1i(βi∈Sn)α−1=(∏i=1tβi)−1=∏i=t1βi−1(βi∈Sn),所以sgn(α−1)=(−1)n−t=sgn(α)sgn(α−1)=(−1)n−t=sgn(α)
(Proposition 2.3)σ∈Sn,定义σ′∈Sn−1: σ′(i)=σ(i)对于一切σ所不固定的i∈[1,n],证明sgn(σ′)=sgn(σ).(Proposition 2.3)σ∈Sn,定义σ′∈Sn−1: σ′(i)=σ(i)对于一切σ所不固定的i∈[1,n],证明sgn(σ′)=sgn(σ).
证明:不妨设σ=αβσ=αβ,αα为所有不固定的轮换之积.则sgn(σ′)=sgn(αβ′)=(−1)n−a1b′=(−1)n−a1b=sgn(σ).sgn(σ′)=sgn(αβ′)=(−1)n−a1b′=(−1)n−a1b=sgn(σ).
(Proposition 2.5)(Proposition 2.5)
(i)(ii)α是r−轮换,证明αr=(1);证明(i)中r是满足的最小正整数.(i)α是r−轮换,证明αr=(1);(ii)证明(i)中r是满足的最小正整数.
证明:
(i)(i)设α=(i0…ir−1):α=(i0…ir−1):
basic step:α1(ik¯¯¯)=ik+1¯¯¯¯¯¯¯¯¯¯(modr).basic step:α1(ik¯)=ik+1¯(modr).
recuisive step:αn(ik¯