Math 541 Archives - Page 3 of 8 - Shawn Zhong

Math 541 - 4/9

$Lemma 58 • Statement ○ If G is an abelian group of order p^n, where p is a prime ○ Let a∈G has maximal order among all the elements of G ○ Then G≅A×Q, where A=⟨a⟩, Q≤G • Proof ○ We argue by induction on n ○ If n=1, then G=A, so we may take Q={1} ○ Now suppose n 1 ○ Case 1: ∃b∈G s.t. b∉A and b^p=1 § Let B≔⟨b⟩⊴G § A∩B={1} □ |b| is prime, since b^p=1 □ Recall: If (x,n)=1, then Z\/nZ=⟨x ̅ ⟩ □ So every non-identity element of B is a generator □ Thus, if x∈A∩B, and x≠1, then B=⟨x⟩⊂A∩B⊂A □ Then b∈A, which contradicts the assumption □ Therefore A∩B={1} § Let G ̅≔G\/B, then |G ̅ | |G| since B≠{1} § aB is an element of maximal order in G ̅ □ ├ |aB|┤ ||a|┤ ® a^|a| =1 ® ⇒a^|a| ∈B ® ⇒(aB)^|a| =1_G ̅ ® ⇒├ |aB|┤ ||a|┤ □ ├ |a|┤ ||aB|┤ ® (aB)^|aB| =1_G ̅ ® ⇒a^|aB| B=B ® ⇒a^|aB| ∈B ® ⇒a^|aB| ∈A∩B={1} ® ⇒a^|aB| =1 ® ⇒├ |a|┤ ||aB|┤ □ So |aB|=|a| □ Therefore aB is an element of maximal order in G ̅ § By induction, ∃Q ̅≤G ̅ s.t. G ̅≅⟨aB⟩×Q ̅ § Apply the Correspondence Theorem, choose Q≤G s.t. Q ̅=Q\/B § Claim: G≅A×Q □ By Lemma 55, we need only show A∩Q={1} and AQ=G □ A∩Q={1} ® Let g∈A∩Q, then g=a^i for some i ® Thus, a^i B∈⟨aB⟩∩Q ̅≤G ̅ ® Since G ̅≅⟨aB⟩×Q ̅, ⟨aB⟩∩Q ̅={1} ® Therefore a^i B=1_G ̅ ® ⇒├ |a|=|aB|┤ |i┤ ® ⇒a^i=1 ® ⇒A∩Q={1} □ AQ=G ® Let g∈G ® Since G ̅=⟨aB⟩×Q ̅, ® gB=a^i ByB for some a^i B∈⟨aB⟩ and yB∈Q ̅, ® Thus gB=a^i yB⟺g(a^i y)^(−1)∈B ® Choose b∈B s.t. ga^(−i) y^(−1)=b ® Then g=⏟(a^i )┬(∈A) ⏟yb┬(∈Q) ® Therefore AQ=G ○ Case 2: ∄b∈G s.t. b∉A and |b|=p § In this case, we need to prove G=A § By way of contradiction, suppose otherwise § Choose x∈G∖A with the smallest order § Recall: If H=⟨z⟩,then |⟨z^m ⟩|=|z|/((|z|,m) ) § |x^p | |x|, so x^p∈A § Choose i s.t. x^p=a^i § Say |a|=p^s § Since a has maximal order, x^(p^s )=1 § ⇒1=x^(p^s )=(x^p )^(p^(s−1) )=(a^i )^(p^(s−1) )=a^(ip^(s−1) ) § It follows that ├ p┤ |i┤ § So x^p=a^i, where ├ p┤ |i┤ § Set y≔a^(−i\/p) x, then y^p=a^(−i) x^p=1 § But y∉A, since ya^(i\/p)=x∉A § This contradicts the assumption that ∄b∈G s.t. b∉A and |b|=p § So G∖A=∅ § Therefore G=A=⟨a⟩, and Q={1} Theorem 59: Fundamental Theorem of Finite Abelian Groups • Statement ○ Every finite abelian group G is a product of cyclic groups • Proof ○ Say |G|=p_1^(m_1 )⋯p_n^(m_n ), where p_i are distinct primes ○ By Corollary 57, and induction G≅P_1×…×P_n where ○ P_i={x∈G│x^(p_i^(m_i ) )=1} and |P_i |=p_i^(m_i ) ○ So, it suffices to show each P_i is a product of cyclic groups ○ By Lemma 58, P_i≅A_i×Q_i, where A_i is cyclic ○ The result immediately follows by induction on m_i • Example ○ How may abelian groups of order 8 are there up to isomorphism ○ There are 3 abelian groups of order 8: Z8ℤ, Z2ℤ×Z4ℤ, Z2ℤ×Z2ℤ×Z2ℤ Corollary 60 • Partition ○ A partition of n∈Z( 0) is a way of writing n as a sum of positive integers ○ Example: 3 has 3 partitions: 3, 2+1, 1+1+1 • Statement ○ If n=p_1^(e_1 )⋯p_n^(e_m ), where p_i are distinct primes ○ Then the number of finite abelian groups of order n is ○ ∏_(i=1)^m▒〖number of partitions of e_i 〗 • Note ○ If (λ^1,…,λ^m ) are partitions of e_1,…,e_m, where λ_i={λ_i^1,…,λ_i^(s_i ) } ○ Then this list of partitions corresponds to the abelian group ○ (Z(p_1^(λ_1^1 ) Z×…×Z(p_1^(λ_1^(s_1 ) ) Z)×…×(Z(p_1^(λ_m^1 ) Z×…×Z(p_1^(λ_m^(s_m ) ) Z) • Example ○ When n=72=2^3⋅3^2 ○ Z2ℤ×Z2ℤ×Z2ℤ×Z3ℤ×Z3ℤ≅Z2ℤ×Z6ℤ×Z6ℤ ○ Z2ℤ×Z2ℤ×Z2ℤ×Z9ℤ ○ Z4ℤ×Z2ℤ×Z3ℤ×Z3ℤ ○ Z4ℤ×Z2ℤ×Z9ℤ ○ Z8ℤ×Z3ℤ×Z3ℤ ○ Z8ℤ×Z9ℤ$

Math 541 - 4/6

$Homework 8 Question 3 • Statement ○ If G is a group with |G|≤11, and ├ d┤ ||G|┤, then G has a subgroup of order d • Proof ○ For |G|=2,3,5,7,11 § |G| is prime, thus cyclic ○ For |G|=4,6,9,10 § |G| is product of two primes, so use the Cauchy s Theorem ○ For |G|=8 § d∈{1,2,4,8} § When d=1,2,8, this is obvious § So assume d=4 § If G contains an element of order 4, then we are done § So, we may assume |g|=2,∀g∈G∖{1}, then G is abelian § Let a,b∈G∖{1}. Let H≔{1,a,b,ab} § H is closed under inverse □ The inverse of every element of G is itself § H is closed under multiplication □ ■8(⋅&1&a&b&ab@1&1&a&b&ab@a&a&1&ab&b@b&b&ab&1&a@ab&ab&b&a&1) Lemma 56 • Statement ○ Let G be a finite abelian group of order mn, where (m,n)=1 ○ If M={x∈G│x^m=1}, N={x∈G│x^n=1}, then ○ M,N≤G and the map α:M×N→G given by (g,h=gh is an isomorphism ○ Moreover, if m,n≠1, then M and N are nontrivial • Proof ○ M,N≤G § It suffices to check M≤G § M≠∅, since 1∈M § If x,y∈M, then (xy^(−1) )^m=x^m (y^m )^(−1)=1. Thus xy^(−1)∈M ○ MN=G § Choose r,s∈Z s.t. mr+ns=1 § Let g∈G, then g=g^(mr+ns)=g^mr g^ns § (g^mr )^n=(g^mn )^r=(g^|G| )^r=1 by Lagrange s Theorem § Similarly, (g^ns )^m=1 § So, g^ns∈M, g^mr∈N, so g∈MN § Therefore MN=G ○ M∩N={1} § Let g∈M∩N, then g^m=1=g^n § Then ├ |g|┤ |m┤ and ├ |g|┤ |n┤ § Since (m,n)=1,|g|=1 § Thus M∩N={1} ○ By Lemma 55, M∩N={1} and MN=G⇒α is an isomorphism ○ M and N are nontrivial § Suppose m≠1 § Let p be a prime divisor of m § Then G contains an element z of order p, by Cauchy s Theorem § z∈M, so M≠{1} § Similarly, if n≠1, N≠{1} Corollary 57 • Statement ○ Let G be a finite abelian group, and p be a prime divisor of |G| ○ Choose m∈Z( 0) s.t. |G|=p^m n and p∤n ○ Then G≅P×T, where P,T≤G, |P|=p^m, and p∤|T| • Intuition ○ If |G|=p_1^(m_1 ) p_2^(m_2 )…p_n^(m_n ) ○ This corollary says G≅P_1×…×P_n, where |P_i |=p_i^(m_i ) ○ This reduces the Fundamental Theorem of Finite Abelian Groups ○ to the case where the group has order given by a prime power • Proof ○ Let P≔{x∈G│x^(p^m )=1}, T≔{x∈G│x^n=1} ○ By Lemma 56, G≅P×T ○ p∤|T| § Suppose, by way of contradiction, that ├ p┤ ||T|┤ § By Cauchy s Theorem, ∃z∈T s.t. |z|=p § Since z∈T, z^n=1, so ├ p┤ |n┤ § This is impossible, thus p∤|T| ○ |P|=p^m § Since |G|=|P|⋅|T|=p^m n, ├ p^m ┤ ||T|┤ § Suppose p^m |P| § Then, ∃ prime q s.t. p≠q and ├ q┤ ||P|┤ § By Cauchy s Theorem, ∃y∈P s.t. |y|=q § This is impossible since y∈P⇒y^(p^m )=1⇒├ q┤ |p^m ┤ § Thus p^m=|P|$

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 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

Math 541

Math 541 - 4/9

Math 541 - 4/6

Math 541 - 4/4

Math 541 - 4/2

Math 541 - 3/23

Search

WeChat Account