下载文档原格式
The 14th National Conference on Algebra
第十四届
全国代数学学术会议
会议程序
扬州大学主办
中国扬州
2016年5月26日——5月31日
注意：各报告场地以本手册为准。
一、会议日程
二、会议日程简表
三、会议日程详单
5月26日（星期四）扬州会议中心报到
5月27日（星期五）开幕式与大会报告地点：杏园楼百畅厅
5月28日（星期六）分组报告(第一组) 地点：群贤楼会议室二
5月28日（星期六）分组报告(第二组) 地点：群贤楼会议室三
5月28日（星期六）分组报告(第三组) 地点：群贤楼贵宾室一
5月28日（星期六）分组报告(第四组) 地点：群贤楼贵宾室二
5月28日（星期六）分组报告(第五组) 地点：群贤楼多功能厅
5月28日（星期六）分组报告(第六组) 地点：群贤楼贵宾室四
5月28日（星期六）分组报告(第七组) 地点：群贤楼会议室六
5月28日（星期六）分组报告(第八组) 地点：群贤楼会议室七
5月29日（星期日上午）分组报告(第二组) 地点：群贤楼会议室三
5月29日（星期日上午）分组报告(第四组) 地点：群贤楼贵宾室二
5月29日（星期日上午）分组报告(第六组) 地点：群贤楼贵宾室四
5月29日（星期日下午）邀请报告(第二组) 地点：群贤楼贵宾室二
5月29日（星期日下午）邀请报告(第三组) 地点：群贤楼贵宾室四
5月30日（星期一）大会报告与闭幕式地点：春晖楼春台庆禧厅。
Difference sets in groups containing subgroups of index2Wai Chee ShiuDepartment of Mathematics,Hong Kong Baptist University,224Waterloo Road,Kowloon,Hong KongE-mail:wcshiu@math.hkbu.hkAbstractIn this paper,difference sets in groups containing subgroups of index2are considered, especially group of order2m where m is odd.The author shows that the only difference setsin groups of order2pαare trivial.The same conclusion is true for some special parameters.1IntroductionLet G be afinite(multiplicative)group of order v 2with identity e.Let D⊆G.We define D(t)=d∈Dd t as an element of the group ring Z[G],where t∈Z,and D=D(1).A(v,k,λ) -difference set D in G is a subset of G such that DD(−1)=ne+λG,where n=k−λ[11,section3].A difference set is a combinatorial object which can be used to construct a balanced incomplete block design[1].Difference sets in groups have been studied by many authors(see,for instance, [3]).Most of them are concerned on abelian difference sets,especially cyclic different sets(see,for instance,[1],[8],[12][13]).Recently,Fan,Siu and Ma(1985)[6,section4]studied difference sets in dihedral groups,which is a non-abelian case.In this paper,the author is concerned difference sets in groups which obtain subgroups of index2.There are some constraints on the parameters v,k,λof a difference set.Two common con-straints are[1,section1]:(1)k(k−1)=λ(v−1)or equivalently,k2=n+λu(1.1)(2)If v is even,then n is a square.From(1.1)it is easy to see thatλis even.When n=0it is easy to obtain that k=0or k=v;when n=1it is easy to see that k=1 or k=v−1.These two cases are called trivial cases.The corresponding difference sets are called trivial difference sets.It is known that if D is a(v,k,λ)-difference set in G then G\D is a(v,v−k,v−2k+λ)-difference set in G.So without loss of generality we may assume k<v2((1.1)shows that k=v2does not occur).Fan,Siu and Ma[6,section4]proved that only trivial difference sets in dihedral group of order2m exist if(m,n)=1and proved that only trivial difference sets in dihedral group oforder2pαexist,where p is an odd prime.They also showed that if D=A∪σB is a(2m,k,λ)-difference set in D m= ρ,σ|ρm=σ2=e=ρσρσ ,the dihedral group of order2m,then D ={ϕ2i|ρi∈A}∪{ϕ2i+1|ρj∈B}is a(2m,k,λ)-difference set in C2m= ϕ|ϕ2m=e ,the cyclic group of order2m.Since no difference sets in a cyclic group of order v are known which have(v,n)>1(or equivalently(v,k)>1).Fan et al.conjectured that there is no non-trivial difference set in a dihedral group.Remark 1.1:If D is above then D =A∪τB is a(2m,k,λ)-difference set in C m×C2= ρ,τ|ρm=τ2=e,ρτ=τρ .In this paper we generalize the result above.We show that there are only trivial difference sets in groups of order2pαwhere p is an odd prime.2General PropertiesIn this section we let G be a group of order2m which contains a(normal)subgroup H of index 2.Then G=H∪σH for someσ∈G\H.Henceσ2=h−1orσh=σ−1for some h∈H.Suppose D=A∪σB is a(2m,k,2λ0)-difference set in G where A,B⊆H.We have,AA(−1)+˜B˜B(−1)=ne+2λ0H(2.1)h˜A˜B(−1)+BA(−1)=2λ0HorhσAB(−1)σ−1+BA(−1)=2λ0H(2.2) where˜A=σAσ−1and˜B=σBσ−1Lemma 2.1.If D=A∪σB is a(2m,k,2λ0)-difference set in G as described,thenσD= h−1B∪σA is a(2m,k,2λ0)-difference set in G.Proof:SinceσD is a shift of DSuppose D is as above.Let k1=|A|and k2=|B|then k1=|˜A|and k2=|˜B|=|h−1B|. Clearly k1,k2≤k<m.By Lemma2.1,from now on,we assume m>k1≥k2.For convenience, the difference set D=A∪σB as above is called a(2m,k1&k2,2λ0)-difference set in G=H∪σH. Note that if D is a(2m,k,2λ0)-difference set in G=H∪σH,then D is a(2m,k1&k2,2λ0)-difference set in G for some k1,k2andλ0.By applying the trivial character of H on(2.1)and(2.2)we havek21+k22=n+2λ0m(2.3)k1k2=λ0m,(2.4) where n=k−2λ0and k=k1+k2.From(2.3)and(2.4)we haven=(k1−k2)2.(2.5) Examples2.2:(1)The dihedral group D m has a subgroup,which is isomorphic to C m,of index2.(2)The symmetric group S m of order m!has a subgroup A m,the alternating group,of index2.(3)Let G be a group of order2m,where m is odd.Since G is even order,there isσ∈G such thatσ2=e.Consider G L,the group of left translations of G,it contains an odd permutation induced fromσ.Thus G L contains a subgroup of index2.Since G L∼=G,G contains a subgroup H of index2.In this case G=H∪σH whereσ2=e.(4)Let G be an abelian group of even order v.By Sylow Theorem[7,Theorem2.12.1],if2γ vthen there is a subgroup H of G having order2γ−1whereγ≥1and for any odd prime factors of v there is a(unique)Sylow p-subgroup of G.Thus G contains a subgroup of index2.Note that not every group of even order contains a subgroup of index2,for example A mwhen m≥4.Let a be a nonzero integer.By v p(a)we denote the power to which prime p enters in the factorization of a into prime factors.For convenience,we set v p(0)=∞[2,p.175].Theorem2.3.If integers k1,k2,λ0,m and n satisfy(2.3),(2.4)and(2.5)where n=k−2λ0and k=k1+k2then v p(k1)≥v p(m)or v p(k2)≥v p(m)for p is an odd prime and v2(k1)+1≥v2(m) or v2(k2)+1≥v2(m).Proof:For p=2,from(2.4)and(2.5)we havev p(m)=v p(k1)+v p(k2)−v p(2λ0)(2.6)v p(n)≥2(min{v p(k1),v p(k2)}).(2.7) Since n=k−2λ0and k=k1+k2and(2.7),v p(2λ0)≥(min{v p(k),v p(n)})≥(min{v p(k1),v p(k2),v p(n)})=(min{v p(k1),v p(k2)}).From(2.6)we have v p(m)≤v p(k1)+v p(k2)−min{v p(k1),v p(k2)}.The lemma holds for p is an odd prime.By a similar proof,the lemma holds for p=2.Corollary 2.4.There are only trivial difference sets in groups of order 2p αwhere p is an odd prime.Proof:The theorem follows from Example 2.2(3)and Theorem 2.3.When p =2there are many results on difference sets in 2-groups.Dillon [5]had classified all groups of order 16containing nontrivial difference sets.In his paper,he showed that there is no nontrivial difference in D 8.Leibler and Smith [10]and Davis and Smith [4]have some constructions of difference sets in high exponent 2-groups.Corollary 2.5.Suppose p and q are distinct odd primes.If there is a (2pq,k 1&k 2,2λ0)-difference set then (pq,(k 1−k 2)2)=1.Proof:By Theorem 2.3and k 2≤k 1<pq ,we may assume v p (k 1)≥1and v q (k 2)≥1.Then q |k 1and p |k 2.By (2.5)we have (pq,(k 1−k 2)2)=1. By Theorem 2.4we see that there are only trivial difference sets in dihedral group of order 2p αexist where p is an odd prime.From the Theorem 4.1in [6]and Corollary 2.5we haveCorollary 2.6.There are only trivial difference sets in D pq where p and q are distinct odd primes.Our results apply to groups containing subgroup of index 2.Of course for certain specific groups,we may use specialized techniques exploiting properties not assume here.For example,our results do not say anything about (66,26,10),(70,24,8)and (154,18,2)-difference sets in general groups.However,if we consider the abelian case then it can be found in the Lander’s table [9]that there are only trivial (66,26,10),(70,24,8)and (154,18,2)-abelian difference sets.For the dihedral groups,Corollary 2.6shows that there are only trivial (66,26,10),(70,24,8)and (154,18,2)-difference sets in dihedral groups.Let us go back to discuss the parameters of the (2m,k 1&k 2,2λ0)-difference sets.From (2.3)and (2.4)we havek 21+k 22−2k 1k 2−k 1−k 2+2λ0=0(k 1≥k 2)(2.8)⇔k 1=2k 2+1+√8k 2+1−8λ02⇔k 2=2k 1+1−√8k 1+1−8λ02Suppose x 0,x 1∈N ,x 0<x 1,satisfy (2.8).Let x j +1=2x j +1+√8x j +1−8λ02for j ≥0.It is easy to see that x j +1−x j =1+x j −x j −1for j ≥1.Similarly,suppose y 0,y 1∈N ,y 0>y 1,satisfy (2.8).Let y j +1=2y j +1−√8y j +1−8λ02for j ≥0and 8y j +1−8λ0≥0.We also have 1+y j −y j +1=y j −1−y j for j ≥1.Finally,it is easy to see that if k 1,k 2∈N satisfy (2.8)and k 1−k 2=1then k 2=x j −bine the discussions above we haveLemma2.7.Let x j=j(j+1)2+λ0,j≥0.If k1,k2∈N satisfy(2.8)then k1=x j,k2=x j−1for some j≥1.Remark2.8:From Lemma2.7,there is only one set of parameters,namely(2(j4−j24λ0+j2+λ0),j(j+1)2+λ0&(j−1)j2+λ0,2λ0),according to the givenλ0≥1and n=k1+k2−2λ0=j2≥1.Theorem2.9.Only(2(j4−j24λ0+j2+λ0),j(j+1)2+λ0&(j−1)j2+λ0,2λ0)-difference set may exist ifj4−j24≡0(modλ0),j≥1.Proof:it follows from(2.4).Let us consider two special cases,namelyλ0=1or2Case1:Forλ0=1.Suppose m is odd andλ0=1.From(2.4)we have k1k2=m.Thus k1,k2are odd.From(2.3) we have n≡0and k≡2(mod4).Thus k1≡k2(mod4)and m≡1(mod4).Hence we have the following theorem.Theorem2.10.There are only trivial(2m,k1&k2,2)-difference sets if m≡3(mod4).Lemma2.11.Suppose m is odd.If there is a(2m,k1&k2,2)-difference set then(m,(k1−k2)2)= 1.Proof:By Theorem2.3and(2.4)whenλ0=1we have k1=P and k2=Q,where m=P Q, (P,Q)=1.Thus(m,(k1−k2)2)=1.Suppose m is odd.By Lemma2.11and Theorem4.1in[6]we haveTheorem2.12.There are no non-trivial(2m,k1&k2,2)-difference sets in D m where m is odd. Case2:Forλ0=2.Theorem2.13.There is no non-trivial(2m,k1&k2,4)-difference set when n=(k1−k2)2≡4 (mod8).Proof:j4−j24≡0(mod2)⇔j4−j2≡0(mod8).The equation does not hold when j≡2 (mod4).Since n=j2,then the equation does not hold when n≡4(mod8).Forλ0=1or2,consider thefirst ten terms of the sequence defined in Lemma2.7.We can prove that there are only trivial difference sets in D m.Among these cases,some of them can be proved by applying Theorem4.1in[6]and the rest which indicated by“† are proved by Corollary 3in[11].We list them as follows:Forλ0=1,the sequence is{1,2,4,7,11,16,22,29,37,46,56,...}.According to thefirst ten terms of this sequence we haven=k−λ14†9162536†496481100†k21247111622293746k124711162229374656k3611182738516683102m282877176352638107317022576(m,n)1411121114Forλ0=2,the sequence is{2,3,5,8,12,17,23,30,38,47,57,...}.According to thefirst ten terms of this sequence we haven=k−λ14916†25364964†81100k22358121723303847k135812172330384757k5813202940536885104m3-2048102-345570893-(m,n)1-1161-121-References[1]L.D.Baumert,Cyclic Difference Sets,Springer-Verlag,New York,1971.[2]Z.I.Borevich and I.R.Shafarevich,Number Theory,Academic Press,New York,1966.[3]R.H Bruck,Difference sets in afinite group,Trans.Amer.Math.Soc.,78(1955),464−481.[4]Davis and Smith,A construction of difference sets in high exponent2-groups using represen-tation theory,to appear in J.Algebraic Combinatorics.[5]J.F.Dillon,Varations on a scheme of McFarland for noncyclic difference sets,b.TheorySer.A,40(1985),9−21.[6]C.T.Fam,M.K.Siu and S.L.Ma,Difference sets in dihedral groups and interlocking differencesets,Ars Combinatoria,20A(1985),99−107.[7]I.N.Herstein,Topics in Algebra,2nd ed.,John Wiley&Sons,1975.[8]E.C.Johnsen,The inverse multiplier for abelian group difference sets,Canad.J.Math.,16(1964),787−796.[9]nder,Symmetric Designs:An Algebraic Approach,Cambridge Univ.Press,1983.[10]Leibeler and Smith,On difference sets in certain2-groups,Coding Theory,Design Theory,Group Theory:Proceedings of the Marshall Hall Conference,John Wiley,1992.[11]K.H.Leung,S.L.Ma and Y.L.Wong,Difference sets in dihedral groups,Designs,Codes andCryptograph,1(1992),333−338.[12]P.K.Menon,Difference sets in Abelian Groups,Proc.American Math.Soc.,11(1960),368−376.[13]E.Spence,A family of difference sets,b.Theory,Ser.A,22(1977),103−106.。
抽象代数⼀、课程⽬的与教学基本要求本课程是在学⽣已学习⼤学⼀年级“⼏何与代数”必修课的基础上，进⼀步学习群、环、域三个基本的抽象的代数结构。
要求学⽣牢固掌握关于这三种抽象的代数结构的基本事实、结果、例⼦。
对这三种代数结构在别的相关学科，如数论、物理学等的应⽤有⼀般了解。
⼆、课程内容第1章准备知识（Things Familiar and Less Familiar）10课时复习集合论、集合间映射及数学归纳法知识，通过学习集合间映射为继续学习群论打基础。
1、⼏个注记(A Few Preliminary Remarks)2、集论(Set Theory)3、映射(Mappings)4、A（S）(The Set of 1-1 Mappings of S onto Itself)5、整数(The Integers)6、数学归纳法(Mathematical Induction)7、复数(Complex Numbers)第2章群(Groups) 22课时建⽴关于群、⼦群、商群及直积的基本概念及基本性质；通过实例帮助建⽴抽象概念，掌握群同态定理及其应⽤；了解有限阿贝尔群的结构。
1、群的定义和例⼦(Definitions and Examples of Groups)2、⼀些简单注记(Some Simple Remarks)3、⼦群(Subgroups)4、拉格朗⽇定理(Lagrange’s Theorem)5、同态与正规⼦群(Homomorphisms and Normal Subgroups)6、商群(Factor Groups)7、同态定理(The Homomorphism Theorems)8、柯西定理(Cauchy’s Theorem)9、直积(Direct Products)10、有限阿贝尔群(Finite Abelian Groups) （选讲）11、共轭与西罗定理(Conjugacy and Sylow’s Theorem)(选讲)第3章对称群(The Symmetric Group) 8课时掌握对称群的结构定理，了解单群的概念及例⼦。
第38卷第4期西南师范大学学报(自然科学版)2013年4月V o l.38N o.4J o u r n a l o f S o u t h w e s t C h i n aN o r m a lU n i v e r s i t y(N a t u r a l S c i e n c eE d i t i o n)A p r.2013文章编号:10005471(2013)04014804G A P 在近世代数教学中的应用①刘建军西南大学数学与统计学院,重庆400715摘要:针对近世代数课程内容比较抽象的特点,以及当前近世代数教学中过多注重理论体系的完整性,忽视学生作为主体地位的现状,介绍了适合近世代数教学的软件G A P及其在教学中的具体运用.通过G A P的使用,可以变革近世代数内容的呈现方式,使得证明清晰化,计算简单化,以及结论直观化,从而使近世代数学习变得更生动㊁更形象㊁更具体,以促进近世代数教学质量的提高.关键词:近世代数;教学;数学软件中图分类号:G420文献标志码:A近世代数(又名抽象代数)是以研究代数系统的性质与构造为中心的一门学科,是现代数学的重要基础,对培养学生严谨的思维方法和数学素养,提高学生的抽象思维能力和逻辑推理能力都具有重要意义.本文试图利用数学软件G A P来辅助近世代数的教学,变革教学内容的呈现方式,使其变得形象直观,达到激发学生的学习兴趣,提高逻辑思维能力,开阔视野的目的.1国内近世代数课程的教学现状近世代数的研究对象是群㊁环㊁域等带有运算的集合,它把集合中运算的共同点抽象出来作为不同的代数结构进行研究,因此近世代数具有高度的抽象性和严密的逻辑性,许多初学者感到这门课程生涩难懂,不具体直观.在我国,一般的近世代数教学都是教师按照教学大纲的要求,对定义㊁引理㊁定理等在课堂上给学生进行理论上的推导和计算,直到学生们理解并记忆下来为止.这种以教师讲授为主的教学方式在传授系统知识时具有比较好的效果,但过多偏重理论体系的完整性,过多强调证明和推理,忽视了学生作为主体的地位,不利于培养学生主动获取知识的能力,使学生缺乏创新能力.因此,学生很难具备用近世代数的基本思想和理论来处理或解决具体问题的能力,从而直接影响了后继课程学习的热情[1-4].随着计算机技术的迅速发展,利用计算机软件(如M a t l a b,M a t h e m a t i c a,M a p l e等)来辅助各门课程的教学已经非常普遍(参见文献[5-9]).然而,这些软件很少被应用到近世代数的教学中.究其原因主要是这门课程研究的对象较抽象,在一般的软件上难以实现.2G A P介绍G A P(G r o u p s,A l g o r i t h m s a n dP r o g r a mm i n g)和M a g m a的出现可以说是一场革命,它们实现了抽象对象的计算机化.由于M a g m a的使用需要收取一定的费用,不是很普及,所以我们这里只介绍可以从其官方网站免费下载的G A P.G A P于1986年由德国RWT H A a c h e n大学的研究团队开发,它是计算离散代数领域内的一个优秀系统,主要专注于计算群论的计算.G A P提供了上千个由G A P语言写成的用于算法①收稿日期:20111001Copyright©博看网. All Rights Reserved.基金项目:中央基本科研业务费(X D J K2012C039);西南大学博士科研启动基金(S WU111052).作者简介:刘建军(1981),男,山西吕梁人,博士,讲师,主要从事有限群论的研究.补充方面的函数库,以及已经计算好的庞大的代数对象数据库.G A P 的软件系统是可扩展的,它支持面向对象的编程,用户可以使用G A P 语言编写需要的程序和建立自己的函数库.G A P 可用于群及其表示㊁环㊁向量空间㊁代数㊁组合结构等的研究.它还有如下的特点:拥有多种数学运算功能,内存自动管理,对一些关键的抽象对象加入了嵌入式数据类型,有灵活的菜单管理和完整的记录保存.国内已有很多学者在群论的研究中用到了G A P (参见文献[10-11]).因此,G A P 比较适合应用于近世代数的教学当中.3 G A P 在近世代数教学中的一些应用我们可以借助G A P 对近世代数的相关内容作较为直观的认识.另外,G A P 给出的结果反过来可以指导我们研究的方向,并能极大地减轻计算负担.下面我们介绍G A P 在近世代数教学中的一些应用.3.1 将繁琐证明清晰化近世代数的很多证明非常繁琐和抽象,可以用G A P 来具体说明,使得某些证明更加清晰和顺理成章.例1 证明凡200阶群都不是单群[1].解析 主要应用S y l o w 第三定理来证明.S y l o w 定理是近世代数教学中的一个难点.学生要理解它的具体用法比较难.课本给出的证明是:判断阶为200的群必然有正规子群.在这个过程中会有阶的分解㊁同余等知识的灵活应用,理解起来有一定的困难,而G A P 的使用可以很好地解决这个问题,降低学生理解的难度.下面我们用G A P 来具体说明.输入程序:g a p >A :=A l l S m a l l G r o u p s (S i z e ,200,I s S i m p l e ,t r u e ); #直接寻找200阶的单群输出:[]这表明不存在这样的群.如果我们想了解更多200阶群的信息,也可以将所有200阶群的结构输出,只需输入命令:g a p >B :=A l l S m a l l G r o u p s (200);L i s t (B ,S t r u c t u r e D e s c r i p t i o n );3.2 将复杂计算简单化几乎所有的数学软件都有非常好的计算功能,G A P 也不例外.在近世代数中出现的某些计算,使用G A P 之后会变得非常简单.3.2.1 共轭类中的G A P 计算由于有限群G 中与它的元素a 共轭的元素个数为|G ʒC G (a )|,因此要计算某个群中元素的共轭类需要计算它的中心化子.当群G 比较大的时候这个计算量将非常的大,但是用G A P 计算就显得非常容易了.例2 计算5次对称群S 5的所有共轭类的代表元及每类所含元素的个数.输入程序:g a p >G :=S y mm e t r i c G r o u p (5); #5次对称群g a p >c l :=C o n j u g a c y C l a s s e s (G );#S 5的共轭类输出:[()ɡG ,(1,2)ɡG ,(1,2)(3,4)ɡG ,(1,2,3)ɡG ,(1,2,3)(4,5)ɡG ,(1,2,3,4)ɡG ,(1,2,3,4,5)ɡG ]g a p >L i s t (c l ,R e p r e s e n t a t i v e );#共轭类代表元输出:[(),(1,2),(1,2)(3,4),(1,2,3),(1,2,3)(4,5),(1,2,3,4),(1,2,3,4,5)]g a p >L i s t (c l ,S i z e );#每个类所含的个数输出:[1,10,15,20,20,30,24]3.2.2 元素相乘及元素统计的G A P 计算例2中的S 5包含有120个元素,若要验证每个元素之间的性质,计算量将比较大.下面的两个例子将更能体现G A P 在计算方面的优势.例3 找一个S 5中的元,与给定的5阶元相乘,得另一个给定的5阶元.输入程序:g a p >G :=S y mm e t r i c G r o u p(5);251西南师范大学学报(自然科学版) h t t p ://x b b jb .s w u .c n 第38卷Copyright ©博看网. All Rights Reserved.g a p >a :=(1,4,3,2,5);b :=(1,2,3)(5,4);g a p >f o r g i nGd o #使用循环语句>i f g *b =a t h e n >P r i n t (g , , );>f i ;>o d ;输出:(1,2,4)(3,5)下面我们介绍如何用G A P 来统计满足某些性质的元素的个数.例4 找某个群的固定阶元并统计个数.输入程序:g a p >G :=G r o u p ((1,2),(1,2,3,4,5,6));g a p >i :=0;A :=[];g a p >f o r a i nGd o >i fO r d e r (a )=2t h e n >A d d (A ,a );>i :=i +1;>f i ;>o d ;g a p >A ;#可以得到群G 的每个2阶元g a p >i ;#群G 的所有2阶元的个数输出:753.3 将抽象结论直观化近世代数中抽象内容多,具体例子少,命题㊁定理多,推理㊁论证各式各样,使得学生较难理解和掌握[12].与抽象的内容相比,学生更易于识记生动㊁形象和有趣的知识和结论.G A P 强大的功能可以做到这一点.例5 4次交错群A 4无6阶子群.解析 在近世代数中这是一个非常基本的结论,常常会在构造反例及基本证明中用到,但是很难让学生有深刻的体会.下面我们用G A P 来具体说明.输入程序:g a p >L o a d P a c k a g e ( s o n a t a );#调用 s o n a t a ,为了使用命令s u b g r o u p s g a p >G :=A l t e r n a t i n g G r o u p (4);#4次交错群g a p >S :=S u b g r o u p s (G );g a p >L i s t (S ,S i z e );#列出所有子群的阶输出:[1,2,2,2,3,3,3,3,4,12]g a p >L i s t (S ,S t r u c t u r e D e s c r i p t i o n );#列出所有子群的结构输出:[ 1 , C 2 , C 2 , C 2 , C 3 , C 3 , C 3 , C 3 , C 2ˑC 2 , A 4 ]其中C 2表示2阶循环群,C 2ˑC 2表示两个2阶循环群的直积.经过软件G A P 的运算之后,4次交错群A 4的所有子群摆在了我们的面前,非常直观.4 小 结从上面的例子可以看到,利用G A P 辅助近世代数教学,可以让抽象的数学理论具体化㊁直观化,复杂的计算简单化.而且这种教学不仅是简单的演示,更大程度上能让学生亲身参与,并培养学生解决实际问题的能力.这在一定程度上实现了从被动接受的学习方式到主动发现和探索的过程,而且能增强数学学习兴趣,提高分析和解决问题的能力,最终达到提高教学质量的目的.351第4期 刘建军:G A P 在近世代数教学中的应用Copyright ©博看网. All Rights Reserved.451西南师范大学学报(自然科学版)h t t p://x b b j b.s w u.c n第38卷G A P作为一个辅助教学工具应用于近世代数教学中,主要功能在于协助教师教学,辅助学生学习.毫无疑问,G A P不能替代逻辑推理,如果一味追求其形象直观这一优势而不考虑其它因素,势必将会淡化主体,影响学习效果.因此,在实践中,教师需要根据每个章节具体的教学内容㊁教学目标以及学生的认知水平来决定是否有必要使用G A P.参考文献:[1]杨子胥.近世代数[M].北京:高等教育出版社,2003.[2]赵建伟,罗敏霞.关于近世代数教学的探讨[J].运城学院学报,2008,26(5):1-2.[3]吕恒,徐海静.关于近世代数中群论学习的探讨[J].西南师范大学学报:自然科学版,2012,37(2):131-133.[4]夏静波,邹庭荣,张四兰. 近世代数 的教学技巧[J].大学数学,2009,25(1):5-8.[5]王小华.基于M a t h e m a t i c a的高等数学教学实践[J].重庆科技学院学报:自然科学版,2010,12(4):14-16.[6]鲍四元,孙洪泉,陈旭元.M a t h e m a t i c a在振动波问题中的应用[J].物理与工程,2010,20(4):22-26.[7]堵秀凤,张水胜,李晓红.在大学数学中开设数学实验的实践研究[J].大学数学,2009,25(3):21-25.[8]戚景南,黄玉明.MA T L A B软件在构建潜流人工湿地水力学模型中的应用[J].西南大学学报:自然科学版,2008,30(5):145-148.[9]宋海珍,卢成,张鸿军.基于M a p l e的理论力学教学实践[J].实验室研究与探索,2011,30(7):11-14.[10]B E D A I W I SA,L I S h a n g-z h i.A n I n v e s t i g a t i o no nt h eP a r a b o l i cS u b g r o u p o f t h eG e n e r a lL i n e a rG r o u p sb y G A P[J].C h i n e s eQ u a r t e r l y J o u r n a l o fM a t h,2004,19(3):221-231.[11]B E D A I W I SA,李尚志.某些子群的正则结构的研究及其诱导特征标的计算[J].中国科学技术大学学报,2006,36(7):704-711.[12]李桃生.怎样克服近世代数学习中的困难[J].高等函授学报,1995(5):9-16.O nE x p l o r a t i o n i nT e a c h i n g A b s t r a c t A l g e b r a A s s i s t e dw i t hG A PL I UJ i a n-j u nS c h o o l o fM a t h e m a t i c sa n dS t a t i s t i c s,S o u t h w e s t U n i v e r s i t y,C h o n g q i n g400715,C h i n aA b s t r a c t:I n t h i s p a p e r,G A P,am a t h e m a t i c a l s o f t w a r e,i s i n t r o d u c e d f o r t e a c h i n g o f a b s t r a c t a l g e b r a.B y m e a n so fG A P,t e d i o u s p r o o fw i l l b e c o m e c l e a r,c o m p l e x c a l c u l a t i o n s s i m p l e a n d a b s t r a c t c o n c l u s i o n s i n t u-i t i v e.I t i s g e n e r a l l y b e l i e v e d t h a t t h eu s e o fG A Pw i l l i m p r o v e t h e q u a l i t y o f a b s t r a c t a l g e b r a t e a c h i n g. K e y w o r d s:a b s t r a c t a l g e b r a;t e a c h i n g;m a t h e m a t i c a l s o f t w a r e责任编辑廖坤Copyright©博看网. All Rights Reserved.。
学习抽象代数的体会Learning the experience of abstract algebrain the sophomore year, I have been studying the foundation of Chinese modern algebra, at that time, I have been deeply realize the algebraic abstract and difficult. Because do not understand and dodging and then entirely should try to learn, just learn the teacher should take an examination of the content of the dead to learn hard to remember, forgotten all natural finish test. But now is different, the school and teacher to our requirements are higher, we should study it on the basis of understanding and in-depth exploration. Now to learn is more abstract and more ugly understand all the English abstract algebra, this is undoubtedly a double challenge for me.For the study of abstract algebra, I must overcome the following three problems:The firstly: the English learning (the expression of the accumulation of vocabulary, English).Just got the book of abstract algebra, I scared by it, the UK's mathematics, I was the first time contact. See the back of the book of the appendix, 16 pages of math English vocabulary, I sighed deeply, I know my problem again. My English is very rotten, level 4 test many times or take an examination of, however, poor vocabulary, learn to contain so many mathematics mathematics of proper nouns, it is a huge challenge for me.Before class, I have to take out before the Chinese version of modern algebra to study hardly ,. Then in contrast to review all the abstract algebra, look up the word; After class, remember the keywords, taking the notes. The learning of English, is the precondition of my study abstract algebra.The Secondly to overcome the fear of proving the heart.In preparing for the college entrance examination, many are proving, inner has fear on the certificate, consciously avoid certificate. Now, most are in the form of certificate of abstract algebra, I want to learn it, must overcome the fear.I first put all certificate of modern algebra learned again and again in learn to contact the UK on the basis of abstract algebra, the heart thinks he has learned, clear, naturally less fear, learning to not so nervous, having no fear, the confidence of learning.The finally , to improve the logic reasoning ability.Abstract algebra, it is a logical reasoning ability is very strong discipline, to learn it well, I must constantly improve their logical reasoning ability.In Class, I listen to the teacher carefully and follow the teacher's footsteps, step by step, the launch of the conclusion, sums up the steps of proving slowly, bit by bit of reasoning.More than half of the semester has passed, also have learned more than half of the book, but I'm learning this subject, is still in its adaptation. I deeply know that time and tide wait for no man, I have to take the time to ask classmates, modestly to spend more time to learn to understand, digest it. Looking forward to learning about it not only can harvest the subject taught us logic, also can harvest I hope for the English.Below said this time I learn the feelings of abstract algebra.Unit 1 is the preface. This chapter tells us the concept and a subset of the set in the first quarter, as well as between the set of operations; The second section is the function of the concept and nature; The third section is to introduce a collection of equivalence relation and classification.For me, this chapter is the most difficult of the third quarter of the equivalence relation and the classification of learning.A relation R on a set A is called on equivalence relation , denoted by ~ , if R satisfies the following three properties : first , R is reflexive if aRa for all a∈A . second, R is symmetric if aRb always implies that bRa .third, R is transitive if aRb and bRc imply that aRc.The second unit is group. This chapter has a total of 13 section, but we just learn the front nine section, they are binary operation, definition and examples for a group , elementary properties of groups , subgroups , cyclicgroup ,symmetric group , cosets and lagrange’s theorem , normal subgroups and quotient groups, hommorphisms .I think the focus of this chapter are the following:A group（G ，*）is啊nonempty set G together with a binary operation * on G such that the following conditions hold :first, closure: for all a,b∈G ,we have a*b∈G ;second ,Associativity: for all a,b,c∈G, we have a*(b*c)＝(a*b)*c; third , identity : thee exists an inverse element a﹣1∈G such that a*a﹣1＝eand a﹣1*a＝e. we will usually write ab for theproduct a*b . An equivalent definition con also be redefind as.If a and n are integers q and n is a positive number , then there exist unique itegers q r such that a＝qn＋r and 0≦r≤n ( q and r represent the quotient and remainder, respectively . )If G is a group and x in G , then x is said to be of finite order if there exists a positive integer n such that nx=e .if such an integer exists ,then the smallest positive n such that nx=e is called the order of x and denoted by 0(x) . if x isnot of finite order , then we say that x is of infinite order and write 0(x )=∞.Let G be a group , and let H be a such of G ,we say that H is a subgroup of G ifthe following condition are satisfied: fiert , the identity element of G is an element of H , that is , if e ∈G , then e ∈H . second, the product of any two elements of H is itself an element of H ,that is , for x,y∈H ,then xy∈H. third ,the inverse of any element of H is itself an element of H ,that is , if x ∈H ,then x﹣1∈H.In the next time, I will spend more time to learn, to master English, learn abstract algebra.。
Vol. 45 No. 2Mar. 2021第45卷第2期2021年3月江西师范大学学报(自然科学版)Journal of Jiangxi Normal University (Natural Science)文章编号：1000-5862 （2021） 02-0198-065次对称群S5的子群占颖，张细苟*（江西师范大学数学与统计学院，江西南昌330022）摘要:利用在群论中一些重要的定理以及在"次对称群中的重要知识，该文通过理论推导得到了对称群s 5的所有子群（共156个），并分析了这些子群的结构.关键词:n 次对称群S ” ；子群；Lagrange 定理；Sylow 定理中图分类号：0 152.1 文献标志码:A DOI ： 10.16357/j. cnki. issnlOOO-5862.2021.02.14o 引言在群论中子群是一个非常重要的概念.由有限群的凯莱（Caylay ）定理⑷知，每一个有限群都同构 于一个S ”的子群，所以从理论上来说，若能够找出S ”的所有子群，则对有限群的结构也就了解清楚了•但事实上，当"较大时,要找出S ”的全部子群就 显得比较困难⑵.近年来，文献［3-14］从理论上和 借用电子计算机辅助的方法研究了某些对称群S ”的子群•本文仅利用有限群的Lagrange 定理、Sylow 第一定理以及在群论中的重要结论，通过理论推导 详细阐述了 5次对称群S,的所有子群,并对它们的结构进行较清晰地描述文中弓I 用的符号参见文献［15］.1群论预备知识1.1群论基本知识引理1 （ Lagrange 定理）⑴ 设G 是一个有限群月是G 的子群（记为日<6）,则|6| = \H\IG--H}. 其中［G ：H ］是左陪集或右陪集的个数，|G|表示在 群G 中元素的个数.引理2口］素数阶群必为循环群.引理3（Sylow 定理）⑴ 设G 是一个有限群，| G| =p"m （心1），其中p 为素数，且（p,m ） = 1,则对每个必含有阶为p ‘的子群.定义设G 为群,N 为G 的正规子群（记为NUG ） ,H<G,若 G = NH 且 {e ［,则 G 是 N和H 的半直积，记作G = N X H.12 "次对称群S ”设X= {1,2,…严}，则X 的全体可逆变换关于变换的合成构成了一个n 次对称群S ”.引理4问 "次对称群S ”的阶是“!•13在5次对称群比中的元素由61 =5! =120知,S5中元素的阶数只能是1、2、3、4、5、6.具体分布如下：1） 1个1阶元即单位元（1）;2） 共有25个2阶元，（i ） 单独的2轮换（C ；=10）；（ii ） 2个2轮换乘积的形式（30=15）；3） 共有20个3阶元（2C ； =20）,所有的3-轮换；4） 共有30个4阶元（6以=30）,所有的4-轮换；5） 共有24个5阶元（£/5 =24），所有的5-轮换；6） 共有20个6阶元（2C ； =20）,此6阶元为2 轮换乘以3轮换的形式.2在S5中子群的个数由 | S 5 | =23 x3 x5 = 120 及 Sylow 第一定理知，S 5中一定含有2阶子群、3阶子群、4阶子群、5阶子 群、8阶子群以及平凡子群.由引理1知，结合120所有可能的因子,子群的阶数还可能是6、10、12、 15、20、24、30、40、60.在下面的叙述中，用符号H 們 表示该子群的阶是m 和序号为n.为方便查阅，把本文的所有结果（即S 5的所有子群）均列在附录A 中.收稿日期：2020-07-08基金项目：国家自然科学基金（11961031）资助项目.通信作者:张细苟（1971-），男，江西余干人，副教授，博士，主要从事代数群与Hecke 代数相关研究.E-mail ：xyzhang@ 第2期占颖，等：5次对称群S5的子群199定理1S5有25个2阶子群.证由第1.3节的讨论知,S5有25个2阶子群.如={(1),(13)丨.详细结果见附录A(b).所有的这些2阶子群均是循环子群.定理2S5有10个3阶子群.证类似地,S5有10个3阶子群.如= {(1),(123),(132)}*缪={(i),(i24),(142)}.类似地，可得到其他的子群(见附录A(c)).所有的这些子群也是循环子群.定理3S5有35个4阶子群.证从同构意义上说,4阶群只有2类：(i)4阶循环群，即元素阶数为1、2、4、4.因为S5含有15对4阶元，以其中1对为例.假如这对4阶元为(1234)和(1432).即(1234)“=(1432).由(1234)2=(13)(24),所以该2阶元为(13)(24).由于每对4阶元都能确定1个4阶循环子群，所以S5共有15个4阶循环子群.如={(1), (1234),(1432),(13)(24)}严={(1),(1243), (1342),(14)(23)).(ii)4阶非循环群，均同构于klein4元群，该群由1个1阶元,3个2阶元组成.这类子群共有35个.如日即={(1),(12),(34),(12)(34)},圧？= 1(1),(12),(35),(12)(35)).定理4S5有6个5阶子群.证S5共有6个5阶子群，这些子群是循环子群.如H卩=〈(12345)>={(1),(12345), (13524),(14253),(15432)).定理5S5有15个8阶子群.证在8阶子群中有1个1阶元、2个4阶元和5个2阶元,其中这5个2阶元中必有2个2轮换的2阶元，且无公共数字;另外3个2阶元为2轮换乘积形式的2阶元.所以,S5有15个8阶子群，且每个子群同构于4阶循环群和2阶循环群的半直积.如琦〉={(1),(1234),(13)(24),(1432),(12)⑶)，(14)(23),(13),(24)}=^4)淤理,其中H严和时分别见附录A(d)和A(b).接下来讨论在S5中可能存在的子群，即子群可能的阶数为6、10、12、15、20、24、30.40,60.定理6S5有30个6阶子群.证从同构的意义来看,6阶群只有2类:一类是6阶循环群，另一类是3次对称群S3.(i)考虑在S5中的6阶循环群.6阶循环群一共有10个，如已°=<(12)(345)>=1(1), (12)(345),(345),(12),(345),(12)(354)},其他9个6阶循环群可类似得到.(ii)考虑6阶非循环群.由于同构保持元素的阶不变，故在该6阶群中含有1个1阶元、3个2阶元和2个3阶元经分析,S5共有20个6阶非循环群.如峦『={(1),(123),(132),(12),(13),(23)},時={(1),(123),(132),(12)(45),(13)(45), (23)(45)}.所以，S5共有30个6阶子群，即10个6阶循环群和20个6阶非循环群(均同构于S3).定理7$5有6个10阶子群.证由引理1知,10阶子群中元素的阶数可能为1、2、5.由于10=2x5,所以在10阶子群中一定存在5阶子群.而这个5阶子群也一定是S5的5阶子群，从而该10阶子群中必有4个5阶元和5个2阶元.可以得到10阶子群共有6个，如H[10)= {(1),(12345),(13524),(14253),(15432),(12)(35), (13)(45),(14)(23),(15)(24),(25)(34)}.且每个子群同构于一个5阶循环群和2阶循环群的半直积.如H[10)=H[5)为，其他5个10阶子群可类似得到.定理8S5有15个12阶子群.证(i)»<S5,而在»中有汕这个12阶子群,所以在S5中一定有5个12阶子群，这些子群均同构于A”如圧⑵={(1),(123),(132),(124), (142),(134),(143),(234),(243),(12)(34), (13)(24),(14)(23)丨.其他4个12阶子群可类似获得.(ii)从6阶循环子群出发可得民功={(1),(12)(345),(354),(12),(345), (12)(354),(12)(34),(12)(35),(12)(45),(34), (35),(45)).其他9个12阶子群可类似得到，且每个子群同构于1个6阶循环群和2阶循环群的半直积.以H严为例，民⑵=民6〉为昭.定理9S5不存在15阶子群.证由引理1知，在15阶子群中元素的阶数可能为1、3、5.假设存在15阶子群则在这个子群H 中一定有3阶子群和5阶子群.而这样的3阶子群和5阶子群也一定是S5的子群.因为任意3阶群里的3阶元和任意5阶群里的5阶元相乘，结果会出现2阶元,但2不是15的因子，矛盾，所以S5不存在15阶子群.定理10S5有6个20阶子群.证经计算，该20阶子群中有1个1阶元、4个5阶元、5个2阶元和10个4阶元.如H[20)={(1),(12345),(13524),(14253), (15432),(12)(35),(13)(45),(14)(23),(15)(24), (25)(34),(1243),(1325),(1452),(1534),(2354),200江西师范大学学报(自然科学版)2021年(1342),(1523),(1254),(1435),(2453)|.其他5个20阶子群可类似获得，且每个子群同构于1个5阶循环群和4阶循环群的半直积.以H严为例，H严=H(S)X H^.定理11$5有5个24阶子群.证显然,S4<S5,且&恰为24阶子群.所以在«5中一定有5个24阶子群，如H严=1(1),(12),(13),(14),(23),(24),(34),(12)(34),(13)(24),(14)(23),(123),(132),(124), (142),(134),(143),(234),(243),(1234), (1432),(1243),(1342),(1324),(1423)}.其他4个24阶子群可类似得到，每个子群均同构于S4.定理12S5不存在30阶子群.类似于定理9可得到.定理13S5不存在40阶子群.类似于定理9可得到.定理14$5有1个60阶子群.证仏是S5的一个子群,且为60阶子群.下证该60阶子群有且只有1个.(反证法)设P是S5的另一个60阶子群.因为[G：B]=2,所以B<S5.又£<S5,则妇B与生介鸟都是S5的正规子群.因为\A5B\^60,所以\A5B\= 60或丨仏引=120.⑴若|生引=60,由引理4得|生"|=60,则B=A4.(ii)若|右B|=120,由引理4得人卩|=30.又因为a5H b仍然是子群，所以A5H b是S5的30阶子群.但S5无30阶子群，矛盾.所以A5是唯一的60阶子群.3总结结合上述讨论,S5共有156个子群.除2个平凡子群外，有25个2阶子群、10个3阶子群、35个4阶子群、6个5阶子群、15个8阶子群、30个6阶子群、6个10阶子群、15个12阶子群、6个20阶子群、5个24阶子群和1个60阶子群.4参考文献[1]韩士安，林磊•近世代数[M].北京:科学出版社,2009.[2]张远达.有限群构造[M].北京:科学出版社」982.[3]孙自行，崔方达.4次对称群&的子群个数及其证明[J].阜阳师范学院学报：自然科学版,2005,22(4)：13-16.[4]班桂宁，吴建平,张中健，等.％的一类子群的一个构造方法[J]-吉首大学学报：自然科学版，2008,29(4)：14[5]Machi A,Siconofi A.A new characterization of A5[J].Arch Math,1977,29(1)：385-388.[6]Arad Z,Chillang D,Hegog M.Classification of finite groupsby a maximal subgroup[J].Journal of Algebra,1981,71(1)：235-244.[7]施武杰，杨文泽.企的一个新刻划与有限质元群[J]-西南师范学院学报：自然科学版,1984,9(1):3640. [8]何立国，张泽，尚玮.计算对称群S]。
