Notes Archives - Page 20 of 84 - Shawn Zhong

Math 541 – 4/4

$Theorem 54: Cauchy s Theorem • Statement ○ If G is a finite group, and p is a prime divisor of |G|, then ∃H≤G of order p • Proof ○ Write |G|=mp. We argue by strong induction on m ○ When m=1, this is trivial, since any non-identity element of G has order p ○ Suppose m 1, and ∀n∈{1,…,m−1} if |G^′ |=np, then ∃H^′≤G^′ of order p ○ If G is abelian § Let x∈G∖{1} § If ⟨x⟩=G □ By the Fundamental Theorem of Cyclic Groups, □ G=⟨x⟩ contains a (unique) subgroup of order p § If ⟨x⟩≠G □ Set H≔⟨x⟩⊴G □ By the Lagrange s Theorem, |G|=|H|[G:H]=|H|⋅|G/H| □ Since ├ p┤ ||G|┤,either ├ p┤ ||H|┤ or ├ p┤ ||G/H|┤ □ If ├ p┤ ||H|┤ ® Since H is cyclic, H contains a (unique) subgroup of order p ® It follows that G contains a subgroup of order p □ If ├ p┤ ||G/H|┤ ® |G/H| |G|, so, by induction, ∃gH∈G\/H s.t. |gH|=p ® So we only need to prove ├ |gH|┤ ||g|┤ ◊ If K ⟶┴f K′ is a group homomorphism, ├ |f(k)|┤ ||k|┤,∀k∈K ◊ Now, take K=G, K^′=G\/H, f the usual surjection g↦gH ® Therefore ├ p┤ ||g|┤ ® Since ⟨g⟩ is cyclic, ⟨g⟩ contains a (unique) subgroup of order p ® It follows that G contains a subgroup of order p ○ If G is not abelian § By the Lagrange s Theorem, |G|=|C_G (g_i )|⋅[G:C_G (g_i )], ∀i∈{1,…,r} § Since ├ p┤ ||G|┤,either ├ p┤ ||C_G (g_i )|┤ or ├ p┤ |[G:C_G (g_i )]┤ § If ├ p┤ ||C_G (g_i )|┤ for some i □ Since G is not abelian, C_G (g_i )≨G for all i □ Apply the induction hypothesis, C_G (g_i ) contains a subgroup of order p □ It follows that G contains a subgroup of order p § If ├ p┤ |[G:C_G (g_i )]┤,∀i □ By the Class Equation, |G|=|Z(G)|+∑_(i=1)^r▒[G:C_G (g_i )] where g_1,…,g_r∈G □ are the representatives of the conjugate classes not contained in Z(G) □ It follows that ├ p┤ |(|G|−∑_(i=1)^r▒[G:C_G (g_i )] )┤=|Z(G)| □ G is not abelian, so Z(G)≨G □ Apply the induction hypothesis, Z(G) contains a subgroup of order p □ It follows that G contains a subgroup of order p Lemma 55: Recognizing Direct Products • Statement ○ Let G be a group with normal subgroups N_1,N_2 ○ The map α:N_1×N_2→G given by (n_1,n_2 )↦n_1 n_2 is an isomorphism ○ if and only if N_1 N_2=G and N_1∩N_2={1} • Proof (⟹) ○ Since α is surjective, N_1 N_2=G ○ Suppose n∈N_1∩N_2 ○ Then α(n,1)=n=α(1,n) ○ Since α is injective, (1,n)=(n,1)⇒n=1 ○ So N_1∩N_2={1} • Proof (⟸) ○ α is surjective § This is true since N_1 N_2=G ○ α is a homomorphism § α((n_1,n_2 ),(n_1^′,n_2^′ ))=α((n_1 n_1^′,n_2 n_2^′ ))=n_1 n_1^′ n_2 n_2^′ § α(n_1,n_2 )α(n_1^′,n_2^′ )=n_1 n_2 n_1^′ n_2^′ § We want show that α((n_1,n_2 ),(n_1^′,n_2^′ )) (α(n_1,n_2 )α(n_1^′,n_2^′ ))^(−1)=1 § (n_1 n_1^′ n_2 n_2^′ ) (n_1 n_2 n_1^′ n_2^′ )^(−1)=n_1 n_1^′ n_2 n_2^′ (n_2^′ )^(−1) (n_1^′ )^(−1) n_2^(−1) n_1^(−1) § =n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) )┬(∈N_2 ) n_2^(−1) n_1^(−1)=n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_2 ) n_1^(−1)∈N_2 § =n_1 n_1^′ ⏟(n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_1 ) n_1^(−1)=n_1 ⏟(n_1^′ n_2 (n_1^′ )^(−1) n_2^(−1) )┬(∈N_1 ) n_1^(−1)∈N_1 § Thus (n_1 n_1^′ n_2 n_2^′ ) (n_1 n_2 n_1^′ n_2^′ )^(−1)∈N_1∩N_2={1} § Therefore α((n_1,n_2 ),(n_1^′,n_2^′ ))=α((n_1,n_2 ),(n_1^′,n_2^′ )) ○ α is injective § If (n_1,n_2 )=1 § ⇒n_1 n_2=1 § ⇒n_1=n_2^(−1) § ⇒n_1∈N_2,n_2∈N_1 § ⇒n_1=n_2=1 § ⇒(n_1,n_2 )=(1,1) § ⇒α is injective$

Math 521 – 4/4

$Series • Given a sequence {a_n } • We associate a sequence of partial sums {s_n } where • s_n=∑_(k=1)^n▒a_k =a_1+a_2+…+a_n • ∑_(k=1)^∞▒a_k is called an infinite series, or simply series • If {s_n } diverges, the series is said to diverge • If {s_n } converges to s, the series is said to converge, and write ∑_(k=1)^∞▒a_k =s • s is called the sum of the series • But it is technically the limit of a sequence of sums Theorem 3.22 (Cauchy Criterion for Series) • Statement ○ ∑_(n=1)^∞▒a_n converges⟺∀ε 0, ∃N∈N s.t. |∑_(k=n)^m▒a_k | ε,∀m≥n≥N • Proof ○ This is Theorem 3.11 applied to {s_n } Theorem 3.23 • Statement ○ In the setting of Theorem 3.22, take m=n ○ We have |a_n | ε for n≥N ○ If ∑_(n=1)^∞▒a_n converges, then (lim)_(n→∞)⁡〖a_n 〗=0 • Note ○ If a_n→0, the series ∑_(n=1)^∞▒a_n might not converge • Example: ∑_(n=1)^∞▒1/n diverges ○ 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+…≥1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+… ○ Therefore ∑_(n=1)^∞▒1/n diverges Theorem 3.24 • Statement ○ A series of nonnegative real numbers converges if and only if ○ its partial sum form a bounded sequence • Proof ○ See Theorem 3.14 (Monotone Convergence Theorem) Theorem 3.25 (Comparison Test) • If |a_n | c_n for n≥N_0∈N and ∑_(n=1)^∞▒c_n converges, then ∑_(n=1)^∞▒a_n converges ○ Given ε 0,∃N≥N_0 s.t. |∑_(k=n)^m▒c_k |=∑_(k=n)^m▒c_k ε for m≥n≥N ○ By the Cauchy Criterion, |∑_(k=n)^m▒a_k |≤∑_(k=n)^m▒|a_k | ≤∑_(k=n)^m▒c_k ε ○ Thus ∑_(n=1)^∞▒a_n converges • If a_n≥d_n≥0 for n≥N_0∈N and ∑_(n=1)^∞▒d_n diverges, then ∑_(n=1)^∞▒a_n diverges ○ If ∑_(n=1)^∞▒a_n converges, then so must ∑_(n=1)^∞▒d_n ○ This is a contradiction, so ∑_(n=1)^∞▒a_n diverges Theorem 3.26 • Statement ○ If 0 x 1, then∑_(n=0)^∞▒x^n =1/(1−x) ○ If x 1, the series diverges • Note ○ {█(S=1+x+x^2+…@xS=x+x^2+…)┤⇒S−xS=1⇒S=1/(1−x) ○ This only works if we know this series converges • Proof ○ If 0 x 1, we have ○ {█(s_n=1+x+x^2+…+x^n@xs_n=x+x^2+…+x^n+x^(n+1) )┤ ○ ⇒s_n−xs_n=1−x^(n+1) ○ ⇒s_n=(1−x^(n+1))/(1−x) ○ Since 0 x 1,lim_(n→∞)⁡〖s_n 〗=lim_(n→∞)⁡〖(1−x^(n+1))/(1−x)〗=1/(1−x) ○ Note if x=1, ∑_(n=1)^∞▒x^n =1+1+…which diverges$

6.3 Permutations and Combinations

Permutations • Definition ○ A permutation of a set of distinct objects is an ordered arrangement of these objects. ○ An ordered arrangement of r elements of a set is called an r-permutation. • Example ○ Let S = {1,2,3}. ○ The ordered arrangement 3,1,2 is a permutation of S. ○ The ordered arrangement 3,2 is a 2-permutation of S. ○ The number of r-permutations of a set with n elements is denoted by P(n,r). ○ The 2-permutations of S = {1,2,3} are 1,2; 1,3; 2,1; 2,3; 3,1; and 3,2 ○ Hence, P(3,2) = 6. A Formula for the Number of Permutations • Theorem 1 ○ If n is a positive integer and r is an integer with 1 ≤ r ≤ n, then there are ○ P(n,r)=n(n−1)(n−2)⋯(n−r+1) ○ r-permutations of a set with n distinct elements. • Proof ○ Use the product rule. ○ The first element can be chosen in n ways. ○ The second in n − 1 ways ○ And so on until there are (n−(r−1)) ways to choose the last element. ○ Note that P(n,0) = 1, since there is only one way to order zero elements. • Corollary 1 ○ If n and r are integers with 1 ≤ r ≤ n, then ○ P(n,r)=n!/(n−r)! • Example ○ How many ways are there to select a first-prize winner, a second prize winner, and a third-prize winner from 100 different people who have entered a contest? ○ P(100,3) = 100 ∙ 99 ∙ 98 = 970,200 Example ○ Suppose that a saleswoman has to visit eight different cities. ○ She must begin her trip in a specified city ○ But she can visit the other seven cities in any order she wishes. ○ How many possible orders can the saleswoman use when visiting these cities? ○ The first city is chosen, and the rest are ordered arbitrarily. ○ Hence the orders are: 7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040 ○ If she wants to find the tour with the shortest path that visits all the cities, ○ she must consider 5040 paths! Example ○ How many permutations of the letters ABCDEFGH contain the string ABC ? ○ We solve this problem by counting the permutations of six objects, ABC, D, E, F, G, H. ○ 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720 Combinations • Definition ○ An r-combination of elements of a set is an unordered selection of r elements from the set. ○ Thus, an r-combination is simply a subset of the set with r elements. ○ The number of r-combinations of a set with n distinct elements is denoted by C(n, r). ○ The notation (█(n@r)) is also used and is called a binomial coefficient. ○ (We will see the notation again in the binomial theorem in Section 6.4.) • Example: ○ Let S be the set {a, b, c, d}. ○ Then {a, c, d} is a 3-combination from S. ○ It is the same as {d, c, a} since the order listed does not matter. ○ C(4,2) = 6 because the 2-combinations of {a, b, c, d} are the six subsets ○ {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}. • Theorem 2 ○ The number of r-combinations of a set with n elements, where n≥r ≥ 0, equals ○ C(n,r)=n!/(n−r)!r! • Proof ○ By the product rule P(n, r)=C(n,r)⋅P(r,r). Therefore, ○ C(n,r)=P(n,r)/P(r,r) =(n!∕(n−r)!)/(r!∕(r−r)!)=n!/(n−r)!r! • Example ○ How many poker hands of five cards can be dealt from a standard deck of 52 cards? ○ Also, how many ways are there to select 47 cards from a deck of 52 cards? ○ Since the order in which the cards are dealt does not matter ○ the number of five card hands is: ○ C(52,5)=52!/5!47!=(52⋅51⋅50⋅49⋅48)/(5⋅4⋅3⋅2⋅1)=26⋅17⋅10⋅49⋅12=2,598,960 ○ The different ways to select 47 cards from 52 is ○ C(52,47)=52!/47!5!=C(52,5)=2,598,960 • Corollary 2 ○ Let n and r be nonnegative integers with r≤n. Then C(n, r) = C(n, n−r). • Proof ○ From Theorem 2, it follows that ○ C(n,n−r)=n!/(n−r)!(n−(n−r))!=n!/(n−r)!r!=C(n,r) ○ Hence, C(n, r)=C(n, n−r) Combinatorial Proofs • Definition ○ A combinatorial proof of an identity is a proof that uses one of the following methods. ○ Double Counting Proof § A double counting proof uses counting arguments to prove that § both sides of an identity count the same objects, but in different ways. ○ Bijective Proof § A bijective proof shows that there is a bijection § between the sets of objects counted by the two sides of the identity. • Example ○ Here are two combinatorial proofs that C(n, r)=C(n, n−r) ○ Bijective Proof § Suppose that S is a set with n elements. § The function that maps a subset A of S to A ̅ is a bijection between § the subsets of S with r elements and the subsets with n−r elements. § Since there is a bijection between the two sets § They must have the same number of elements. ○ Double Counting Proof § By definition the number of subsets of S with r elements is C(n, r). § Each subset A of S can also be described by § specifying which elements are not in A, i.e., those which are in A ̅. § Since the complement of a subset of S with r elements has n−r elements § There are also C(n,n−r) subsets of S with r elements. More Examples • How many words can you formed by rearranging the letters in the word: ○ Combine § 7! ○ Permutation § Pick where t s go § Arrange remaining 9 letters § (█(11@2))⋅9!=11!/2! ○ Rearrange § Pick where r s go § Pick where e s go § Pick where a s go § Arrange remaining 2 letters § (█(9@3))⋅(█(6@2))⋅(█(4@2))⋅2!=9!/3!2!2! • In a game of cards a hand consists of 13 cards. How many possible hands are there with ○ Exactly one ace § Pick which ace § Pick the rest § 4⋅(█(52−4@12)) ○ At least one ace § 1 Ace + 2 Aces + 3 Aces + 4 Aces § 4⋅(█(52−4@12))+(█(4@2))⋅(█(52−4@11))+(█(4@3))⋅(█(52−4@10))+(█(4@4))⋅(█(52−4@9)) ○ Exactly one ace and two diamonds § Case 1: We pick one diamond and a diamond ace □ 1⋅12⋅(█(52−4−12@11)) § Case 2: We pick two diamonds and a different ace □ 3⋅(█(12@2))⋅(█(52−4−12@10)) § So the total number of hands is 1⋅12⋅(█(52−4−12@11))+3⋅(█(12@2))⋅(█(52−4−12@10))

Math 541 – 4/2

$Conjugacy Class • Definition ○ If G is a group, G acts on itself by conjugation: g⋅h=ghg^(−1) ○ The orbits under this action are called conjugacy classes ○ Denote a conjugate class represented by some element g∈G by conj(g) • Example 1 ○ If g∈G, and g∈Z(G), then conj(g)={g} ○ Since hgh(−1)=hh(−1) g=g, ∀h∈G ○ The converse is also true: If conj(g)={g}, then g∈Z(G) • Example 2 ○ Let G=S_n ○ If σ∈S_n, then conj(g)={all permutations of the same cycle type as σ} ○ For instance § If σ is a t-cycle, then conj(σ)={all t-cycles} ○ More generally § Let σ=(a_1^((1) )…a_(t_1)^((1) ) )⋯(a_1^((m) )…a_(t_m)^((m) ) ) be a product of disjoint cycles § Then conj(σ)={all products of disjoint cycles of length t_1,…,t_m } Theorem 51: The Class Equation • Statement ○ Let G be a finite group ○ Let g_1,…g_r∈G be representatives of the conjugacy classes of G that ○ are not contained in the center Z(G) of G ○ Then |G|=|Z(G)|+∑_(i=1)^r▒[G:C_G (g_i )] • Recall: C_G (g_i )={g∈G│gg_i=g_i g} • Proof ○ G is the disjoint union of its disjoint conjugate classes ○ Then G=Z(G)∪⋃24_(i=1)^r▒conj(g_i ) ○ ⇒|G|=|Z(G)|+∑_(i=1)^r▒|conj(g_i )| ○ ⇒|G|=|Z(G)|+∑_(i=1)^r▒|orb(g_i )| (under conjugacy action) ○ ⇒|G|=|Z(G)|+∑_(i=1)^r▒[G:stab(g_i )] by Proposition 48 ○ ⇒|G|=|Z(G)|+∑_(i=1)^r▒[G:C_G (g_i )] Corollary 52 • Statement ○ If p is a prime, and P is a group of order p^α (α 1), then |Z(P)| 1 • Note: Group of order p^α is called a p-group • Proof ○ By the class equation, |Z(P)|=|P|−∑_(i=1)^r▒[P:C_P (p_i )] , where p_1,…p_r∈P ○ are representatives of the conjugate classes of P not contained in Z(P) ○ g_i∉Z(P)⇒C_P (g_i )≠P⇒[P:C_P (g_i )]≠1 ○ By Lagrange s Theorem, ├ [P:C_P (g_i )]┤ |p^α ┤ ○ Combing previous two results, ├ p┤|├ [P:C_P (g_i )]┤ ○ Thus, ├ p┤ |(|P|−∑_(i=1)^r▒[P:C_P (g_i )] )┤=|Z(P)|, since ├ p┤ ||P|┤ ○ ⇒|Z(P)|≠1 Corollary 53 • Statement ○ If p is a prime, and P is a group of order p^2, then P is abelian. ○ In fact, either P≅Z\/p^2 Z or P≅Z\/pZ×Z\/pZ • Proof ○ By corollary 52, |Z(P)|=p or p^2 ○ Suppose |Z(P)|=p § |P/Z(P)|=[P:Z(P)]=|P|/|Z(P)| =p^2/p=p § By Corollary 26, P\/Z(P) is cyclic § By HW6 Q2, P is abelian § In this case Z(P)=P⇒|Z(P)|=p^2, which is impossible ○ Suppose |Z(p)|=p^2 § We have |Z(p)|=|P|⇒Z(P)=P § So P is abelian ○ If P is cyclic, then clearly P≅Z\/p^2 Z ○ If P is not cyclic, we need to show P≅Z\/pZ×Z\/pZ § Let z∈P∖{1}, then |z|=p. Let y∈P∖⟨z⟩ § Set H≔⟨z⟩,K≔⟨y⟩, then H∩K={1} § Since any non-identity element of H or K is a generator § For instance, if 1≠y^k∈H for some k, then y∈H, which is impossible § |HK|=(|H|⋅|K|)/|H∩K| =|H|⋅|K|=p^2=|P|⇒HK=P § By HW6 Q1, there exists an isomorphism P ⟶┴≅ P\/H×P\/K § |P/H|=[P:H]=|P|/|H| =p^2/p=p⇒P\/H≅Z\/pZ § Similarly for P\/K § Therefore P=HK≅Z\/pZ×Z\/pZ$

Math 541 – 3/23

$Proposition 47 • Statement ○ Let G act on a set X ○ The relation x~x^′⟺∃g∈G s.t. gx=x′ is an equivalence relation • Proof ○ Reflexive § 1⋅x=x ○ Symmetric § Suppose x~x^′, then ∃g∈G s.t. gx=x^′⇒x=g^(−1) x^′ ○ Transitive § Suppose x~x^′ and x^′~x^′′ § Choose g,h∈G s.t. gx=x′ and hx^′=x′′ § Then ghx=hx^′=x^′′ • Note ○ The equivalence classes are the orbits of the group action ○ Thus, the orbits partition X Proposition 48 • Statement ○ If G acts on X, and x∈X, then |orb(x)|=[G:stab(x)] • Proof ○ Define a function § F:orb(x)→{left costs of stab(x)} § gx↦g stab(x) ○ F is injective § g stab(x)=g′ stab(x) § ⟺(g^′ )^(−1) g∈stab(x) § ⟺(g^′ )^(−1) gx=x § ⟺gx=g^′ x ○ F is surjective § This is clear ○ So orb(x)≅{left costs of stab(x)} ○ Therefore |orb(x)|=[G:stab(x)] Proposition 49: Lemma for Cayley s Theorem • Statement ○ Let G be a group acting on a finite set X={x_1,…,x_n } ○ Then each g∈G determines a permutation σ_g∈S_n by ○ σ_g (i)=j⟺g⋅x_i=x_j • Proof ○ The map f:X→X, given by x↦g⋅x is bijection ∀g∈G § Injectivity: g⋅x=g⋅x^′⇒(g^(−1) g)⋅x=(g^(−1) g)⋅x^′⇒x=x^′ § Surjectivity: f(g^(−1)⋅x)=(gg^(−1) )⋅x=x ○ So each g∈G determines a permutation σ_g∈S_n where § σ_g (i)=j⟺g⋅x_i=x_j • Statement ○ The map Φ:G→S_n,g↦σ_g is a homomorphism • Proof ○ Let g,h∈G,i∈{1,…,n} ○ Suppose σ_gh(i)=j for some j ○ Then (gh x_i=x_j ○ Write hx_i=x_k for some k, then σ_h(i)=k ○ (gh x_i=x_j⟺gx_k=x_j⟺σ_g (k)=j⟺σ_g (σ_h(i))=j ○ Therefore σ_gh(i)=σ_g σ_h(i),∀i∈{1,…,n} Theorem 50: Cayley s Theorem • Statement ○ Every finite group is isomorphic to a subgroup of the symmetric group • Proof ○ Let G={g_1,…,g_n } act on itself by left multiplication g⋅h=gh ○ By Proposition 49, this action determines a homomorphism § Φ:G→S_n § g↦σ_g § where σ_g (i)=j⟺g⋅g_i=g_j ○ Φ is injective § Φ(g)=Φ(h⟺σ_g=σ_ℎ⟺ggi=hgi,∀i⟺g=h ○ Thus G≅im(Φ)≤S_n • Example ○ Klein 4 group K={1,a,b,c} ○ where a^2=b^2=c^2=1⟺ab=c,bc=a,ac=b 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 ○ Label the group elements with 1, 2, 3, 4 ○ 1↦σ_1=(1) since § σ_1 (1)=1 § σ_2 (2)=2 § σ_3 (3)=3 § σ_4 (4)=5 ○ a↦σ_a=(1 2)(3 4) since § σ_a (1)=2 § σ_a (2)=1 § σ_a (3)=4 § σ_a (4)=3 ○ b↦σ_b=(1 3)(2 4) since § σ_b (1)=3 § σ_b (2)=4 § σ_b (3)=1 § σ_b (4)=2 ○ c↦σ_c=(1 4)(2 3) since § σ_c (1)=4 § σ_c (2)=3 § σ_c (3)=2 § σ_c (4)=1 ○ Therefore K≅{(1),(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}≤S_4$

Shawn Zhong

Shawn Zhong

Notes

Math 541 – 4/4

Math 521 – 4/4

6.3 Permutations and Combinations

Math 541 – 4/2

Math 541 – 3/23

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