4.4 Solving Congruences Mar 12, 2018 Shawn Math 240 No comments yet Math 541 – 3/12 Mar 12, 2018 Shawn Math 541 No comments yet Math 541 – 3/9 Mar 12, 2018 Shawn Math 541 No comments yet Math 541 – 3/7 Mar 08, 2018 Shawn Math 541 No comments yet Math 521 – 3/7 Mar 08, 2018 Shawn Math 521 No comments yet 1 … 25 26 27 28 29 … 87
![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)
![Theorem 3.4 • Statement (a) ○ Suppose (x_n ) ⃗=(α_(1,n),α_(2,n),…,α_(k,n) )∈Rk where n∈N, then ○ {(x_n ) ⃗ } converges to (α_1,α_2,…,α_k )⟺(lim)_(n→∞)〖α_(j,n) 〗=α_j (1≤j≤k) • Proof (a) ○ Assume (x_n ) ⃗→x ⃗ § Given ε 0, there exists N∈N s.t. |(x_n ) ⃗−x ⃗ | ε for n≥N § Thus, |α_(j,n)−α_j |≤|(x_n ) ⃗−x ⃗ | for n≥N,1≤j≤k § Therefore lim_(n→∞)〖α_(j,n) 〗=α_j for 1≤j≤k ○ Assume lim_(n→∞)〖α_(j,n) 〗=α_j for 1≤j≤k § Given ε 0, there exists N∈N s.t. |α_(j,n)−α_j | ε/√k for n≥N § |(x_n ) ⃗−x ⃗ |=|√(∑_(i=1)^k▒|α_(j,n)−α_n |^2 )|=√(∑_(i=1)^k▒|α_(j,n)−α_n |^2 ) √(∑_(i=1)^k▒ε^2/k)=ε § Therefore (x_n ) ⃗→x ⃗ • Statement (b) ○ Suppose § {(x_n ) ⃗ } and {(y_n ) ⃗ } are sequences in Rk, {β_n } is a sequence in R § (x_n ) ⃗→x ⃗, (y_n ) ⃗→y ⃗, β_n→β ○ Then § (lim)_(n→∞)〖(x_n ) ⃗+(y_n ) ⃗ 〗=x ⃗+y ⃗ § (lim)_(n→∞)〖(x_n ) ⃗⋅(y_n ) ⃗ 〗=x ⃗⋅y ⃗ § (lim)_(n→∞)〖β_n⋅(x_n ) ⃗ 〗=β⋅x ⃗ • Proof (b) ○ This follows from (a) and Theorem 3.3 (Algebraic Limit Theorem) Compact Sets • Intuition for Rk: Closed and bounded • Open cover ○ An open cover of a set E in a metric X is ○ a collection of open sets {G_α } in X s.t. E⊂⋃8_α▒G_α • Compact ○ A set K in a metric space X is compact if ○ every open cover of K has a finite subcover • Example 1 ○ Let E=(0,1), X=R ○ E is a open cover of itself, but E is not compact ○ Let G_α=(α/2,1) for α∈(0,1), then E has {G_n } as an open cover ○ We cannot take a finite collection of these G_α and still have an open cover ○ So it has no finite subcover ○ Therefore E=(0,1) is not compact • Example 2 ○ Let K=[0,1], X=R ○ Consider {G_α }∪{G_0 }∪{G_1 }, where § G_α=(α/2,1) for α∈(0,1) § G_0=(−ε,ε) § G_1=(1−ε,1+ε) for some ε 0 ○ Then {G_α }∪{G_0 }∪{G_1 } is an open cover of [0,1] ○ It has finite subcover {G_0,G_1,G_ε } where G_ε=(ε/2,1) ○ Therefore K=[0,1] is compact](https://shawnzhong.com/wp-content/uploads/2018/04/img_5ac84e84386bb.png)
