ALIGARH MUSLIM UNIVERSITY MATHEMATICS DEPARTMENT
A Simple Course In Abstract Algebra
Dr. NADEEMUR REHMAN
Department of Mathematics AligarhMuslim University Aligarh-202002 (U.P.), India
nu.rehman.mm@amu.ac.in
November 4, 2020 [C:R] = 2
0 = 1
xy= 0 xn = 0 Zp
Chapter 1
Normal Subgroups and Quotient Groups
IfH is a subgroup of a groupG, then right cosets are not always the same as left cosets; that is, it is not always the case thatgH =Hg for all g∈G. The subgroups for which this property holds play a critical role in group theory:
they allow for the construction of a new class of groups, called factor or quotient groups. Factor groups may be studied by using homomorphisms, a generalization of isomorphisms.
1.1 Normal Subgroups
Definition 1.1.1 (Normal Subgroup). LetN be a subgroup of a groupG. ThenN isnormalifgng−1 ∈N for all g∈Gandn∈N. or equivalentlygN g−1⊆N. We writeNEG.
Example 1.1.1. For any group G, it is clear that GEG. Also,{e}EG, sincegeg−1=gg−1=efor allg∈G.
Example 1.1.2. (i) LetG=S3 andN =h(1,2)i={e,(1,2)}. Since (1,2,3)(1,2)(1,2,3)−1= (2,3)∈/N, thenN 6CG.
(ii) LetG=S3andN1={e,(1,2,3),(1,3,2)} Since
(1,2)(1,2,3)(1,2)−1= (1,3,2)∈N1, (1,2)(1,3,2)(1,2)−1= (1,2,3)∈N1, (1,3)(1,2,3)(1,3)−1= (1,3,2)∈N1, (1,3)(1,3,2)(1,3)−1= (1,2,3)∈N1, (2,3)(1,2,3)(2,3)−1= (1,3,2)∈N1, (2,3)(1,3,2)(2,3)−1= (1,2,3)∈N1. Thus,N1EG.
Theorem 1.1.1. (a) LetG be an abelian group, then every subgroup ofGis normal in G.
(b) If Gis a group, then every subgroup of the centerZ(G)is normal in G. In particular,Z(G)EG.
Proof. (a) Since gN =N g for all g ∈G. Then gng−1 =ngg−1 =ne= n∈N for all g ∈ Gand n ∈N. That is, NEG.
(b) Let N be any subgroup ofZ(G). Letx∈Z(G) and g ∈G. The we find thatgxg−1=gg−1x=ex =x∈Z(G) and hence it shows that ifN ≤Z(G), thenNEG.
In the abelain group case, N g=gN is satisfied trivially, forng=gnfor alln∈N andg∈G. You should notice, however,N g=gNdoes not mean thatgcommutes with every element ofN. This is an equation between certain sets, so is equivalent to the inclusionsN g⊆gN and gN ⊆N g. The first inclusion means for alln∈N, there is n1 ∈N such thatng=gn1. Heren6=n1in general and thereforeng=gn16=gn. The second inclusion has a similar meaning.
N g =gN means that, when we multiply the elements of N by g on the right and on the left, we get the same collection of elements. It does not mean that, when we multiply any element ofN byg on the right and on the left, we get the same product.
In the following theorem, we give some conditions onN ≤Gequivalent to normality.
Theorem 1.1.2. Let Gbe a group and N be a subgroup ofG. Then the following statements are equivalent.
1.1. NORMAL SUBGROUPS N. REHMAN
(a) NEG
(b) For allg∈G,g−1N g⊆N. (c) For allg∈G,gN g−1=N. (d) For allg∈G,gN=N g.
Proof. (a)⇒(b). SinceNEG, sogN g−1⊆N for allg∈G. Replacinggbyg−1yields thatg−1N g=g−1N(g−1)−1⊆ N for allg∈Gand (b) holds.
(b)⇒(c). Letg∈G. SincegN g−1⊂N, we need only showN ⊂gN g−1. Forn∈N,g−1ng=g−1n(g−1)−1∈N. Hence,g−1ng=n0 for somen0∈N. Therefore,n=gn0g−1∈gN g−1.
(c) ⇒ (d). We have gN g−1 =N for all n ∈N. For giveng ∈G andn ∈N, there exists ann0 in N such that gng−1=n0. This implies thatgn=n0g. Thus,gN ⊆N g. Similarly,N g⊆gN. This proves thatgN =N g.
(d)⇒(a). For anyg ∈Gand anyn∈N, gn∈gN =N g. Hencegn=n0g for somen0 ∈N. This implies that gng−1=n0∈N and henceNEG.
Example 1.1.3. (a) LetG=S4 andN ={e,(1,2),(3,4),(1,2)(3,4)}. IsNES4? We compare the right and left cosets ofN inS4. We see that
(1,3)(2,4)N ={(1,3)(2,4),(1,4,2,3),(1,3,2,4),(1,4)(2,3)}
and
N(1,3)(2,4) ={(1,3)(2,4),(1,3,2,4),(1,4,2,3),(1,4)(2,3)}.
This is of course not enough to conclude NES4. We must examine the other cosets also. We see (1,3)N = {(1,3),(1,2,3),(3,4,1),(1,2,3,4)}andN(1,3) ={(1,3),(1,3,2),(1,4,3),(1,4,3,2)}. This shows that (1,3)N 6=
N(1,3). HenceN 6CS4.
(b) LetG=S4andN ={e,(1,2),(1,3)(2,3),(1,2,3)(1,3,2)}. IsNES4? we observe that
(1,4)N(1,4)−1= (1,4)N(1,4) ={e,(2,4),(3,4),(2,3),(2,3,4),(2,4,3)} 6=N, and so N6CS4.
Example 1.1.4. LetG=GL2(R), the group of 2×2 matrices with nonzero determinant and N =SL2(R), 2×2 matrices with determinant 1. ThenSL2(R)EG2(R).
LetA∈GL2(R),B∈SL2(R). By the basic property of determinants we have det(ABA−1) = (det(A))(det(B))(det(A)−1)
=det(A)·1·(det(A)−1)
=det(AA−1) =det(I2) = 1.
So,ABA−1∈N. Therefore,AN A−1⊆N and it follows that SL2(R)EGL2(R).
Theorem 1.1.3. If Ga group andN ≤G, then NEGwhen any of the following holds:
(a) N is finite and is the only subgroup in Gof order|N|.
(b) [G:N] = 2.
Proof. (a) Since N ≤G, so by Proposition?? gN g−1≤Gand|N|=|gN g−1|. Clearly,ϕ(n) =gng−1 is a bijection fromN togN g−1 whose inverse is Φ(gng−1) =g−1(gng−1)g= (g−1g)n(g−1g) =n. (Exercise in 8 *2.2 CL)Thus by (a)N =gN g−1 so Theorem??implies thatNEG.
(b) Letg∈G. Ifg∈N, then N g=N =gN, thenNEG, by Theorem??. Letg∈G\N. Since [G:N] = 2,N and gN are the right costs of N inG, and G=N∪gN, a disjoint union. Hence,gN =G\N. Similarly,N and N gare the left cosets ofN in G, and G=N∪N gand N∩N g=∅. Thus,N g=G\N. Therefore,gN =G\N =N gand hence by Theorem??, NEG.
Example 1.1.5. AnESn, where An is the alternating group. The alternating group An is of index 2 in Sn, and henceAnESn by Theorem??.
Proposition 1.1.1. Let Gbe group,NEGandKEG. IfN∩K={e}, thennk=kn for alln∈N andk∈K.
1.2. QUOTIENT GROUP(OR FACTOR GROUP) N. REHMAN
Proof. Considerα=nkn−1k−1. Thinking ofα=n(kn−1k−1). SinceNEG, we see thatkn−1k−1∈kN k−1=N and henceα∈N. Similarly, writing α= (nkn−1)k−1 shows thatα∈K because KEG. Henceα∈N∩K={e} by our hypothesis, soα=e. Therefore,nkn−1k−1={e}, which gives thatnk=kn for alln∈N andk∈K.
.
Theorem 1.1.4. Let Gbe a group,N ≤GandK≤G. Then (a) If NEGandK≤G, thenN∩KEK.
(b) If both NEG andKEG, thenN∩KEG.
(c) If NEGorKEG, thenN K =KN is normal in G.
(d) If both NEG andKEG, thenN KEG.
Proof. (a) Letx∈N∩Kand anyk∈K. Thenkxk−1∈N sincex∈N andNEG. Also,kxk−1∈Kbecausex∈K andK≤G. Thus,kxk−1∈N∩Kfor allx∈N∩K andk∈K, and henceN∩KEK.
(b) Since the intersection of subgroups is a subgroup, soN∩K≤G. Considerg(N∩K)g−1for allg∈G. Thengag−1 be any element ofg(N∩K)g−1, wherea∈N ∩K. Sincea∈N∩K, we havea∈N anda∈K. Hence,gag−1∈N andgag−1∈K. Thus,gag−1∈N∩K. This shows thatg(N∩K)g−1⊆N∩K. Hence,N∩K is a normal subgroup by Theorem??.
(c) Suppose KEG. If nk ∈N K, wheren ∈N and k ∈K. Since KEG and n∈ G, we havenK =Kn. Thus, nk∈nK=Kn. SinceKn⊆KN, we have nk∈KN. Hence, N K⊆KN. Similarly, KN⊆N K and soN K =KN.
SinceN ≤GandK≤GandN K=KN and henceN K≤Gby Lemma??.
A similar argument works ifNEG.
(d) To show thatN KEG, letg∈G. ThengN g−1⊆N andgKg−1⊆KsinceN andK are normal subgroups. Now g(N K)g−1=g(N g−1gK)g−1= (gN g−1)(gKg−1)⊆N K.Therefore,N KEGby Theorem??.
1.2 Quotient Group(or Factor Group)
LetGbe a group andNEG. Is there a natural generalization of the operation ofGthat makesG/N a group? By a
“natural” generalization, we mean something likeG/N is the collection of all right cosets of N inG.
G/N ={N a|for anya∈G}.
AsNEG, each right cosetsN ais also a left cosetsaN. Thus, we also have G/N ={aN|for any a∈G}.
Define a product on the right cosets inG/N by
N aN b=N ab.
The problem is that the multiplication defined above depends upon the representativesaandband we know that cosets can have many different representatives. We need to ensure our multiplication is well-defined: it depends only upon the cosets concerned and not on the choice of representatives for the cosets. More precisely, we have the following theorem.
Theorem 1.2.1. Let N be a normal subgroup of a group G and let G/N be the set of all right cosets of N in G.
Define the operation onG/N by
(N a)(N b) =N(ab).
Then(G/N,·)is a group
Proof. First we show that the above operation is well defined. Let a, a1, b, b1 ∈ G such that N a = N a1 and N b=N b1. This implies thata1=n1aandb1=n2nfor some n1, n2∈N
a1b1 = n1a.n2b
= n1an2a−1ab
= n1(an2a−1)ab
SinceN is normal in G,an2a−1∈N, for anya∈G,n2∈N. Also Sincen1∈N, we findn1(an2a−1)∈N. N a1b1=N(n1an2a−1)ab
N a1b1=N ab
1.2. QUOTIENT GROUP(OR FACTOR GROUP) N. REHMAN
sincen1an2a−1∈N. Hence we have a well-defined multiplication onG/N. Associative: For anyN a, N b, N c∈G/N
(N aN b)N c= (N ab)N c=N(ab)c and
N a(N bN c) =N a(N bc) =N a(bc).
But the multiplication onGis associative, so (ab)c=a(bc) and hence (N aN b)N c=N a(N bN c).
Identity: Ifeis the identity of group, thenN =N eis the identity ofG/N. For anyN a∈G/N N aN=N aN e=N ae=N a.
Similarly
N N a=N eN a=N ea=N a.
This implies thatN N a=N aN=N a, for anyN a∈G/N and henceN =N eis the identity elemnt ofG/N. Inverse: For anyN a∈G/N there existsN a−1∈G/N such that
N aN a−1=N aa−1=N e=N Also,
N a−1N a=N a−1a=N e=N.
Hence,G/N is a group.
Definition 1.2.1 (Quotient Group or Factor Group). Let Gbe a group, NEG. Then the group G/N with cosets multiplication is called thequotient grouporfactor groupofGbyN.
Remark 1.2.1. IfGis additive group andNEG, then operation onG/N is defied as (N+a) + (N+b) =N+ (a+b)
Note that for a finite group G, the number of elements ofG/N is just the index ofN in G, i.e. [G:N]. That is,
|G/N|= [G:N]. By Lagrange’s theorem,|G|= [G:N]|N|so that [G:N] =|N|G||. Hence,|G/N|=|N||G|.
Remark 1.2.2. IfN is not a normal subgroup ofG, then the operation defined in the Theorem?? will not be well- defined. To see this, consider the subgroupN =h(1,2)i={e,(1,2)}ofS3. Since (1,2,3)(1,2)(1,2,3)−1= (2,3)∈/N thenN is not a normal subgroup ofS3. However, N(1,2,3) =N(2,3) ={(1,2,3),(2,3)} andN(1,3,2) =N(1,3) = {(1,3),(1,3,2)}. ButN(1,2,3)(1,3,2)6=N(2,3)(1,3) sinceN(1,2,3)(1,3,2) =N andN(2,3)(1,3) =N(1,2,3).
Example 1.2.1. LetG=S3andN =h(1,2,3)i={e,(1,2,3),(1,3,2)}. One can verify easily thatNEGby Example
??(i),G/N ={N, N(1,2)}.
Example 1.2.2. Forn >1, (Z,+)/nZ= (Zn,+n).
A typical coset isnZ+afora∈Z. Thus,{nZ+a|a∈Z}={nZ+i|0≤i≤n}=Zn. The group operation inZ/nZ is (nZ+a) + (nZ+b) =nZ+ (a+b), which is exactly the addition in Zn. Consequintly, the group (Z,+)/nZand (Zn,+n) are the same.
Example 1.2.3. LetG=Zandh5i={0,±5,±10,±15,· · · }=N.One can easily verify that NEG. Then we find the Quotient groupZ/h5ias follows:
N ={0,±5,±10,±15,· · · }. Leta∈Z. Then by division algorithma= 5b+cwhere 0≤c <5. Thus,a−c= 5b∈N andN+ (a−c) =N that isN+a=N+c, wherec= 0,1,2,3 or 4 and henceN+ 0, N+ 1, N+ 2, N+ 3 orN+ 4.
Therefore,Z/h5i={N, N+ 1, N+ 2, N+ 3, N+ 4}.
Example 1.2.4. Quotient groups can be used to show thatA4has no subgroup of order 6 and thus showing that the converse of Lagrange’s Theorem is false in general. To see this,assume thatN is a subgroup ofA4 of order 6. Then [G:N] = 2 and thereforeNEA4by Theorem??. Hence,A4/Nmakes sense. Moreover, for eacha∈N, (N a)−1=N a so thatN a2 =N and hencea2∈N. One can show that (1,2,3)2= (1,3,2), (1,3,2)2= (1,2,3), (1,2,4)2 = (1,4,2), (1,4,2)2 = (1,2,4), (1,3,4)2 = (1,4,3), (1,4,3)2= (1,3,4), (2,3,4)2 = (2,4,3), and (2,4,3)2 = (2,3,4). This yields more than six different elements ofA4 inN. That is,|N|>6, a contradiction. Thus,A4has no subgroup of order 6.
Now, we gave some properties of quotient groups.
Theorem 1.2.2. Let Gbe a group and NEG.
1.3. COMMUTATOR SUBGROUP (DRIVED SUBGROUP) N. REHMAN
(a) If Ga abelian , theG/N is abelain.
(b) if g∈Gandk∈Z, thenN gk= (N g)k.
(c) If G=hziis cyclic, then G/N =hN ziis cyclic.
Proof. (a) For any N a, N b ∈ G/N. We have N aN b = N ab. Now, since G is abelain then ab = ba. Thus N aN b=N ab=N ba=N bN a.That is,G/N is abelain.
(b) Ifg∈N andk >0 ,N gk=N gN gN g· · ·N g= (N g)k by the definition of powers, the definition of multiplication in G/N, and induction cleary N = N g0 = (N g)0. When n = −m for m > 0, N gn = N(g−1)m = (N g−1)m = ((N g)−1)m= (N g)−m= (N g)n.
(c) When G=hziand A ∈G/N, then A=N g for g =zk ∈ hzi. Thus using (b), A=N zk = (N z)k ∈ hziand it follows thatG/N =hN ziis cyclic.
Theorem 1.2.3. If Gis a group andG/Z(G) is cyclic, thenGis abelain.
Proof. First we show thatZ(G)EG. Ifg∈Gandx∈Z(G), then gxg−1=gg−1x=ex=x∈Z(G). SinceG/Z(G) is cyclic thenG/Z(G) =hZ(G)gifor someg∈G. Ifa, b∈G, thenZ(G)a, Z(G)b∈G/Z(G). Thus,Z(G)a=Z(G)gn and Z(G)b = Z(G)gm for somem, n ∈ Z. Hence a = xgn and b = ygm for some x, y ∈ Z(G). This implies that ab= (xgn)(ygm) =xygn+m=yxgm+n= (ygm)(xgn) =ba. HenceGis abelain.
Theorem 1.2.4. Let Gbe a group of order p2 where pis a prime number. If Z(G), the center ofG, is nontrivial, thenGis Abelian.
Proof. LetGbe a group such that|G|=p2. Sincee∈Z(G) thenZ(G)6=∅. By Lagrange’s theorem either|Z(G)|=p2 or|Z(G)|=p. If|Z(G)|=p2 thenZ(G) =Gand soGis Abelian. If|Z(G)|=pthen|G/Z(G)|=pand soG/Z(G) is cyclic by Corollary??. Thus, by Theorem??,Gis Abelian.
1.3 Commutator subgroup (Drived subgroup)
Definition 1.3.1. LetGbe a group anda, b∈G. Then
aba−1b−1∈G is called the commutator ofaandb and is denoted by [a, b].
Lemma 1.3.1. Let Gbe a group and a, b∈GThen (i) [a, b]−1= [b, a]
(ii) [a, b] =e if and only ifGis abelain.
Proof. (i) we have [a, b] =aba−1b−1. Then
[a, b]−1= (aba−1b−1)−1= (b1)−1(a−1)−1(b−1)(a−1) =bab−1a−1= [b, a].
(ii) If [a, b] = e, then aba−1b−1 = e , and multiplying ba on the right gives that ab =ba and hence G is abelian.
Conversely, ifGis abelain that isab=ba such that [a, b] =aba−1b−1=ba(a−1b−1) =b(aa−1)b−1=bb−1=e.
LetGbe a group. We define the set
C={x1x2· · ·xn|n≥1, eachxi is a commuttaor inG}.
In other words,Cis the collection of all finite products of commutators inG. Then we have Theorem 1.3.1. If Gis any group, then CEG.
Proof. First, we havee=eee−1e−1∈C, soC is nonempty and contains the identity. Ifα, β∈C, then we haveα= x1x2· · ·xn andβ=y1y2· · ·ym, where eachxiand eachyj is a commutator inG. Thenαβ=x1x2· · ·xny1y2· · ·ym∈ C, since this is just another finite product of commutators. We also haveα−1= (x1x2· · ·xn)−1=x−1n · · ·x−12 x−11 .
Ifxi=aibia−1i b−1i , thenx−1i =biaib−1i a−1i , which is also a commutator. Thusα−1∈C, and henceC≤G.
To proveC is a normal subgroup ofG, letg∈G, andα=x1x2· · ·xn∈C. Then we have
gαg−1=gx1x2· · ·xng−1= (gx1g−1)(gx2g−1)· · ·(gxng−1), (1.3.1)
1.4. SIMPLE GROUP N. REHMAN
where we have just insertedgg−1=ebetweenxi andxi+1 for eachi < n. Now, ifxi =aibia−1i b−1i , then we have gxig−1=gaibia−1i b−1i g−1= (gaig−1)(gbig−1)(ga−1i g−1)(gb−1i g−1).
Now note that (gaig−1)−1= (g−1)−1a−1i g−1=ga−1i g−1, and we have the analogous statement if we replaceai bybi. So, we have
gxig−1= (gaig−1)(gbig−1)(gaig−1)−1(gbig−1)−1,
which is a commutator. Now, from??, we havegαg−1 is a product of commutators, and sogαg−1∈C. ThusCEG.
Definition 1.3.2. LetGbe a group. Then the subgroupCofGis called thecommutaor subgroup ofGand it is denoted byC=G0 orC= [G, G], that is
[G, G] =h[a, b]|a, b∈Gi and is also called thederived subgroupofG.
Remark 1.3.1. The product of two commutators need not itself be a commutator, and so the set of all commutators in Gis not necessarily a subgroup. To put this another way, the derived group G0 may contain elements which are not commutators
IfGis Abelian, then we haveC={e}, so in one sense the commutator subgroup may be used as one measure of how far a group is from being Abelian. Specifically, we have the following result.
Theorem 1.3.2. LetGbe a group, and letG0 be its commutator subgroup. Suppose thatNEG. ThenG/N is Abelian if and only ifG0⊆N. In particular,G/G0 is Abelian.
Proof. First assume thatG/N is Abelian. Let a, b∈G. Since we are assuming that G/N is Abelian, then we have (N a)(N b) = (N b)(N a), and soN ab=N baby the definition of coset multiplication in the factor group. Now, we know N ab=N baimplies ab(ba)−1 ∈N, whereab(ba)−1 =aba−1b1, and so aba−1b−1∈N. Sincea andb were arbitrary, any commutator inGis an element ofN, and sinceN is a subgroup of G, then any finite product of commutators in Gis an element ofN. ThusG0⊆N.
Now suppose thatG0 ⊆N, and leta, b∈G. Thenaba−1b−1∈N, and soab(ba)−1∈N. This impliesN ab=N ba, or that (N a)(N b) = (N b)(N a). Sinceaandbwere arbitrary, this holds for any elementsN a, N b∈G/N, and thusG/N is Abelian.
1.4 Simple Group
Lagrange’s theorem shows that the cyclic groupsGof prime order have no subgroups except{e}andG. More generally, ifGis a group then Such groups are calledsimple groups.
Definition 1.4.1. A groupGis said to besimple if it has no nontrivial normal subgroup, that is, the only normal subgroups ofGis either the trivial group (N =hei) orGitself.
Example 1.4.1. LetG= (Z,+) and N= (Zp,+p), wherepis prime.Then the quotient groupG/N is simple group.
By the Lagrange’s Theorem|N| divides the|G|. Since pis prime, it has only two divisor 1 and p. This means the only two subgroups are the trivial group and the entire group. Since there are no other subgroups the integer mod pare simple group. And because there are an infinite number of prime numbers, there are infinite number of simple groups.
The example of simple groupsZp , wherepis prime are trivially simple since they have no proper subgroups other than the subgroup consisting solely of the identity. Other examples of simple groups are not so easily found.
Theorem 1.4.1. An abelian groupG6={e} is simple if and only if it is cyclic of prime order.
Proof. SupposeGis simple and abelian. Ifg∈G, be a nonidentity element ofG. Then the grouphgigenerated by g is a subgroup ofG. SinceGis an abelian group, every subgroup is a normal subgroup. SinceGis simple, we must have hgi=G. If the order ofg is not finite, then hg2iis a proper normal subgroup of hgi=G, which is impossible sinceGis simple. Thus the order ofg is finite, and henceG=hgiis a finite group, say ordero(g) =n≥2. Ifp|nfor some primep, thenhgn/piis a subgroup of orderpby Theorem??. HenceG=hgn/piis cyclic of prime order.
Let us now suppose that the order of G is a prime. Let g ∈ G be a nonidentity element. Then the order of the subgrouphgimust be a divisor of the order ofG, hence it must bep. Therefore we haveG=hgi, and Gis a cyclic group and in particular an abelian group. Since any normal subgroup N of G has order 1 or p, N must be either trivial{e}orGitself. Hence Gis simple. Thus, Gis a simple abelian group.
We can, however, show that the alternating group,An, is simple forn≥5. The proof of this result requires several preliminary results.
1.4. SIMPLE GROUP N. REHMAN
Lemma 1.4.1. The alternating group An is generated by 3-cycles forn≥3.
Proof. To show that the 3-cycles generateAn, we need only show that any pair of transpositions can be written as the product of 3-cycles. Since (a, b) = (b, a), every pair of transpositions must be one of the following:
(a, b)(a, b) =e
(a, b)(c, d) = (a, c, b)(a, c, d) (a, b)(a, c) = (a, c, b).
Lemma 1.4.2. Let N be a normal subgroup of An, wheren≥3. If N contains a3-cycle, thenN =An.
Proof. We will first show that An is generated by 3-cycles of the specific form (i, j, k), where i and j are fixed in {1,2,· · · , n}and letkvary. Every 3-cycle is the product of 3-cycles of this form, since
(i, a, j) = (i, j, a)2 (i, a, b) = (i, j, b)(i, j, a)2 (j, a, b) = (i, j, b)2(i, j, a)
(a, b, c) = (i, j, a)2(i, j, c)(i, j, b)2(i, j, a).
Now suppose that N is a nontrivial normal subgroup of An for n ≥ 3 such that N contains a 3-cycle of the form (i, j, a). Using the normality ofN, we see that
[(i, j)(a, k)](i, j, a)2[(i, j)(a, k)]−1= (i, j, k)
is inN. Hence,N must contain all of the 3-cycles (i, j, k) for 1≤k≤n. By Lemma ??, these 3-cycles generateAn and hence,N =An.
Proposition 1.4.1. Forn≥5, every nontrivial normal subgroupN of An contains a 3-cycle.
Proof. Letπbe an arbitrary element in a normal subgroup N. There are several possible cycle structures for π.
• πis a 3-cycle.
• πis the product of disjoint cycles,π=π1(a1, a2· · ·ar)∈N, wherer >3.
• πis the product of disjoint cycles,π=π1(a1, a2, a3)(a4, a5, a6).
• π=π1(a1, a2, a3), whereπ1 is the product of disjoint 2-cycles.
• π=π1(a1, a2)(a3, a4), where π1 is the product of an even number of disjoint 2-cycles.
Ifπis a 3-cycle, then we are done.
If N contains a product of disjoint cycles, π, and at least one of these cycles has length greater than 3, say π = π1(a1, a2,· · ·ar), then (a1, a2, a3)π(a1, a2, a3)−1∈N.sinceN is normal; hence,π−1(a1, a2, a3)π(a1, a2, a3)−1is also in N. Since
π−1(a1, a2, a3)π(a1, a2, a3)−1=π−1(a1, a2, a3)π(a1, a3, a2)
= (a1, a2,· · ·, ar)−1π−11 (a1, a2, a3)π1(a1, a2,· · ·, ar)(a1, a3, a2)
= (a1, ar, ar−1,· · ·a2)(a1, a2, a3)(a1, a2,· · ·, ar)(a1, a3, a2)
= (a1, a3, ar), N must contain a 3-cycle; hence, N =An.
Now suppose thatNcontains a disjoint product of the formπ=π1(a1, a2, a3)(a4, a5, a6).Thenπ−1(a1, a2, a4)π(a1, a2, a4)−1∈ N since (a1, a2, a4)π(a1, a2, a4)−1∈N.
π−1(a1, a2, a4)π(a1, a2, a4)−1= [π1(a1, a2, a3)(a4, a5, a6)]−1(a1, a2, a4)π1(a1, a2, a3)(a4, a5, a6)(a1, a2, a4)−1
= (a4, a6, a5)(a1, a3, a2)π−11 (a1, a2, a4)π1(a1, a2, a3)(a4, a5, a6)(a1, a4, a2)
= (a4, a6, a5)(a1, a3, a2)(a1, a2, a4)(a1, a2, a3)(a4, a5, a6)(a1, a4, a2)
= (a1, a4, a2, a6, a3).
SoN contains a disjoint cycle of length greater than 3, and we can apply the previous case.
SupposeN contains a disjoint product of the form π = π1(a1, a2, a3), where π1 is the product of disjoint 2-cycles.
Sinceπ∈N,π2∈N, and
π2=π1(a1, a2, a3)π1(a1, a2, a3) = (a1, a3, a2).
1.5. EXERCISE N. REHMAN
SoN contains a 3-cycle.
The only remaining possible case is a disjoint product of the formπ=π1(a1, a2)(a3, a4), whereπ1 is the product of an even number of disjoint 2-cycles. Butπ−1(a1, a2, a3)π(a1, a2, a3)−1∈N since (a1, a2, a3)π(a1, a2, a3)−1∈N; and so
π−1(a1, a2, a3)π(a1, a2, a3)−1=π−11 (a1, a2)(a3, a4)(a1, a2, a3)π1(a1, a2)(a3, a4)(a1, a2, a3)−1
= (a1, a3)(a2, a4).
Sincen≥5, we can findb∈ {1,2,· · ·, n}such thatb6=a1, a2, a3, a4. Letβ= (a1, a3, b). Thenβ−1(a1, a3)(a2, a4)β(a1, a3)(a2, a4)∈ N and
β−1(a1, a3)(a2, a4)β(a1, a3)(a2, a4) = (a1, b, a3)(a1, a3)(a2, a4)(a1, a3, b)(a1, a3)(a2, a4)
= (a1, a3, b).
Therefore,N contains a 3-cycle. This completes the proof of the proposition.
Theorem 1.4.2. The alternating group,An, is simple forn≥5.
Proof. LetN be a normal subgroup of An. Then by Proposition ?? , N contains a 3-cycle and hence N =An, by Lemma??. Therefore,An contains no proper nontrivial normal subgroups forn≥5.
1.5 Exercise
Exercise 1.5.1. LetN be a subgroup ofG. Ifg2∈N for allg∈G, prove thatNEGandG/H is commutative.
Exercise 1.5.2. For each of the following groups G, determine whether N is a normal subgroup of G. IfN is a normal subgroup, write out a Cayley table for the factor groupG/N.
(i) G=S4 andN =A4
(ii) G=A5 andN ={e,(1,2,3),(1,3,2)}
(iii) G=Q8andN ={±1,±i}
(iv) G=ZandN = 5Z
Exercise 1.5.3. Let M(R) be the group of nonsingular upper triangular 2×2 matrices with entries in R; that is, matrices of the form
M(R) =
a b 0 c
|a, b, c∈Randac6= 0
. LetU consist of matrices of the form
U =
1 x 0 1
| x∈R
. 1. Show thatU is a subgroup ofT.
2. Prove thatU is abelian.
3. Prove thatU is normal inT.
4. Show thatT /U is abelian.
5. Is T normal inGL2(R)?
Exercise 1.5.4. Show thatN ={I,−I} is a normal subgroup ofGL2(R), whereI is the 2×2, identity mtrix.
Exercise 1.5.5. Find all the subgroups of the quaternion group, Q8. Which subgroups are normal?
Exercise 1.5.6. Suppose thatNEG. Then
(a) Prove that if [G:N] is prime, thenG/N is cyclic.
(b) Prove or disprove the converse of the statement in part (a)
Exercise 1.5.7. LetG=h6iandN =h24ibe subgroups ofZ. Show thatNEG. Write the cosets ofN in G. Also, write the Cayley table forG/N.
Exercise 1.5.8. LetG=Z48 andN=h8i.
(i) What is the order of the factor groupG/N?
(ii) What is the order of 2 +h8iin the factor groupG/N?
Exercise 1.5.9. Prove or disprove: IfN andG/N are cyclic, thenGis cyclic.
Exercise 1.5.10. Prove or disprove: IfNEGsuch thatN andG/N are abelian, thenGis abelian.
1.5. EXERCISE N. REHMAN
Exercise 1.5.11. Prove or disprove: If N is a normal subgroup of Gsuch thatN andG/N are abelian, then Gis abelian.
Exercise 1.5.12. LetG=A4 andN ={e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)}EG. Write out the distinct elements of G/N and make a Cayley table forG/N.
Exercise 1.5.13. IfKEN andNEG, then show thatgKag−1EN for allg∈G.
Exercise 1.5.14. Prove that ifNEGandg∈Gthen|N a|/o(g).
Exercise 1.5.15. IfH is an abelian group withHEGandKbe any subgroup of G, then show that H∩KEHK.
Exercise 1.5.16. IfN is a subgroup ofGsuch that the product of two right cosets ofN in Gis again a right coset ofN in G, prove thatNEG.
Exercise 1.5.17. Determine the quotient groups of (a) (2Z,+) in (Z,+),
(b) (Z,+) in (Q,+), (c) (4Z,+) in (Z,+).
Exercise 1.5.18. Let X be a nonempty subset of a group G. The setN(X) ={g ∈G|gXg−1 =X} is called the normalizer ofX in G.
(a) Prove thatN(X) is a subgroup ofG. SupposeH is a subgroup ofG.
(b) Prove thatX is normal inGif and only ifN(X) =G.
(c) Prove that X is normal inN(X).
(d) Prove that N(X) is the largest subgroup of G in which X is normal. That is, if X EK and K ≤ G, then K⊆N(X).
Exercise 1.5.19. LetNEGand letg∈G. If|N g|= 5 and|N|= 4, then what are the possiblilities for the order of g.
Exercise 1.5.20. LetGbe a group such that|G|=pqfor some prime numberspandq. Prove that either|Z(G)|= 1 orGis abelian.
Exercise 1.5.21. LetNEGsuch that |N|= 2. Show thatN ⊆Z(G). Is this true when|N|= 3?
Exercise 1.5.22. LetN ≤K≤Gbe finite groups withNEG, Show thatK/N≤G/N and [G/N :K/N] = [G:K].
Exercise 1.5.23. LetGbe a finite group andNEGsuch that|N|= 7 and|N g|= 4 inG/N for someg∈G. Show thatGhas a subgroup of order 28.
Exercise 1.5.24. Give an example of nonabelain groupGwhich has a normal subgroup N and G/N is cyclic.
Exercise 1.5.25. IfKEGhas indexr. Then (i) show thatgr∈K for allg∈G.
(ii) ifgcd(r, n) = 1, show thatK contains every element ofGof ordern.
Exercise 1.5.26. LetKEG.
(i) Show that [Ka, Kb] =K[a, b] for alla, b∈G.
(ii) IfKEG0, then show that (G/K)0 =G0/K.
Chapter 2
Homomorphisms and Isomorphisms of Groups
In this chapter, we consider certain mappings between groups. These mappings will be defined in such a way as to preserve the algebraic structure of the groups involved. More precisely, If G and G1 are any groups, and there is a function ϕwhich transforms Ginto G1, we say that G1 is a homomorphic image of G. The function ϕ is called HomomorphismsfromGtoG1.
2.1 Group Homomorphisms
Definition 2.1.1. Let (G,∗) and (G1,) be any two groups andϕa function fromGinto G1, ϕ1:G→G1. Then ϕis said to be ahomomorphism(or operation preserving function) fromGintoG1 if and only if
ϕ(x∗y) =ϕ(x)ϕ(y) for every pair of elementsa, b∈G.
Remark 2.1.1. Notice that on the left hand side of the above equation, the product x∗y is computed in G, while on the right side the productϕ(x)·ϕ(y) is that of elements ofG1. The functions indicated in this definition have the charaeteristic property of carrying products into products. A simple image of this shows in Fig. 1.
x
x∗y
y
ϕ(x)
ϕ(x)ϕ(y)
ϕ(y) Fig. 1
Example 2.1.1. A trivial example of a homomorphism, but an important one, is theidentity function. For an abritrary group (G,∗) define the function iG : G→G byiG(x) =xfor all x∈G. It should be clear that this is a homomorphism, since for allx, y∈Gwe have
iG(x∗y) =x∗y=iG(x)∗iG(y).
Example 2.1.2. Suppose that (G,∗) and (G1,·) are two groups with identity elements eand e1, respectively. The functionϕ:G→G1 given byϕ(x) =e1for eachx∈Gis a homomorphism:
ϕ(x∗y) =e1=e1·e1=ϕ(x)·ϕ(y).
This particullar mapping is the only constant function which satisfies Definition??. This homomorphism is called the trivial homomorphism.
2.1. GROUP HOMOMORPHISMS N. REHMAN
Example 2.1.3. Let (Z,+), and G ={2n|n ∈ Z}, (G, .) is a group under multiplication. Ifϕ : Z →G define by ϕ(n) = 2n, thenf is a homomorphism fromZtoG.
ϕ(m+n) = 2m+n= 2m.2n =ϕ(m)·ϕ(n).
Example 2.1.4. Define a map ϕ : (Z,+) → (2Z,+) as f(n) = 2n for any n ∈ Z is a homomorphism. For any m, n∈Z
ϕ(m+n) = 2(m+n) = 2m+ 2n=ϕ(m) +ϕ(n).
Example 2.1.5. The mapping ϕ(x) = x2 from R∗, the set of nonzero real numbers (R∗ = R\ {0}) to itself is a homomorphism, since
ϕ(ab) = (ab)2=a2b2=ϕ(a)ϕ(b) for alla, b∈R∗.
Example 2.1.6. Define a fiunctionϕ: (R,+)→(R,+) asϕ(x) =x2 for anyx∈Ris not a homomorphism, since ϕ(a+b) = (a+b)2=a2+ 2ab+b26=ϕ(a) +ϕ(b)
sinceϕ(a) + (b) =a2+b2.
Example 2.1.7. Let (Z,+) be the group of integers under addition and (Zn,+n) be the group of integers modulon.
Defineϕ:Z→Zn by lettingϕ(x) = [x]. Since
ϕ(x+y) = [x+y] = [x] +n[y] =ϕ(x) +nϕ(y).
Example 2.1.8. LetGLn(R) denote the general linear group ofn×ninvertible matrices overR. Define a function ϕ:GLn(R)→R∗byϕ(A) = det(A) for all A∈GLn(R).
For any two square matricsAandB, det(AB) = det(A) det(B). Thus,
ϕ(A·B) = det(A·B) = det(A)·det(B) =ϕ(A)·ϕ(B) implying thatϕis homomorphism of groups.
Definition 2.1.2. (i) A homomorphismϕ:G→G1is called amonomorphismifϕis one-one.
(ii) A homomorphismϕ:G→G1 is called anepimorphismifϕis onto.
Example 2.1.9. 1. Defineϕ:Z→Zbyϕ(x) = 2x. Then
ϕ(x+y) = 2(x+y) = 2x+ 2y=ϕ(x) +ϕ(y),
so that ϕ is a homomorphism. Note that ϕ is not onto since there is no integer n that satisfies ϕ(n) = 3.
However, ϕis one-to-one sincev(n) =ϕ(m) implies 2n= 2mand this in turn implies thatn=m. Hence,ϕis monomorphism.
2. In Example ??, ϕ(n) = ϕ(2n) with n 6= 2n so that ϕ is not one-to-one. However, ϕ is onto. Hence, ϕ is epimorphism.
Definition 2.1.3(Kernel of a Homomorphism). Letϕ:G→G1 be a homomorphism. Thekernel of a homomor- phismϕfrom a group Gto a groupG1 is the set
kerϕ={x∈G|f(x) =e1, e1 is the identity ofG1}.
The kernel ofϕis denoted by kerϕ.
Example 2.1.10. 1. In Example??,x∈kerϕif and only if 2x=ϕ(x) = 0, that is, iffx= 0. Hence, kerϕ={0}.
2. In Example??,x∈kerϕif and only if [x] =ϕ(x) = [0], that is iffx=nqfor someq∈Z. Thus kerϕ={nq|q∈ Z}.
3. In Example ??, A∈kerϕif and only if detA=ϕ(A) = 1, since 1 is the identity of R∗, that is iff det(A) = 1.
Thus, the kerϕof this homomrphism is alln×nmatrices having determinant one and hence kerϕ=SLn(R).
Example 2.1.11. LetGbe a group. Suppose thatg∈Gandϕis the homomorphism fromZtoGgiven byϕ(k) =gk. If the order ofg is infinite, then the kernel of this homomorphism is {0} sinceϕmaps Zonto the cyclic subgroup of Ggenerated byg. However, if the order ofgis finite, say n, then the kernel ofϕisnZ.
2.1. GROUP HOMOMORPHISMS N. REHMAN
2.1.1 Properties of Homomorphisms
Theorem 2.1.1. Let ϕ: (G,∗)→(G1,)be a homomorphism. Then 1. ϕ(e) =e1, wheree ande1 are identities ofG andG1 respectively.
2. ϕ(x−1) = (ϕ(x))−1, for anyx∈G.
3. kerϕis a subgroup ofG, i.e., kerϕ≤G.
4. kerϕis a normal subgroup inG, i.e., kerϕEG.
5. ϕ(xk) = (ϕ(x))k, wherek∈Z.
6. if x∈G witho(x) =n, theno(ϕ(x))|n.
Proof. 1. Ifeis the identity ofG, thene∗e=e
ϕ(e) =ϕ(e∗e) =ϕ(e)ϕ(e).
Sincee∈Gso,ϕ(e)∈G1. The identity ofG1 ise1. Hence
e1ϕ(e) =ϕ(e) =ϕ(e)ϕ(e).
Therefore, by cancellation lawe1=ϕ(e) 2. It is first necessary to show that
ϕ(x)ϕ(x−1) =e1=ϕ(x−1)ϕ(x).
We can then conclude from the uniqueness of the inverse ofϕ(x) in (G1,) thatϕ(x−1) = (ϕ(x))−1. To obtain this result, we make use of part 1.to get
ϕ(x)ϕ(x−1) =ϕ(x∗x−1) =ϕ(e) =e1. Similarly
ϕ(x−1)ϕ(x) =ϕ(x−1∗x) =ϕ(e) =e1.
3. First, we shall show that kerϕis a subgroup ofG. Sinceϕ(e) =e1, whereeis the identity ofGande1is identity ofG1. Hence kerϕ6=∅. Now, leta, b∈kerϕ. Thenϕ(a) =e1 andϕ(b) =e1. Thus,
ϕ(a∗b−1) =ϕ(a)ϕ(b−1) =ϕ(a)(ϕ(b))−1=e1(e1)−1=e1e1=e1. Hence ab−1∈kerϕ. Then by Theorem??kerϕis a subgroup ofG.
4. By 3.kerϕis a subgroup inG. Now, we only show that for anya∈kerϕandg∈G gag−1∈kerϕ.
Leta∈kerϕandg∈G. Then, by 1.and 2., we get
ϕ(gag−1) =ϕ(g)ϕ(a)ϕ(g−1) =ϕ(g)e1(ϕ(g))−1=ϕ(g)(ϕ(g))−1=ϕ(gg−1) =ϕ(e) =e1. Thus,gag−1∈kerϕ.
5. We consider the cases,k= 0, k >0, andk <0. The casek= 0 is just1.. For the casek >0, we use induction onk≥1. The result is true fork= 1. Assume it is true up tok−1. Then
ϕ(xk) =ϕ(xk−1x) =ϕ(xk−1)ϕ(x) = (ϕ(x))k−1ϕ(x) = (ϕ(x))k. Hence, the result is true for allk >0. Ifk <0 then
ϕ(xk) =ϕ((x−1)−k) = (ϕ(x−1))−k = ((ϕ(x))−1))−k= (ϕ(x))k. 6. Sinceo(x) =n, xn=e. Then
e1=ϕ(e) =ϕ(xn) = (ϕ(x))n. By Theorem??(i)o(ϕ(x))|n.
2.1. GROUP HOMOMORPHISMS N. REHMAN
Example 2.1.12. Consider two groups (Z,+) and (R∗,·). The mappinfgϕ:Z→R∗defined by ϕ(x) =
(1 ifxis even
−1 ifxis odd.
is a homomorphism, as the reader may verifiy by checking the various cases that could arise. In the situation considered, kerϕ={x∈Z| ϕ(x) = 1}= 2Z
while the direct image
ϕ(Z) ={1,−1}.
It is not particularly difficult to show that (2Z,+) is a normal subgroup of (Z,+) and that ({1,−1},·) is a subgroup of (R∗,·).
Theorem 2.1.2. Let ϕ:G→G1 be a homomorphism from GtoG1. Then 1. If H is a subgroup ofG, then H1=ϕ(H)is a subgroup ofG1.
2. If H is normal in Gandϕis onto, then H1=ϕ(H)is normal in G1. 3. If H1 is a subgroup ofG1, thenH =ϕ−1(H1)is a subgroup ofG.
4. If H1 is normal in G1, then H=ϕ−1(H1)is normal in G.
Proof. 1. Leta1, b1∈H1. Then there exist a, b∈H such thata1=ϕ(a) andb1=ϕ(b) a1b−11 =ϕ(a)(ϕ(b))−1=ϕ(a)ϕ(b−1) =ϕ(ab−1).
This implies that ϕ(ab−1)∈H1 (H is a subgroup iffab−1∈H for anya, b∈H) and hencea1b−11 ∈H1. Thus, H1is a subgroup of G1.
2. Letg1 ∈G1 andh1∈H1=ϕ(H). then there existsh∈H such that h1=ϕ(h). Since ϕis onto and g1∈G1, there existsg∈Gsuch thatϕ(g) =g1.
g1h1g1−1=ϕ(g)ϕ(h)(ϕ(g))−1=ϕ(g)ϕ(h)ϕ(g−1) =ϕ(ghg−1)
(H is normal inGiffghg−1∈H, for allg∈G, h∈H). This implies thatg1h1g−11 =ϕ(H) =H1 and henceH is normal inG1.
3. Leta, b∈H =ϕ−1(H1). Sinceϕ−1(H1) ={x∈G|ϕ(x) =H1}, ϕ(a),ϕ(b)∈H1. but H1 is a subgroup ofG1, we find thatϕ(a)(ϕ(b))−1 ∈H1 and hence ϕ(ab−1) =ϕ(a)ϕ(b−1) =ϕ(a)(ϕ(b))−1 ∈H1. Thus, ϕ(ab−1)∈H1 that is,ab−1∈H and henceH is a subgroup of G.
4. Letg∈Gandh∈H=ϕ−1(H1). Thenϕ(g)∈G1andϕ(h)∈H1. SinceH1 is normal inG1. ϕ(g)ϕ(h)(ϕ(g))−1=ϕ(g)ϕ(h)ϕ(g−1) =ϕ(ghg−1)∈H1,
and henceghg−1∈ϕ−1(H1) =H for anyg∈G, h∈H. Therefore,H is normal inG.
We have seen that every homomorphism determinnes a normal subgroup by means of its kernal. Though at first it is not evident that factor groups correspond exactly to homomorphic images, we can use factor groups to study homomorphisms. We already know that with every group homomorphism ϕ : G → G1 we can associate a normal subgroup of G, kerϕ; the converse is also true. The following theorem will show that every normal subgroup of a groupGgives rise to homomorphism, then so-callednaturalorcanonical homomorphism
Theorem 2.1.3. Let Gbe a group and N a normal subgroup ofG, then the mappingϕ:G→G/N defined by η(x) =N x
is homomorphism fromGontoG/N; the kernal of ϕis preciesly the set ofN.
2.2. ISOMORPHISM N. REHMAN
Proof. The fact that the mapping η is homomrphism follows directly from the manner in which multiplication is defined in the quotient group. For anyx, y∈G;
η(xy) = N(xy)
= (N x)(N y) =η(x)η(y)
Moreover,η is onto, because for anyN z∈G/N, there existsz∈Gsuch that,η(z) =N z. This implies that G/N is homomorphic image ofG.
As the cosetN serves as the identity element forG/N, we must have ker(η) ={x∈G|η(x) =N}
={x∈G|N x=N}=N.
Theorem 2.1.4. Letϕ:G→G1be a homomorphism fromGtoG1. Thenϕis a one-to-one if and only ifkerϕ={e}, whereeis identity of G.
Proof. Suppose the functionϕis one-to-one. We already know that e∈ kerϕ. Our aim is to show that this is the only element in the kernel. If there existed another element a∈kerϕ a6=e, then we would have ϕ(a) = e1=ϕ(e).
That is,ϕ(a) =ϕ(e) buta6=e. This would contradict the hypothesis thatϕis one- to-one.
Conversely, suppose that ker ϕ={e}. Leta, b∈Gandϕ(a) =ϕ(b). To proveϕis one-to-one, we must show that a=b. But ifϕ(a) =ϕ(b) then
ϕ(ab−1) =ϕ(a)ϕ(b−1) =ϕ(a)(ϕ(b))−1=ϕ(a)(ϕ(a))−1=e1.
Which implies thatab−1∈kerϕ. But, kerϕ={e} Therefore,ab−1=eora=band henceϕis one-to-one.
Theorem 2.1.5. Let ϕ:G→G1 be a homomorphism from a group Gonto a groupG1. 1. If Gis cyclic, thenG1 is also cyclic.
2. If Gis abelian, thenG1 is also abelian.
Proof. 1. Given thatGis cyclic. Suppose thatGis generated byathat isG=< a >. Ify∈G1, then there exists x∈Gsuch thatϕ(x) =y(as ϕis onto). Since x∈GandG=< a >,x=am, for some m.
y=ϕ(x) =ϕ(am) =ϕ(a·a· · ·.a)
| {z }
m−times
=ϕ(a)·ϕ(a)· · ·ϕ(a)
| {z }
m−times
= (ϕ(a))m. That is,G1=hϕ(a)iand henceG1 is also cyclic.
2. Leta1, b1 ∈G1. Since ϕis onto, there exists a, b∈Gsuch thata1 =ϕ(a), b1 =ϕ(b) and ab=ba, since G ia abelian. Then
a1b1=ϕ(a)ϕ(b) =ϕ(ab) =ϕ(ba) =ϕ(b)f(a) =b1a1, for anya1, b1∈G1. Hence ,G1 is abelian.
2.2 Isomorphism
We start this section by constracting the Cayley tables of the groups (Z3,+3) and (h1,2,3i,◦).
+3 [0] [1] [2]
[0] [0] [1] [2]
[1] [1] [2] [0]
[2] [2] [0] [1]
◦ e (1,2,3) (1,3,2) e e (1,2,3) (1,3,2) (1,2,3) (1,2,3) (1,3,2) e (1,3,2) (1,3,2) e (1,2,3)
By setting the correspondence [0] ⇐⇒ e, [1] ⇐⇒ (1,2,3), [2] ⇐⇒ (1,3,2) we see that the two tables differ only in the names of the symbols, and not in their positions. If we name the mentioned correspondence by the letterϕ, i.e.
by writingϕ([0]) =e,ϕ([1]) = (1,2,3) andϕ([2]) = (1,3,2), and then constructing a Venn diagram of this mapping we see thatϕis a bijection mapping with the propertyϕ([a] +n[b]) =ϕ([a])◦ϕ([b]).
[0]
[1]
[2]
e (1,2,3) (1,3,2)
2.2. ISOMORPHISM N. REHMAN
This example, leads to the following definition.
Definition 2.2.1. Two groups (G,∗) and (G1,) are said to beisomorphic, denoted byG∼=G1, if there exists a bijective homomorphism ϕ:G→G1, that isϕ(G) =G1. Such a homomorphismϕis called an isomorphisms, or isomorphic mapping, of (G,∗) onto (G1,). If (G,∗) = (G1,), then we callϕanautomorphismonG.
Example 2.2.1. In Example??, clearly,ϕis onto andϕone-to-one becauseϕ(n) =ϕ(m) implies 2n= 2mand this in turn implies thatn=m. Hence,ϕis homomorphism and both one-to-one and onto. Hence (Z,+)∼= (2Z,+).
Example 2.2.2. Consider the two groups (Z4,+4) and (G,·), whereG={1,−1, i,−i}.
We wish to prove that the groups (Z4,+4) and (G,·) are abstractly ”equal.” To do so, we must produce a one- to-one homomorphism ϕ from Z4 onto G. Since the preservation of identity clcments is a general feature of any homomorphism,ϕmust be such thatϕ([0]) = 1. Let us suppose for the moment that we were to defineϕ([1]) =−1.
The image of an inverse clement must equal the inverse of the image. We would then have ϕ([3]) =ϕ([1]−1) = (ϕ([1]))−1= (−1)−1=−1,
or ϕ([3]) =ϕ([1]). This, however, would preventϕ from being one-to-one. A more appropriate choice, in the sense that it avoids the above difficulty, is to take ϕ([1]) =i. The condition on inverses then impliesϕ([3]) =−i. Sinceϕ is further required to preserve modular addition,
ϕ([2]) =ϕ([1] +4[1]) =ϕ([1])·ϕ([1]) =i·i=−1.
Wo are thus led in a natural way to consider the function defined by
ϕ([0]) = 1, ϕ([1]) =i, ϕ([2]) =−1, ϕ([3]) =−i
Clearly this function is a one-to-one mapping of the set Z4. onto the set G. Furthermore, ϕ actually preserves the operations of the groups. we observe that
ϕ([1] +4[2]) =ϕ([1 +42]) =ϕ([3]) =−i=i· −1 =ϕ([1])·ϕ([2]).
Consequently, we have (Z4,+4)∼= (G,·).
Example 2.2.3. Two groups (R,+) and (R∗,·) are not isomorphic.
To see this, suppose that there exists a one-to-one onto functionϕ: (R,+)→(R∗,·) with the property ϕ(x+y) =ϕ)(x)·ϕ(y)
for allx, y ∈ R. The identity element in R is 0 ∈R and the identity element in R∗ is 1 ∈ R∗. If x∈R such that ϕ(x) =−1, then
ϕ(2x) =ϕ(x+x) =ϕ(x)·ϕ(x) = (−1)·(−1) = 1.
By the Theorem??, the identity elemnt of (R,+) is the unique clement ofRcorresponding to the identity of (R∗,·), so that 2x= 0, or x= 0. Consequently, bothϕ(0) = 1, andϕ(0) =−1, contradicting the fact that the functionϕis one-to-one. This argument shows that (R,+) cannot be isomorphic to (R∗,·),for no function satisfying Definition??
can exist.
Example 2.2.4. Two groups (R,+) and (R+,·) are isomorphic.
To prove this, we must produce a one-to-one homomorphismϕfrom (R,+) onto (R+,·) . Defineϕ(x) =ex, where exis the exponentiat function.
ϕ(x+y) =ex+y=ex·ey=ϕ(x)·ϕ(y),
for allx, y∈R. This shows thatϕis homomorphism. Now, to showϕis one-to-one, letϕ(x) =ϕ(y), for allx, y∈R. Thenex=ey, so if lnrdenoted the natural logarithm ofr,x= ln(ex) = ln(ey) =y. Thus,ϕis one-to-one. Ita∈R+, thena >0 , soa∈Randϕ(lna) =elna=a. Hence,ϕis onto. This implies that (R,+)∼= (R+,·).
2.2.1 Properties of Isomorphims
Theorem 2.2.1. Let ϕ:G→G1 andΘ :G1→K be two group isomorphisms. Then (a) ϕ−1:G1→Gis an isomorphism.
(b) Θ◦ϕ:G→K is also an isomorphism.
2.2. ISOMORPHISM N. REHMAN
Proof. (a) First we show that ϕ−1 is one-to-one. Ifϕ−1(g1) =ϕ−1(g2) then ϕ(ϕ−1(g1)) =ϕ(ϕ−1(g2)) or eG1(g1) = eG1(g2). Hence, g1 = g2. To see that ϕ−1 is onto, let g ∈ G. Then ϕ(g) ∈ G1 and ϕ−1(ϕ(g)) = g. Finally, we show that ϕ−1 is a homomorphism. If x, y ∈ G1 then x =ϕ(a) and y = ϕ(b) for some a, b ∈ G, since ϕ is onto.
Hence, ifz=ϕ−1(xy) then ϕ(z) =xy=ϕ(a)ϕ(b) =ϕ(xy). Thus,ϕ−1(xy) =ab=ϕ−1(x)ϕ−1(y). Hence,ϕ−1 is an isomorphism.
(b) To see that Θ◦ϕ is one-to-one, suppose that Θ◦ϕ(x) = Θ◦ϕ(y). Then Θ(ϕ(x)) = Θ(ϕ(y)). Since ϕ is one-to-one then Θ(x) = Θ(y). Since Θ is one- to-one thenx=y. To see that Θ◦ϕis onto, letk∈K. Since Θ is onto then we can findg1∈G1 such that Θ(g1) =k. Sinceϕis onto then we can find ag∈Gsuch thatϕ(g) =g1. Hence, Θ(ϕ(g)) = Θ(g1) =k. To see that Θ◦ϕis a homomorphism, let x, y∈G. Since bothϕand Θ are homomorphisms then
(Θ◦ϕ)(xy) = Θ(ϕ(xy)) = Θ(ϕ(x)ϕ(y)) = Θ(ϕ(x))Θ(ϕ(y)) = (Θ◦ϕ)(x)(Θ◦ϕ)(y).
Theorem 2.2.2. Let GandG1 be groups such thatϕ:G→G1 is an isomorphism.
(a) |G|=|G1|.
(b) If G=haithen G1=hϕ(a)i. That is, ifGis cyclic, thenG1 is also cyclic.
(c) If H is a subgroup ofGof ordernthenϕ(H) is a subgroup ofG1 of ordern.
(d) If a∈Gando(a) =ntheno(ϕ(a)) =n.
Proof. (a) Asseration (a) follow from the fact thatϕis bijective.
(b) Suppose that Gis cyclic with generatora. We will show thatG1=hϕ(a)i. Clearly, since ϕ(a)∈G1 and G1 is a group thenhϕ(a)i ⊆G1. Now suppose thatg1∈G1 then there is ang∈Gsuch thatϕ(g) =g1 (sinceϕis onto).
But theng = an for some n ∈Z. Hence, by Theorem ?? 5, ϕ(g) =ϕ(an) = (ϕ(a))n. That is,g1 ∈ hϕ(a)i and so G1⊆ hϕ(a)i.
(c) LetH is a subgroup ofG. Sinceϕ(e) =e1, thene1∈ϕ(G1), soϕ(H)6=∅. Now, ifϕ(x), ϕ(y)∈ϕ(H) then ϕ(x)(ϕ(y))−1=ϕ(x)ϕ(y−1) =ϕ(xy−1)∈ϕ(G1)
since by closure, xy−1 ∈ H, Hence, by Theorem ?? ϕ(H) is a subgroup G1. The mapping ϕ restricted to H is a one-to-one mapping fromH ontoϕ(H). Thus,|H|=|ϕH|.
(d) If o(a) = n then an =e. By Theorem ?? 5, we have (ϕ(a))n = ϕ(an) = ϕ(e) = e1 . By Theorem ?? (i), o(ϕ(a))|n. On the other hand, since (ϕ(a))o(ϕ(a)) =e1 then ϕ(ao(ϕ(a))) =e1 =ϕ(e). Thus, ao(ϕ(a)) =e(since ϕis one-to-one) and by Theorem??(i),n|o(ϕ(a)). Hence,o(ϕ(a))|nandn|o(ϕ(a)) gives thato(ϕ(a)) =n.
Theorem 2.2.3. (i) IfGis a cyclic group of order nthen G∼=Zn. (ii) If Gis an infinite cyclic group thenG∼=Z.
Proof. (i) Suppose thatG=hai. Defineϕ:G→Zn byϕ(ak) = [k] where 0≤k < n. We show thatϕis well-defined.
Indeed, ifak =amthenak−m=eso that by Theorem??(i),n|(k−m). Thus,a≡m( mod n) and hence [k] = [m].
Next, we show thatϕis one-to-one. Ifϕ(ak) =ϕ(am) then [k] = [m] and this implies thatn|(k−m). Thus,k−m=nq for someq∈Z.Thereforeak−m=anq = (an)q =e. Henceak=am.
Next, we show that ϕ is onto. Let x ∈ Z. Then by the Division Algorithm, x = nq+r, 0 ≤ r < n. Hence, ax= (an)qar=ar ∈G and ϕ(ax) =ϕ(ar) = [r]. Since x−r =nq thenn|(x−r so that x≡r( mod n). Hence, [x] = [r]. Finally, it remains to show thatϕis a group homomorphism. Indeed,
ϕ(akam) =ϕ(ak+m) = [k+m] = [k] +n[m] =ϕ(ak)ϕ(am).
(ii) LetG be a cyclic group with infinite order and suppose thatais a generator of G. Define a map ϕ:Z→G by ϕ(n) = an. The ϕ is well-defined mapping. Indeed, if n = m then an = am. We will next show that ϕ is momomorphism that is,
ϕ(m+n) =am+n=aman=ϕ(m)ϕ(n).
To show thatϕis one-to-one, suppose thatmandnare two elements inZ, wherem6=n. We can assume thatm > n.
We must show thatam6=an. Let us suppose the contrary; that is,am=an. In this caseam−n =e, wherem−n >0, which contradicts the fact thatahas infinite order. Its remains to show thatϕonto, pick anz∈G. Thenz=an for somen∈Zandϕ(n) =an=z.
Corollary 2.2.1. If |G|=p, wherepis a prime number, thenG∼=Zp.
Proof. SinceGhas a prime order then by Corollary ??,Gis cyclic. By Theorem?? (i),G∼=Zp. Corollary 2.2.2. Any two cyclic groups of the same order are isomorphic.
