2. normal subgroup and quotient group We begin by stating a couple of elementary lemmas. 2.1. Lemma. Let A and B be sets and f : A → B be an onto function. For b ∈ B,recall that f −1(b)={a ∈ A: f(a)=b}.LetF = {f −1(b): b ∈ B}.ThenF is a partition of A and there is a natural one to one correspondence between B and F given by b #→ f −1(b). 2.2. Lemma. Let A and B be subsets of a group G.Ifg,h ∈ G,theng(hA)=(gh)A.If A ⊆ B and g ∈ G,thengA ⊆ gB. The proofs of the two lemmas above will be left as exercises. We often use easy results like these without mentioning explicitly. 2.3. Lemma. Let φ : G → H be a group homomorphism and let N =ker(φ).Then φ−1(φ(g)) = gN = Ng for all g ∈ G. Proof. Take g ∈ G.Letx ∈ φ−1(φ(g)) or in other words, φ(x)=φ(g). Then −1 −1 −1 φ(g x)=φ(g) φ(x)=φ(g) φ(g)=eH, so g−1x ∈ N, or in other words x ∈ gN.Thusφ−1(φ(g)) ⊆ gN. Conversely, if x ∈ gN, −1 then x = gn for some n ∈ N,soφ(x)=φ(g)φ(n)=φ(g)eH = φ(g), so x ∈ φ (φ(g)). Thus gN ⊆ φ−1(φ(g)). This proves φ−1(φ(g)) = gN. A similar argument shows that φ−1(φ(g)) = Ng (Exercise: Write out this argument). ! 2.4. Definition. Let G be a group. A subgroup N of G is called a normal subgroup if gN = Ng for all g ∈ G. In this case we need not distinguish between left and right cosets. So we simply talk of cosets of N. 2.5. Lemma. Let G be a group and N be a subgroup of G.Thefollowingareequivalent. (a) gN = Ng for all g ∈ G. (b) gNg−1 = N for all g ∈ G. (c) If g ∈ G and n ∈ N,thengng−1 ∈ N. Proof. Suppose (a) holds. Then gNg−1 =(gN)g−1 =(Ng)g−1 = N(gg−1)=N.Thus(a) implies (b). Clearly (b) implies (c). Assume (c). Let x ∈ gN.Thenx ∈ gn for some n ∈ N. So x = gng−1g ∈ Ng since gng−1 ∈ N by our assumption. Thus gN ⊆ Ng. Similarly show that Ng ⊆ gN.SogN = Ng. Thus (c) implies (a). ! 2.6 (Motivating the construction of a quotient). Lemma 2.3 says that kernels of ho- momorphisms are normal subgroups. Conversely, we shall see that all normal subgroups appear as kernels of homomorphisms. Let G be a group and let N be a normal subgroup of G. We shall construct a onto group homomorphism π : G → Q such that ker(π)=N. This group Q will be called the quotient of G by the normal subgroup N and will be denoted by Q = G/N. To motivate the construction, suppose we have an onto group homomorphism π : G → Q and let ker(π)=N.Letq ∈ Q. Since π is onto, we can pick an element g ∈ G such that π(g)=q. Then Then lemma 2.3 gives us π−1(q)=gN. In words, this says that the preimages of elements of Q are just the cosets of N. So the elements of Q are in one to one correspondence with the set of cosets of N in G. Also verify that if q1 and q2 are two elements −1 −1 −1 of Q such that π (q1)=g1N and π (q2)=g2N,thenπ (q1q2)=g1g2N (Exercise: verify this). This suggests the definition that follows. 3 2.7. Definition. Let G be a group and N be a normal subgroup. Let G/N denote the set of cosets of N in G.LetC and D be two cosets of N in G. We say that an element c ∈ G is a representative of the coset C if c ∈ C. Choose a representatitve c of the coset C and a representative d of the coset D. Claim: The coset cdN only depends on the cosets C and D and does not depend on the choice of the representatives c and d. Proof. Let c1 be another representative of C and d1 be another representative of D.Then −1 −1 −1 −1 c1c ∈ N and d1d ∈ N. Since N is normal, c(d1d )c ∈ N as well. So −1 −1 −1 −1 −1 −1 (c1d1)(cd) = c1d1d c =(c1c )c(d1d )c ∈ N. So c1d1 ∈ Ncd = cdN.Soc1d1N = cdN. ! Define a binary operation on G/N as follows: Given two cosets C and D,wechoosecoset representatives c and d of C and D respectively and let let (C.D)tobethecosetcdN.The claim we just proved imply that this procedure gives a well defined binary operation on G/N. Let us reiterate the definition of the binary operation in slightly different words. If c ∈ C and d ∈ D,thenC = cN and D = dN. So our definition says (cN).(dN)=cdN. Let us also reiterate that we have already argued this is a well defined operation on G/N and this argument required the fact that N is a normal subgroup. Next one verifies that G/N with the binary operation just defined becomes a group where the identity is the coset N and inverse of gN is g−1N. This group G/N is called the quotient of G by N. There is a canonical onto function from π : G → G/N given by π(g)=gN. Verify that π is a homomorphism and ker(π)=N. 2.8. Example. Verify that Zn = Z/nZ. 3. Constructing homomorphisms 3.1. Lemma. Let G be a group and a ∈ G.Thenthereexistsahomomorphismφ : Z → G given by φ(n)=an for all n ∈ Z.Thisφ is the unique homomorphism from Z to G such that φ(1) = a. sketch of proof. Define φ(n)=an for all n ∈ Z. Verify that φ : Z → G is a homomorphism such that φ(1) = a.Letφ′ : Z → G be any homomorphisms such that φ′(1) = a.Then φ′(n)=an = φ(n) for all n ∈ Z.Soφ′ = φ. ! The above lemma says that constructing homomorphisms from Z to any group is easy. You can send 1 to anything you want and the homomorphism is completely determined once you decide where 1 goes. 3.2. Corollary. The map φ(k)=k mod n is the unique homomorphism φ : Z → Zn such that φ(1) = 1 mod n.ThekernelofthishomomorphismisnZ. 3.3. Lemma. Let φ1 : G → H1 and φ2 : G → H2 be two homomorphisms. Then there is a homomorphism φ : G → H1 × H2 given by φ(G)=(φ1(G),φ2(G)). 4 3.4. Theorem. Let A, B, C be three groups and let f : A → B and g : A → C be two homomorphisms. (i) if there exists a homomorphism h : B → C such that h ◦ f = g,thenker(f) ⊂ ker(g). (ii) Suppose f is onto and ker(f) ⊂ ker(g). (a) Then there exists a unique homomorphism h : B → C such that h ◦ f = g. (b) If g is onto, then so is h. (c) If ker(g)=ker(f),thenh is one to one. (d) If g is onto and ker(g)=ker(f),thenh is an isomorphism. Proof. (i) Exercise. (ii) (a) Uniqueness: Suppose h, h′ be two homomorphisms such that h ◦ f = g = h′ ◦ f. Pick b ∈ B. Since f is onto there exists a ∈ A such that f(a)=b.thenh(b)=h(f(a)) = g(a)=h′(f(a)) = h′(b). Fix b ∈ B.Supposea, a′ ∈ A such that f(a)=f(a′)=b,soh = h′. Thus there can be atmost one h with the given properties. Existence: Let b ∈ B.Supposea, a′ ∈ A such that f(a)=f(a′)=b.Thenf(a−a′)=0, so (a − a′) ∈ ker(f), so (a − a′) ∈ ker(g), so g(a − a′)=0,sog(a)=g(a′). Thus we observe that “g takes the same value on all the elements of A that map to b”. Given b ∈ B,wecan pick a ∈ A such that f(a)=b (since f is onto) and then define h(b)=g(a). Because of our observation, this procedure gives us a well defined function h : B → C. Verify that h is a homomorphism. Now suppose a ∈ A.Letb = f(a). So a is an element of A such that f(a)=b.Soh(b)=g(a), or h(f(a)) = g(a). So h ◦ f = g. This proves part (a). (b) Now suppose g is onto. Let c ∈ C. Then there exists a ∈ A such that g(a)=c.It follows that h(f(a)) = c,soh is onto. This proves part (b). (c) Now suppose g is onto and ker(f)=ker(g). To show h is one to one, it is enough to verify that ker(h)=eB.Letb ∈ ker(h), that is, h(b)=eC . Pick a ∈ A such that f(a)=b. Then g(a)=h(f(a)) = h(b)=eC ,soa ∈ ker(g)=ker(f), so b = f(a)=eB.Soker(h)=eB. This proves part (c). Part (d) follows from part (b) and (c). ! 3.5. Corollary. Let m, n be positive integers. Then there exists a unique homomorphism h : Zmn → Zm such that h(1 mod mn)=1modm. sketch of proof. Let f : Z → Zmn and g : Z → Zm be the homomorphisms f(k)=k mod mn and g(k)=k mod m.Thenker(f)=mnZ ⊆ mZ =ker(g).