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![Corollary 31 • Recall (Proposition 30) ○ If H,K≤G, then HK≤G⟺HK=KH • Statement ○ If H,K≤G, and either H or K is normal in G, then HK≤G • Proof ○ Without loss of generality, assume K⊴G ○ We will prove HK=KH ○ Let h∈H,k∈K ○ hk=hk(h(−1) h=⏟(h�h(−1) )┬(∈K) h∈KH⇒HK≤KH ○ kh=(h^(−1) )kh=h ⏟(h(−1) kh┬(∈K)∈HK⇒KH≤HK ○ Therefore HK=KH Theorem 32 (The First Isomorphism Theorem) • Statement ○ If f:G→H is a hormorphism, then f induces an isomorphism § f ̅:G⁄kerf ⟶┴≅ im(f) § f ̅(g kerf )=f(g) • Intuition ○ This is an analogue of the Rank-Nullity Theorem ○ Given vector space V,W and a linear transformation A:V→W ○ dim(V⁄kerA )=dim〖im A〗 ○ ⇒dimV−nullity A=rank A • Proof ○ f ̅ is well-defined and injective § Let g_1,g_2∈G § g_1 kerf=g_2 kerf § ⟺g_2^(−1) g_1∈kerf § ⟺f(g_2^(−1) g_1 )=1 § ⟺f(g_2 )^(−1) f(g_1 )=1 § ⟺f(g_1 )=f(g_2 ) § ⟺f ̅(g_1 kerf )=f ̅(g_2 kerf ) § Thus f is well-defined and injective ○ f ̅ is surjective § Let h∈im f § Choose g∈G s.t. f(g)=h § Notice f ̅(g kerf )=h ○ f ̅ is a hormorphism § If g_1 kerf,g_2 kerf∈G\/kerf § f ̅(g_1 kerf⋅g_2 kerf ) § =f ̅(g_1 g_2 kerf ) § =f(g_1 g_2 )=f(g_1 )f(g_2 ) § =f ̅(g_1 kerf ) f ̅(g_2 kerf ) § Thus, f ̅ is a hormorphism Corollary 33 • Statement ○ [G:kerf ]=|im f| • Example ○ Let m,n∈Z, and assume (m,n)=1 ○ Then any homomorphism f:Z\/mZ→Z\/nZ is trivial ○ i.e. f(n ̅ )=0 ̅, ∀n ̅ • Proof ○ Let f be such a homomorphism ○ By the First Isomorphism Theorem ○ |(ZnZ⁄kerf |=|im f| ○ ⇒n/|kerf | =|im f|, where § n/|kerf | is a divisor of n § |im f| is a divisor of m by Lagrange^′ s Theorem ○ Thus, |im f|=1, so im f={0 ̅ } • Note ○ The same proof tells us that ○ If G,H are finite groups such (|G|,|H|)=1, then ○ All homomorphism between them are trivial Theorem 34 (The Second Isomorphism Theorem) • Statement ○ Let A,B≤G, and assume B⊴G ○ Then A∩B⊴A, and AB⁄B≅A⁄(A∩B) • Intuition • Note ○ By Corollary 31, AB≤G ○ And since B⊴G, B⊴AB ○ So, AB⁄B make sense • Proof ○ We have homomorphisms § α:A→AB a↦a § β:AB→AB⁄B x↦xB ○ Set f≔βα, so § f:A→AB⁄B a↦aB ○ f(a)=1_(AB⁄B)=B⟺a∈B ○ Thus, kerf=A∩B ○ Since kernels are normal, A∩B⊴A ○ The First Isomorphism Theorem gives an isomorphism ○ f ̅:A⁄(A∩B) ⟶┴≅ AB⁄B](https://shawnzhong.com/wp-content/uploads/2018/03/img_5aa7549722b67.png)
![Lagrange s Theorem • If G is finite group, and H≤G, then |G|=|H|⋅[G:H] • In particular, ├ |H|┤|├ |G|┤ Corollary 26 • Statement ○ If G is a group, and |G| is prime ○ Then G is cyclic, hence, G≅Z\/pZ • Proof ○ If g∈G, then |g|=|⟨g⟩| ○ By Lagrange s Theorem, ├ |g|┤|├ |G|┤ ○ Thus, |g|∈{1,|G|} ○ It follow, that if g∈G∖\{1}, then |g|=|G| ○ Therefore ⟨g⟩=G ○ i.e. G is cyclic • Groups of small order Order Property 2 Cyclic 3 Cyclic 4 Cyclic or Z\/2Z×Z\/2Z 5 Cyclic 6 Cyclic or S_3 Corollary 27 • Statement ○ If G is a finite group, and g∈G, then g^|G| =1 • Proof ○ ├ |g|=|⟨g⟩|┤|├ |G|┤ ○ Thus, g^|G| =^|g|m for some integer m ○ It follows that g^|G| =1 Corollary 28 • Statement ○ If G is a finite cyclic group, then there is a bijection ○ {positive divisors of |G|}⟷{subgroups of G} • Proof ○ The map from left to right sends a divisor m of |G| to ○ the unique subgroup G with order m ○ The map from right to left sends a subgroup H to |H| Proposition 29 • Definition ○ Let G be a group and H,K≤G ○ Define HK≔{hk│h∈H,k∈K} • Statement ○ If H,K are finite subgroups of a group G, then |HK|=(|H|⋅|K|)/|H∩K| • Proof ○ HK=⋃8_(hH)▒h� ○ In the proof of Lagrange s Theorem, we showed |h�|=|K| ○ Want to show that there are (|H|⋅|K|)/|H∩K| cosets of the form hK, h∈H ○ Let h1,h2∈H ○ h1 K=h2 K ○ ⟺h2^(−1) h1∈K ○ ⟺h2^(−1) h1∈H∩K ○ ⟺h1 (H∩K)=h2 (H∩K) ○ Thus the number of distinct cosets of the form hK,h∈H is ○ [H:H∩K]=|H|/|H∩K| by Lanrange^′ s Theorem ○ Thus HK consists of |H|/|H∩K| distinct cosets of K ○ Therefore, |HK|=(|H|⋅|K|)/|H∩K| • Note: HK is not always a subgroup ○ Let G=S_3, H=⟨(1 2)⟩, K=⟨(1 3)⟩ ○ |HK|=(|H|⋅|K|)/|H∩K| =(2×2)/1=4 ○ But |HK| is not a divisor of S_3 ○ By Lagrange s Theorem, HK is not a subgroup of S_3 Proposition 30 • Statement ○ If H,K≤G, then HK≤G iff HK=KH • Note ○ HK=KH is not equivalent to hk=kh, ∀h∈H,k∈K ○ It implies that every product hk is of the form k^′ h′ and conversely • Proof (⟹) ○ H≤HK, K≤HK⇒KH⊆HK ○ Let hk∈HK ○ Set a≔(h�)^(−1), then a∈HK ○ So, a=h′ k′ for some h′∈H,k^′∈K ○ hk=a^(−1)=(h′ k^′ )^(−1)=(k^′ )^(−1) (h′ )^(−1)∈KH ○ Thus HK⊆KH ○ Therefore HK=KH • Proof (⟸) ○ HK≠∅, since 1⋅1=1∈HK ○ Let hk,h′ k^′∈HK ○ We must show hk(h′ k^′ )^(−1)∈HK ○ hk(h′ k^′ )^(−1)=h ⏟(k(k^′ )^(−1) (h′ )^(−1) )┬(∈KH) ○ Choose h′′,k^′′ s.t. ⏟(k(k^′ )^(−1) (h′ )^(−1) )┬(∈KH)=⏟(h′′ k′′)┬(∈HK) ○ Then hk(h′ k^′ )^(−1)=⏟(h^′′ )┬(∈H) ⏟(k′′)┬(∈K)∈HK ○ Therefore HK≤G • Example ○ Let G=S_3, H=⟨(1 2)⟩, K=⟨(1 3)⟩ ○ HK={(1),(1 2),(1 3),(1 3 2)} ○ KH={(1),(1 2),(1 3),(1 2 3)} ○ Thus HK≠KH ○ Therefore HK is not a subgroup of S_3](https://shawnzhong.com/wp-content/uploads/2018/03/img_5aa70c522205c.png)
![Proposition 24: Quotient Group • Statement ○ If G is a group, and N⊴G, then ○ the set of left costs of N, denoted as G\/N (say "G mod N") ○ is a group under the operation (g_1 N)(g_2 N)=g_1 g_2 N ○ We call this group quotient group or factor group • Proof ○ Check G\/N×G\/N→G\/N, where (g_1 N,g_2 N)↦g_1 g_2 N is well-defined § Suppose g_1 N=g_1^′ N, and g_2 N=g_2^′ N § We must show g_1 g_2 N=g_1^′ g_2^′ N § g_1 N=g_1^′ N⟺(g_1′)^(−1) g_1∈N § g_2 N=g_2^′ N⟺(g_2′)^(−1) g_2∈N § We must show that (g_1^′ g_2^′ )^(−1) g_1 g_2∈N⟺g_1 g_2 N=g_1^′ g_2^′ N § (g_1^′ g_2^′ )^(−1) g_1 g_2 § =(g_2^′ )^(−1) (g_1^′ )^(−1) g_1 g_2 § =(g_2^′ )^(−1) (g_1^′ )^(−1) g_1 [g_2^′ (g_2^′ )^(−1) ] g_2 § =(g_2^′ )^(−1) ⏟((g_1^′ )^(−1) g_1 )┬(∈N) g_2^′ ⏟((g_2^′ )^(−1) g_2 )┬(∈N) § =⏟((g_2^′ )^(−1) (g_1^′ )^(−1) g_1 g_2^′ )┬(∈N) ⏟((g_2^′ )^(−1) g_2 )┬(∈N) § Thus (g_1^′ g_2^′ )^(−1) g_1 g_2∈N § Therefore, the operation is well-defined ○ Existence of Identity § 1⋅N=N ○ Existence of Inverse § (gN)^(−1)=g^(−1) N § Since (gN)(g^(−1) N)=gg^(−1) N=N ○ Associativity § (g_1 Ng_2 N)(g_3 N) § =(g_1 g_2 N)(g_3 N) § =g_1 g_2 g_3 N § =g_1 N(g_2 g_3 N) § =g_1 N(g_2 Ng_3 N) • Note ○ If N⊴G, then there is a surjective homomorphism ○ f:G→G\/N given by g→gN ○ kerf=N, since gN=N⟺g∈N ○ This shows that, if H≤G, then H⊴G iff ○ H is the kernel of a homomorphism from G to some other group • Example 1 ○ Let H be a subgroup of Z ○ H⊴Z since Z is abelian ○ Since Z is cyclic, H is also cyclic ○ So write H=⟨n⟩ ○ Then there is isomorphism § Z\/⟨n⟩→Z\/nZ § a+⟨n⟩→a ̅ • Example 2 ○ If G is a group, then {1_G }⊴G and G⊴G § G\/{1_G }≅G § G\/G≅ ∗ , where ∗ is the trivial group of order 1 ○ Intuition: The bigger the subgroup, the smaller the quotient Index • Definition ○ If G is a group, and H≤G, then ○ The index of H is the number of distinct left cosets of H in G ○ Denote the index by [G:H] • Note ○ If N⊴G, then [G:N]=|G/N| • Example ○ [Z⟨n⟩]=|ZnZ=n Theorem 25: Lagranges Theorem • Statement ○ If G is finite group, and H≤G, then |G|=|H|⋅[G:H] ○ In particular, ├ |H|┤|├ |G|┤ • Notice ○ If in the setting of Lagranges Theorem, H⊴G, then ○ |G|=|H|⋅|G/H|⇒|G/H|=|G|/|H| • Proof ○ Let n≔|H|, and k≔[G:H] ○ Let g_1,…,g_k be the representatives of the distinct cosets of H in G ○ (In other words: if g∈G, then gH∈{g_1 H,g_2 H,…,g_k H}) ○ By proposition 22, left costs are either equal or disjoint ○ So, G=g_1 H∪g_2 H∪…∪g_k H ○ Let g∈G, then there is a function f:H→gH, defined by h↦gh ○ f is certainly surjective ○ f is also injective since if gh1=gh2, then h1=h2 ○ Thus, |gH|=|H| ○ |G|=|g_1 H|+…+|g_k H|=⏟(n+n+…+n)┬(k copies)=kn ○ Therefore |G|=|H|⋅[G:H]](https://shawnzhong.com/wp-content/uploads/2018/04/img_5ad9fa8c77dde.png)
