广西自然科学基金(0447038)资助项目

1、1Application ofThe Fundamental Homomorphism Theoremof GroupLI Qia n-qia n LIU Zhi-ga ng YANG Li-yi ng(Department of Mathematics and Computer Science, Guangxi Teachers Education University,Nanning Gua ngxi 530001, P.R.Ch in a)Abstract:The fun dame ntal homomorphism theorem is very importa nt con sequ

2、e nee in group theory, by using it wecan resolve many problems. In this paper we researches mainly about the fun dame ntal homomorphism theoremapplied to direct products of groups and group of inner automorphisms of a group G.Keyword：The Fun dame ntal Homomorphism Theorem; Direct Products; Inner Aut

3、omorphismsMR(2003) Subject Classification: 16WChinese Library Classification:O153.3Document code: AIn the realm of abstract algebra, group is one of the basic and importa nt con cept, have exte nsive applicati on inthe math itself and many side of modern science technique. For example Theories physi

4、cs, Quantum mechanics, Quantum chemistry, Crystallography applicati on are clear certificati ons. So that, after we study abstract algebra course, godeep into a ground of theories of research to have the necessity very much more. In the contents of group, the fun damental homomorphism theorem is ver

5、y importa nt theorem, we can use it prove many problems about group theory, in thispaper to prove several con clusi ons as follow with the fun dame ntal homomorphism theorem: These contents are allstandard if we not to the special provision and explained.Definition 1. LetNbe a subgroup of a groupGwi

6、th symbolN .The kernel of is denoted byKer .Definition 3. LetG1,G2J|,Gmbe a collection of groups. The external direct product ofG,G2,11(, Gm,广西自然科学基金( (0447038)资助项目written asG1G211 ( Gm, is the set of all m-tuples for which the its component is an element ofGi, and the operati on iscomp onen twise.

7、In symbolsG G2Gm=( g1,g2l,gm)Gi,where(g,g2,11(, gm)(g1, g2,HI, gm)isdefined to be(gg1, gzg?1( ,gmgm)Notice that it is easily to verify that the external direct product of groupsis itself a group.Lemma 1.1( The fun dame ntal homomorphism theorem ) Let be a group homomorphism fromGtoG.Then theN=Ker is

8、 the normal subgroup ofG, andG. N (G).To simplify matters, we call the theorem as the FHT.(1)IfHis a subgroup ofG, thenH(H)is a subgroup ofG;(2)IfHis a normal inG, thenH= (H)isa normal inG;1. -(3)IfNis a subgroup ofG, thenN=(N)is a subgroup ofG;(4)IfNis a normal subgroup ofG, then:：_(N) is a normals

9、ubgroup ofGcalled the left coset ofHi nGcon tai ninga. An alogouslyHa = ha h His called the right coset of HDefinition 4. LetGbe a group andHbe a subgroup ofG. For anyinGcontaininga.a三G, the setaH =、ah h HisLemma 2.2Let -：be a group homomorphism fromGtoG. Then we have the following properties:3Lemma

10、 3.3Let be a homomorphism from a groupGto a groupG, an dNG,N = *(N). ThenG N =G N.4Lemma 4. Let H be a subgroup of G and letabelong to G, then:(1)aH = Hif and only ifa H ；(2)aH = Haif and only ifH = a Ha.By using the above lemmas we can obtain the following mainly results.Theorem 1. Let G and H be t

11、wo groups. Suppose J 1 G and K 1 H, the n(J K)(G H)andGHJK二GJ HK.Proof. First we will prove(J K)(G H ).For anyg,h庄G Hand everyj,kj三J K. We have:g,h j,k g,h，二g,h j.k g,h二gjg hkh.11.iJiJ1jSinceJ GandK H, we can getgjg ,hkh 三J K, i.e.g, h j, k g,h- iJ K.Thus(J K) 0(G H ).We make use of the FHT to prove

12、 thatGHi iJ Kis isomorphic toG. Jr iH.K. Therefore wemust look for a group homomorphism fromG HontoG JrH Kand determine the kernel of it. Infact one can define correspondencef : G，G； JrG.Kdefined byf (x, y) = (Jx, Ky). Clearly,;：Jx, Ky三G JH. K, there must bex, y三G Hto satisfyf x, y = Jx, Ky. Thus,fi

13、s on to.Because of J G, we haveJx=xJfor_x G, similarly,Ky = yKfor- yH.Whenx, y F G H , x, y庄G H, there arex, yix, y二x x, y y.For anyy, x?, y?G H,we have4f (xi, yi)(X2, y2) = f(X1X2, yy)=(JX1X2, Kyy)=(xJJx, yiKKy2)=( JxJ, KyiKy?)=(Jxi, KyJ(Jx2, Ky?) = f(X1, yj f(X2, y?).Hence：ix1, y1i ix2, y2= f x1,

14、y1f x2, y2. Thereforefis group a homomorphism fromG HontoG.JiiH .KandJ, Kis the identity ofG.Jr iH K.For anyx,yi=G H, thenf x, y = Jx, Ky, according to the property of coset, we can get:5f x, y = Jx, Ky = J,Kif and only ifx Jandy K, i.e.ker f - x,y x J, y K ?= J K.Now let we look at our proof:(J K)(

15、G H),fis a group homomorphism fromG HontoG J ! iH . Kand the kernel offisJ K. According to the FHT, we can getG H J K三GJ H K.Theorem 2. Let:is a group homomorphism fromGon toGfH Gandker、“ H, the nG H三G Hwhere H = H：締hh H?.2 Proof:According to Lemma 2.(2), we knowH G.To establishG H三G H, we firstly n

16、 eed to con struct a mapp ingfand provefis a groupdefined byf g二g Hwhereg=(g).For_g H G H, since is a surjection fromGtoG, we must be foundg：= Gsuch thatf g = g H.Thusfison to.For arbitrarya, b G,f ab = abCHabCHH = (a 7 H )(b：H) = f a f bThereforefis a group homomorphism.4, we can get that forg H, t

17、henf g(g)CH=H, sayg ker f, so that H ker f.On the other hand ,g - ker f,f gg：H(g)C H = H, that is to say ,(g) H.Moreoverg4(H ) = ker H = H, because ofker- H, thereforeker fH. That isker f = H.According to the FHT, we can obtainG. H -G H.Theorem 1 and Theorem 2 apply Exercise 1 and Exercise 2.homomor

18、phism fromGontoG H.We give the mapp ingWe will now showker f = H, in fact we know thatHis ide ntity ofG H, according to Lemma6Exercise 1.3is normal subgroup ofU 16,4is a normal subgroup of.So that for anyx U 16andy Z8, for a fun ctionf:fx, y= 3 x, 4 ywe havef : U 16 Z834- U 16 / 3：Z8/ 4is a group is

19、omorphism, so thatU 16 Z*二U 16 3Z*4readily to verify that they are in deed group with ordi nary multiplicati on.b| a,b,C, d一R, ad - be = 0be general linear group of 2 x 2dmatrices overRunder ordinary matrix multiplication . Then the mappingdet Ais a group#homomorphism fromGon toR. ThegroupS = SL 2,

20、Rof2 2matrices withdeterm inant 1 overRis a normal subgroup ofG. MoreoverDefinition 5. An automorphism of groupGis just a group isomorphism fromGto itself. The set of all_ 1automorphisms of groupGis denoted byAut(G). For anya,x G,a：x axais called an inner automorphism ofGandI (G) =% Gis the set of a

21、ll inner automorphism ofG.Theorem 3: LetGbe a group and the mappi ngh:G“ I Gdefi ned byha =a. The nI GAut(G)a ndI G二G ker h.51Proof. It is clearly thatAut(G)SG.To showI G - Aut(G), suffice it to prove that is an automorphism ofGfor anya G. 1 11) (one-to-one) For anyx,yG, ifaxa=aya, thenx = yby using

22、 cancellation law of group. Thusaisone-to-one.2) (onto) For anyx G, we takey = axa G, thena(a(y)a = a(a_lxa)aJ= x, so thatais on to.3) (O.P.) For anyx,y G, we have x y = axya4= axa* aya二axay. Thereforeais isomorphism fromGtoG.According to the definition of automorphism. We knowais an automorphism of

23、G.Assume#RandRare sets of all the non zero real nu mbers and positive real nu mbers respectively, it isExercise 2. LetG =GL2,R =a c- r_ _GS= R#71Notice that for anya,bG, we have 忙 沖b=aband (忙 厂=a丄.In fact for anyx三G, it is clearlyex = exe = x. Alsoa bx二a bx二abxb二a bxb aJ= abxbJaJ二abx abJ=bx,Thusa b

24、= ab.Since(a丄a) x =a丄axi：冷丄axa =x =exe=込x, sayai =aa = e, wehave kn ow n (0a) =札 We can obtai n ($a) 4a=a丄, i e.(a) =a丄.He nce the proof ofI GAut(G)is complete.1 _It is easy to see that for everyx G,axa xif and only ifa C (G)whereC(G)二g G gx = xg ,X/xGis the center ofG(short forC).Leth : G r I Gbe t

25、he mapping defined byh a =a, we will prove thathis a grouphomomorphism from G on to I(G) and that C is its kern el.For everya=1 G, we can readily find thatha =a, that is to say,his onto. For anya,b：=G,sinceh(ab=% =h(a )h(b), so thathis a group homomorphism fromGonto1(G).Notice that for anyb Cand everyx G, we havebx=xb,i.e.,bxb = x = exe，that isb= e. We obtainh(b) =e, henceC二ker h. Next