
則β,γ∈J2且α=βγ.
引理2對2≤k≤r-1, 2≤r≤n-1, 有Jk?Jk+1·Jk+1.
證明: 對任意的α∈Jk, 設(shè)α的標準表示為
這里每個Ai(i=1,2,…,k-1,k)都是凸集, 并且A1A2>…>Ak-1>Ak,a1由于2≤k≤r-1≤n-2, 因此必存在i∈{1,2,…,k-1,k}, 使得Ai≥2. 若α是保序的, 則記x=minAi; 若α是反序的, 則記x=maxAi. 下面分3種情形證明存在β,γ∈Jk+1, 使得α=βγ.
情形1) 若a1≠1, 令
則β,γ∈Jk+1且α=βγ.
情形2) 若ak≠n, 令
則β,γ∈Jk+1且α=βγ.
情形3) 若a1=1且ak=n, 結(jié)合2≤k≤n-2知, 存在j∈{2,3,…,k-1,k}, 使得aj-aj-1>1.
① 如果i則β,γ∈Jk+1且α=βγ.
② 如果i=j, 令
則β,γ∈Jk+1且α=βγ.
③ 如果i>j, 令
則β,γ∈Jk+1且α=βγ.
2.1 定理1的證明
由引理1和引理2可知, 對任意的α∈LD(n,r)都可以表示為LD(n,r)的頂端J-類Jr中秩為r的元素的乘積或α∈Jr. 即Jr是LD(n,r)的生成集,LD(n,r)=〈Jr〉.
引理3設(shè)α,β∈LD(n,r), 若(α,β),(α,αβ)∈J, 則(αβ,β)∈L, (α,αβ)∈R.
證明: 設(shè)α,β∈LD(n,r), 若(α,β),(α,αβ)∈J, 則Imα=Imβ=Im(αβ). 再由Im(αβ)?Imβ, Kerα?Ker(αβ)與Xn的有限性知, Im(αβ)=Imβ, Kerα=Ker(αβ), 即(αβ,β)∈L, (α,αβ)∈R.


注意到當(dāng)r=1時, J1中共有n個L-類和1個R-類, 且每個H=R∩L僅有一個保序的元素, 因此, 有:
推論2設(shè)自然數(shù)n≥3, 則rank(LD(n,1))=n.

2) 這m個冪等元都是保序變換.



其次, 對任意的α∈Jr, 分兩種情形驗證α∈〈M〉, 即Jr?〈M〉.
1) 若存在i,j∈{1,2,…,m-1,m}, 使得Kerα=Kerαi, Imα=Imαj.
① 若α是保序的, 則當(dāng)iα=αiαi+1…αm-1αmα1α2α3…αi-1αi…αm-1αm;
當(dāng)i=j=m時, 有α=αmα1α2α3…αm-1αm; 當(dāng)i=jj時, 有
α=αiαi+1…αm-1αmα1α2α3…αi-1αiαi+1…αm-1αmα1α2…αj-1αj.
② 若α是反序的, 則當(dāng)iα=αiαi+1…αj-1αj…αm-1αmα1α2…αi-1αiαi+1…αj-1αj;
當(dāng)iα=αiαi+1…αm-1αmα1α2…αi-1αi;
當(dāng)i>j時, 有α=αiαi+1…αm-1αmα1α2…αj-1αj.
① 若α是保序的, 則當(dāng)j=1時, 有β=βi; 當(dāng)2≤j≤m時, 有
α=αjαj+1…αm-1αmα1α2…αj-1αjαj+1…αm-1αmβi.
② 若α是反序的, 則當(dāng)1≤j≤m時, 有α=αjαj+1…αm-1αmβi.
2.2 定理2的證明

2.3 定理3的證明
當(dāng)1≤l
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