2. Normal Subgroup and Quotient Group We Begin by Stating a Couple of Elementary Lemmas

Home , Group homomorphism, Quotient group

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 homomorphisms 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). Now apply the above theorem. ! It is convenient to represent the statement of theorem 3.4 in a diagrammatic language. Let us denote an onto homomorphism f : A → B by a double headed arrow f : A " B. The theorem 3.4 says that g / A ❅ C ❅❅ ❅❅ ❅❅ f ❅ B with ker(f) ⊆ ker(g), there exists a unique h : B → C such that the following diagram commutes: g / A ❅ ? C (2) ❅❅ ⑦? ❅❅ ⑦ ❅❅ ⑦ f ❅ ⑦ h B 5 Saying that the diagram in 2 commutes is a pictorial way of saying that g = h ◦ f.

6 3.6. Theorem (The ﬁrst isomorphism theorem). Let ψ : G → H be a group homomorphism. Let K =ker(ψ).Letφ : G → G/K be the natural quotient homomorphism: φ(g)=gK. Then there is a unique isomorphism η : G/K → ψ(G) such that ψ = η ◦ φ. Proof. Recall that kernels of homomorphisms are normal subgropus, so K is normal in G. From given data, we get the following diagram of homomorphisms

ψ G / / ψ(G) ❈❈ ❈❈ ❈❈ ❈❈ φ ❈! ! G/K Note that the horizontal arrow is simply the given map ψ, but we consider it as a map to the subgroup ψ(G)ofH.Soψ : G → ψ(G) is an onto homomorphism. From the deﬁnition of the quotient map, we have ker(φ)=K =ker(ψ). So by theorem 3.4(d), we get an isomorphism η : G/K → ψ(G), such that the following diagram commutes:

ψ G / / ψ(G) ❈❈ ; ❈❈ ✇✇; ❈❈ ≃ ✇✇ ❈❈ ✇✇η φ ❈! ! ✇✇ G/K Saying that the above diagram commutes is just a pictorial way of saying ψ = η ◦ φ. ! Let G be a group and N be a normal subgroup. The quotient homomorpshim G → G/N collapses the normal subgroup N to the identity in G/N. In this way, considering the quotient group G/N lets us forget the structure inside N.Thecorrespondencetheorem below says that taking quotient by N “retains the subgroup structure in G above N”. So passing to the quotient group G/N lets us disregard what was happening inside N and focus solely on what happens “above” N. Thus, the quotient construction becomes an important simplifying tool. 3.7. Theorem (the correspondence theorem). Let N be a normal subgroup of G.Thenthere is a natural inclusion preserving bijection between {set of subgroups of G containing N }→{set of subgroups of G/N} Under this bijection, a subgroup H of G such that H ⊃ N corresponds to the subgroup H/N of G/N.Undertheabovecorrespondence,normalsubgroupsontheleft hand side correspond to normal subgroups on the right hand side. The proof of the correspondence theorem is left as a routine exercise. We want to move on to the second and the third isomorphism theorems. These theorems let us compare the subgroup structures in groups with subgroup structures in various quotients. Before stating them we need a lemma. The proof of the lemma is left as an exercise. 3.8. Lemma. Let H and K be two subgroups of a group G.ThenHK is a subgroup of G if and only if HK = KH. 3.9. Theorem (The second isomorphism theorem). Let N be a normal subgroup of G.Let H be a subgroup of G.ThenHN is a subgroup of G and H ∩ N is a normal subgroup of H and H/H ∩ N ≃ HN/N. 7 sketch of proof. Since N is normal in G,wehavgN = Ng for all g ∈ G.SoHN = NH.By lemma 3.8, it follows that HN is a subgroup of G. Verifying H ∩ N is a normal subgroup of H is left as a routine exercise. Now let ψ be the composition of homomorphisms: quotient map H −− −inclusion − − − − −→ HN −− − − − − − − − −→ HN/N. So ψ : H → HN/N is given by ψ(h)=hN.Notethatψ is a homomorphism since both the inclusion and the quotient map are homomorphisms. Now verify that ψ is onto and ker(ψ)=H ∩ N. Then the second isomorphism theorem follows from the ﬁrst isomorphism theorem. ! 3.10. Theorem (The third isomorphism theorem). Let H and N be normal subgroups of G G/N G with N ⊂ H ⊂ G.ThenthereisanaturalisomorphismH/N ≃ H . Proof. Consider the following diagram of homomorphisms:

φH G / / G/H ❇❇ ❇❇ ❇❇ ❇❇ φN ❇! ! G/N where both the arrows φN and φH are the natural quotient homomorphisms. Note that ker(φN )=N ⊂ H =ker(φH ). So by theorem 3.4(c), we have a onto homomorphism ψ : G/N → G/H such that the following diagram commutes:

φH G / / G/H ❇❇ ; ; ❇❇ ✇✇; ; ❇❇ ✇✇ ❇❇ ✇✇ φN ❇! ! ✇✇ ψ G/N

If g ∈ G,thenonehasψ(gN)=ψ ◦ φN (g)=φH (g)=gH. Now verify that ker(ψ)=H/N and apply the ﬁrst isomorphism theorem. !

Let G1 and G2 be two groups. Let us consider the product group G1 × G2.Theproduct comes with natural projection maps π1 : G1 × G2 → G1 and π2 : G1 × G2 → G2 given by π1(g1,g2)=g1 and π2(g1,g2)=g2. It is easy to verify that π1 and π2 are onto homomorphisms. The next theorem tells us how to construct homomorphism toaproduct.

3.11. Theorem. Let G1,G2,G be groups. Given homomorpshims f1 : G → G1 and f2 : G → G2,thereexistsauniquehomomorphismf : G → G1 × G2 such that π1 ◦ f = f1 and π2 ◦ f = f2 (where π1 and π2 are the natural projections. sketch of proof. Deﬁne f : G → G1 × G2 by f(g)=(f1(g),f2(g)) and verify that f is a homomorphism and that π1 ◦ f = f1 and π2 ◦ f = f2 ! 3.12. Theorem. Let H, K be two normal subgroups of a group G such that H ∩ K = {e}. Then show that G is isomorphic to a subgroup of G/H × G/K. sketch of proof. Let φH : G → G/H and φK : G → G/K be the natural quotient homomorphisms. By 3.11, we get a homomorphism φ : G → G/H × G/K, given by

φ(g)=(φH (g),φK(g)) = (gH,gK). 8 Let g ∈ Ker(φ). Then (gH,gK) is the identity in G/H × G/K. This means gH is the identity in G/H and gK is the identity in G/K.TosaygH is the identity in G/H,means gH = H,sog ∈ H. Similarly, we get g ∈ K.Sog ∈ H ∩ K = {e}.Soker(φ)={e} and hence φ is a one to one homomorpshim. So φ : G → G/H × G/K gives an isomorphism φ : G → φ(G)andφ(G) is a subgroup of G/H × G/K. ! 3.13. Theorem. Let H and K be two subgroups of a finite group G such that K is normal in G, H ∩ K = {e} and HK = G.Thenshowthat|G| = |H|·|K|. Proof. By the second isomorphism theorem, we know that HK is a subgroup of G and H ∩K is a normal subgroup of H and that H/H ∩ K ≃ HK/K. Now since H ∩ K = {e} and HK = G,weget H ≃ H/{e} = H/H ∩ K ≃ HK/K ≃ G/K. So |H| = |G/K| = |G|/|K|. ! Infact, given the hypothesis of 3.13, it follows that G ≃ H × K (Exercise: show this). In this situation, one says that G is the internal direct product of H and K. 3.14. Theorem. Let H, K be two normal subgroups of a finite group G such that H∩K = {e} and HK = G.ThenshowthatG is isomorphic to G/H × G/K. Proof. By 3.12, we have an one to one homomorphism φ : G → G/H × G/K given by φ(g)=(gH,gK). By 3.13 we have |G| = |K|·|H|.So|G/H| = |G|/|H| = |K| and |G/K| = |G|/|K| = |H|.So |G/H × G/K| = |G/H|·|G/K| = |K|·|H| = |G|. By 3.12, we have an one to one homomorphism φ : G → G/H × G/K given by φ(g)= (gH,gK). Since G and G/H × G/K have the same size, the map φ must be onto as well. So φ : G → G/H × G/K is an isomorphism. ! The following result, known as the Chinese remainder theorem is a special case of the above considerations, but we find it convenient to say it directly. 3.15. Corollary. Let m, n be two nonzero integers. If m and n are relatively prime, then Zmn ≃ Zm × Zn.

Proof. By 3.5, we have a natural homomorphisms Zmn → Zm and Zmn → Zn given by taking (k mod mn)to(k mod m)andto(k mod n) respectively. By 3.11, this gives a homomorphism f : Zmn → Zm × Zn given by f(k mod mn)=(k mod m, k mod n) Let k ∈ Z such that (k mod mn) ∈ ker(f), then (k mod m, k mod n)=(0modm, 0modn). In other words m and n divide k. Since m and n are relatively prime, this means mn divides k,sok mod mn is the identity of Zmn.Soker(f) is just the identity of Zmn,hence f : Zmn → Zm × Zn is one to one. Since |Zmn| = mn = |Zm|·|Zn| = |Zm × Zn|,themapf must be onto as well. So f is an isomorphism. ! 3.16. Corollary. Let m and n be relatively prime non-zero integers. Then (a) U(mn) ≃ U(m) × U(n). 9 (b) For a positive integer n let φ(n) be the Euler Totient function deﬁned as the number of integers k such that 1 ≤ k

sketch of proof. Consider the isomorphism f : Zmn → Zm × Zn given in 3.15. Since f is onto, given any (α,β) ∈ U(m) × U(n), there exists k ∈ Z such that f(k mod mn)=(α,β). This means (k mod m)=α and (k mod n)=β are relatiely prime to m and n respectively, so k is relatively prime to mn, that is, (k mod mn) ∈ U(mn). Conversely, if (k mod mn) ∈ U(mn), then f(k mod mn)=(k mod m, k mod n) ∈ U(m) × U(n). So the restriction of f to U(mn) gives an onto map f : U(mn) → U(m) × U(n). Since f is one to one, so is the restriction. Finally verify that f is also a homomorphism for the multiplicative groups U(mn) to U(m) × U(n). This proves part (a). Part (b) follows immediately, since |U(n)| = φ(n). Part (c) follows from part (b) and the easy observation that φ(pr)=pr(1 − p−1) for a prime number p. !