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6.The Homomorphism Theorems

In this section, we investigate maps between groups which preserve the group- operations. Definition.LetGandHbe groups and let?:G→Hbe a mapping fromGto H. Then?is called ahomomorphismif for allx,y?Gwe have: ?(xy) =?(x)?(y). A homomorphism which is also bijective is called anisomorphism. A homomorphism fromGto itself is called anendomorphism. An isomorphism fromGto itself is called anautomorphism, and the set of all automorphisms of a groupGis denoted by Aut(G). Before we show that Aut(G) is a group under compositions of maps, let us prove that a homomorphism preserves the group structure. Proposition6.1.If?:G→His a homomorphism, then?(eG) =eHand for all x?G,?(x-1) =?(x)-1. Proof.Since?is a homomorphism, for allx,y?Gwe have?(xy) =?(x)?(y). In particular,?(y) =?(eGy) =?(eG)?(y), which implies?(eG) =eH. Further, ?(eG) =?(xx-1) =?(x)?(x-1) =eH, which implies?(x-1) =?(x)-1.? Corollary6.2.If?:G→His a homomorphism, then the image of?is a subgroup ofH. Proof.Letaandbbe in the image of?. We have to show that alsoab-1is in the image of?. Ifaandbare in the image of?, then there arex,y?Gsuch that ?(x) =aand?(y) =b. Now, by Proposition 6.1 we get ab -1=?(x)?(y)-1=?(x)?(y-1) =?(xy-1). Proposition6.3.For any groupG, the set Aut(G) is a group under compositions of maps. Proof.Let?,ψ?Aut(G). First we have to show that?◦ψ?Aut(G): Since?and ψare both bijections,?◦ψis a bijection too, and since?andψare both homomor- phisms, we have (?◦ψ)(xy) =??ψ(xy)?=??ψ(x)ψ(y)?= ??ψ(x)???ψ(y)?= (?◦ψ)(x)(?◦ψ)(y). Hence,?◦ψ?Aut(G). Now, let us show that?Aut(G),◦?is a group: (A0) Let?1,?2,?3?Aut(G). Then for allx?Gwe have

??1◦(?2◦?3)?(x) =?1??2◦?3)(x)?=?1??2??3(x)??=??1◦?2???3(x)?=?(?1◦?2)◦?3?(x),

which implies that?1◦(?2◦?3) = (?1◦?2)◦?3, thus, "◦" is associative. (A1) The identity mappingιonGis of course a bijective homomorphism fromG to itself, and in fact,ιis the neutral element of?Aut(G),◦?. 22
23
(A2) Let??Aut(G), and let?-1be such that for everyx?G,???-1(x)?=x. It is obvious that?◦?-1=ιand it remains to show that?-1is a homomorphism:

Since?is a homomorphism, for allx,y?Gwe have

-1(xy) =?-1??(?-1(x))???? =x?(?-1(y))???? =y? =?-1????-1(x)?-1(y)??=?-1(x)?-1(y), which shows that?-1?Aut(G).? Definition.If?:G→His a homomorphism, then?x?G:?(x) =eH?is called thekernelof?and is denoted by ker(?). Theorem6.4.Let?:G→Hbe a homomorphism, then ker(?)?G. Proof.First we have to show that ker(?)?G: Ifa,b?ker(?), then ?(ab-1) =?(a)?(b-1) =?(a)?(b)-1=eHe-1 H=eH, thus,ab-1?ker(?), which implies ker(?)?G. Now we show that ker(G)?G: Letx?Ganda?ker(?), then ?(xax-1) =?(x)?(a)?(x)-1=?(x)eH?(x)-1=?(x)?(x)-1=eH, thus,xax-1?ker(?), which implies ker(?)?G.?

Let us give some examples of homomorphisms:

(1) The mapping x?→ex is an isomorphism, and?-1= ln. (2) Letnbe a positive integer. Then ?: (O(n),·)→?{1,-1},·?

A?→det(A)

is a surjective homomorphism and ker(?) = SO(n). Further, forn= 1,?is even an isomorphism. (3) The mapping ?:?3→?2 (x,y,z)?→(x,z) is a surjective homomorphism and ker(?) =?(0,y,0) :y???. (4) Letn≥3 be an integer, letCn={a0,... ,an-1}, and letρ?Dnbe the rota- tion through 2π/n. Then?:Cn→Dn, defined by?(ak) :=ρkis an injective homomorphism fromCnintoDn. Thus,Cnis isomorphic to a subgroup of D n. 24
(5) Letn≥3 be an integer. For anyx?Dn, let sg(x) =?

1 ifxis a rotation,

-1 ifxis a reflection, then?:Dn→?{1,-1},·? x?→sg(x) is a surjective homomorphism. (6) The mapping ?: (?12,+)→(?12,+) x?→4x is an endomorphism of (?12,+), where ker(?) ={0,3,6,9}and the image of ?is{0,4,8}. (7) For everyr???, the mapping q?→rq is an automorphism of (?,+). (8) LetC2×C2={e,a,b,c}, then every permutation of{a,b,c}is a bijective homomorphism fromC2×C2to itself. Hence, Aut(C2×C2) is isomorphic to S

3(or toD3).

In order to define an operation on the setG/N, whereN?G, we need the following: Fact6.5.IfN?G, then for allx,y?G, (xN)(yN) = (xy)N.

Proof.SinceNis a normal subgroup ofG, we have

(xN)(yN) =?x(yNy-1???? =N)?(yN) = (xy)(NN) = (xy)N .

This leads to the following:

Proposition6.6.IfN?G, then the setG/N={xN:x?G}is a group under the operation (xN)(yN) := (xy)N. 25
Proof.First we have to show that the operation (xN)(yN) is well-defined: If (xN) = (˜xN) and (yN) = (˜yN), then, by Lemma 3.6(d),x-1˜x,y-1˜y?N. Now, sinceNis a normal subgroup ofG, (xy)-1(˜x˜y) =y-1(x-1˜x???? ?N) ˜y?y-1N˜y=y-1N(y???? =Ny -1)˜y=N(y-1˜y) =N , which implies (xN)(yN) = (xy)N= (˜x˜y)N= (˜xN)(˜yN).

Now, let us show thatG/Nis a group:

(A0) (xN)?(yN)(zN)?=?x(yz)?N=?(xy)z?N=?(xN)(yN)?(zN). (A1) For allx?Gwe have (eN)(xN) = (ex)N=xN , therefore,eN=Nis the neutral element ofG/N. (A2) For allx?Gwe have (xN)(x-1N) = (xx-1)N=eN=N= (x-1x)N= (x-1N)(xN), therefore, (xN)-1= (x-1N).? For example, letCbe the cube-group and letNbe the normal subgroup ofCwhich is isomorphic toC2×C2. Then, by Proposition 6.6,C/Nis a group, and in fact,C/N is isomorphic toS3(see Hw9.Q41).

Lemma6.7.IfN?G, then

π:G→G/N

x?→xN is a surjective homomorphism, called the natural homomorphism fromGontoG/N, and ker(π) =N. Proof.For allx,y?Gwe haveπ(xy) = (xy)N= (xN)(yN) =π(x)π(y), thus,πis a homomorphism. Further, letxN?G/N, thenπ(x) =xN, which shows thatπis surjective. Finally, by Lemma 3.6(c), ker(π) ={x?G:xN=N}=N.? By Theorem 6.4 we know that if?:G→His a homomorphism, then ker(?)?G. On the other hand, by Lemma 6.7, we get the following: Corollary6.8.IfN?G, then there exists a groupHand a homomorphism ?:G→Hsuch thatN= ker(?). Proof.LetH=G/Nand let?be the natural homomorphism fromGontoH.? Theorem6.9 (First Isomorphism Theorem).Letψ:G→Hbe a surjective homo- morphism, letN= ker(ψ)?Gand letπ:G→G/Nbe the natural homomorphism fromGontoG/N. Then there is a unique isomorphism?:G/N→Hsuch that ψ=?◦π. In other words, the following diagram "commutes": G ??ψ??H 26
Proof.Define?:G/N→Hby stipulating?(xN) :=ψ(x) (for everyx?G). Then ψ=?◦πand it remains to be shown that?is well-defined, a bijective homomorphism and unique. ?is well-defined: IfxN=yN, thenx-1y?N(by Lemma 3.6(d)). Thus, since N= ker(ψ),ψ(x-1y) =eHand sinceψis a homomorphism we haveeH=ψ(x-1y) = ψ(x)-1ψ(y), which impliesψ(x) =ψ(y). Therefore,?(xN) =ψ(x) =ψ(y) =?(yN). ?is a homomorphism: LetxN,yN?G/N, then ??(xN)(yN)?=??(xy)N?=ψ(xy) =ψ(x)ψ(y) =?(xN)?(yN). ?is injective: ?(xN) =?(yN)??ψ(x) =ψ(y)?? ??eH=ψ(x)-1ψ(y) =ψ(x-1)ψ(y) =ψ(x-1y)?? ??x-1y?N??xN=yN . ?is surjective: Sinceψis surjective, for allz?Hthere is anx?Gsuch that

ψ(x) =z, thus,?(xN) =z.

?is unique: Assume towards a contradiction that there exists an isomorphism ˜?: G/N→Hdifferent from?such that ˜?◦π=ψ. Then there is a cosetxN?G/N such that ˜?(xN)?=?(xN), which implies

ψ(x) = (˜?◦π)(x) = ˜??π(x)?= ˜?(xN)?=?(xN) =??π(x)?= (?◦π)(x) =ψ(x),

a contradiction.? For example, letmbe a positive integer and letCm={a0,... ,am-1}be the cyclic group of orderm. Further, letψ:?→Cm, whereψ(k) :=ak. Thenψis a surjective homomorphism from?toCmand ker(ψ) =m?. Thus, by Theorem 6.9, ?/m?andCmare isomorphic and the isomorphism?:?/m?→Cmis defined by ?(k+m?) :=ak. Let us consider some other applications of Theorem 6.9: (1) Letnbe a positive integer. Then

ψ: (O(n),·)→?{1,-1},·?

A?→det(A)

is a surjective homomorphism with ker(ψ) = SO(n), and thus, O(n)/SO(n) and{1,-1}are isomorphic (where{1,-1}≂=C2). (2) Letnbe a positive integer and let GL(n)+={A?GL(n) : det(A)>0}.

Thenψ: (GL(n)+,·)→(?+,·)

A?→det(A)

is a surjective homomorphism with ker(ψ) = SL(n), and thus, GL(n)+/SL(n) and?+are isomorphic. 27
(3) The mapping z?→ |z| is a surjective homomorphism with ker(ψ) =?={z??:|z|}, and thus, ?/?and?+are isomorphic. (4) The mapping

ψ:?3→?2

(x,y,z)?→(x,z) is a surjective homomorphism with ker(ψ) =?(0,y,0) :y???≂=?, and thus,?3/?and?2are isomorphic. (5) The mapping

ψ: (?12,+)→(?3,+)

x?→x(mod 3) is a surjective homomorphism with ker(ψ) ={0,3,6,9}= 3?12, and thus,

12/3?12and?3are isomorphic.

Theorem6.10 (Second Isomorphism Theorem).LetN?GandK?G. Then (1)KN=NK?G. (2)N?KN. (3) (N∩K)?K. (4) The mapping ?:K/(N∩K)→KN/N x(N∩K)?→xN is an isomorphism.

Proof.(1) This is Theorem 5.8.

(2) SinceKN?GandN?KN,N?KN. Hence, sinceN?G,N?KN. (3) Letx?Kanda?N∩K. Thenxax-1belongs toK, sincex,a?K, but also to

N, sinceN?G, thus,xax-1?N∩K.

(4) Letψ:K→KN/Nbe defined by stipulatingψ(k) :=kN. Thenψis a surjective homomorphism and ker(ψ) ={k?K:k?N}=N∩K.

Consider the following diagram:

K ??ψ??KN/N Sinceψis a surjective homomorphism, by Theorem 6.9,?is an isomorphism.? 28
For example, letmandnbe two positive integers. Thenm?andn?are normal sub- groups of?, and by Theorem 6.10,m?/(m?∩n?) and (m?+n?)/n?are isomorphic. In particular, form= 6 andn= 9 we havem?∩n?= 18?andm?+n?= 3?. Thus, 6?/18?and 3?/9?are isomorphic, in fact, both groups are isomorphic toC3. Theorem6.11 (Third Isomorphism Theorem).LetK?G,N?G, andN?K. Then

K/N?G/Nand

?:G/K→G/N? K/N xK?→(xN)(K/N) is an isomorphism. Proof.First we show thatK/N?G/N. So, for anyx?Gandk?K, we must have (xN)(kN)(xN)-1?K/N: (xN)(kN)(xN)-1=xNkNx-1N=xNkx-1xNx-1???? =NN= =xNkx-1N=xNx-1???? =Nxkx -1???? =:k??KN=Nk?N=k?NN=k?N?K/N . Let

ψ:G→G/N?

K/N x?→(xN)(K/N) Thenψis a surjective homomorphism and ker(ψ) ={x?G:xN?K/N}=K.

Consider the following diagram:

G ??ψ??G/N? K/N Sinceψis a surjective homomorphism, by Theorem 6.9,?is an isomorphism.? For example, letmandnbe two positive integers such thatm??n. Thenm?and n?are normal subgroups of?,n??m?, and by Theorem 6.11, ?/m?≂=?/n?? m?/n?.

In particular, form= 6 andn= 18,

6≂=?18?

6?/18?,

and in fact, both groups are isomorphic toC6.quotesdbs_dbs12.pdfusesText_18

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