资源描述:
《一类非富足全变换半群.pdf》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、学校代码:10663学号:4201210000328贵贵贵州州州师师师范范范大大大学学学硕硕硕士士士学学学位位位论论论文文文一一一类类类非非非富富富足足足全全全变变变换换换半半半群群群Onnon-abundanceofcertainsemigroupsoffinitefulltransformations专业名称:基础数学专业代码:070101研究方向:半群代数申请人姓名:李旺威导师姓名:徐波(教授)二零一五年四月二十日目录摘摘摘要要要································································
2、······························IAbstract······························································································III1Introduction······················································································12Preliminaries··································
3、···················································33Green’sstarrelationsandabundance··················································83.1?−(?)isnon-abundant···································································8?3.2Green’srelationsandGreen’sstarrelationsof??−.·········
4、·····················10?4Regularpartof??−··········································································17?5Therankof??−················································································21?References····························································
5、·······························24Appendix·····························································································26Acknowledgements···············································································27DeclarationofOriginality···························
6、··········································281摘要设?−是全变换半群上的降序子半群.假设?−−??(?)={?∈??:(∀?∈?)??},其中?⊆?∖{1},那么?−(?)是?−的子半群.我们在此证明了?−(?)是非富足半群,其????中?̸=?和?̸={?}.同时,我们考虑一个特殊情形,即?是偶数,且?={?∈??:2
7、?},记为??−.我们考察了它的格林关系及其广义格林关系,正则子半群和秩.主要内容如?下:第二部分,我们介绍了?−(?)和??−的概念.我们证明了??−中的元素除了幂等???元外都是非正则元.同时
8、,我们给出了??−中的幂零元基数和??−的基数。??引引引理理理2.1.半群?−(?)是R-平凡的.?命命命题题题2.10.若?=2?,则∏︁?−22
9、???
10、=ℎ=[(2?−1)!!];ℎ=1,2-ℎ进而,
11、?−
12、(2?)!!√?=s??(?→∞).
13、???−
14、(2?−1)!!第三部分,我们刻画了?−(?)(??−)的格林关系及它们的推广形式,进而证明?−(?)???(??−(?≥4))是非富足半群.?引引引理理理3.1.2.设?,?∈?−(?).?(1)若2∈?,(?,?)∈L*⇐⇒??(?)∖{1,2}=??(?)∖{1,2}.(2)若2∈/?,(?,?)
15、∈L*⇐⇒??(?)=?