Math 541 Archives - Page 6 of 8 - Shawn Zhong

Math 541 - 2/23

$Theorem 20 • Let G=⟨g⟩ be a cyclic group • Statement (1) ○ Every subgroup of G is cyclic ○ More precisely, if H≤G, then either H={1} or ○ H=⟨g^d ⟩, where d is the smallest positive integer s.t. g^d∈H • Proof (1) ○ Assume H≠{1} ○ Choose a≠0 s.t. g^a∈H, then (g^a )^(−1)=g^(−a)∈H ○ Thus, H contains some positive power of g ○ Let S≔{b∈Z( 0) |g^b∈H}, then S≠∅ ○ By the Well-Ordering Principle, S contains a minimum element d ○ Thus, ⟨g^d ⟩⊆H; we must show H⊆⟨g^d ⟩ ○ Let h∈H, then h=g^a for some a∈Z ○ Choose q,r∈Z s.t. a=qd+r, 0≤r d ○ g^d∈H⇒g^(−qd)∈H⇒g^a g^(−qd)∈H⇒g^r∈H ○ If r 0, then r∈S, which is impossible, since r d ○ Therefore r=0 ○ So g^a=g^qd∈⟨g^d ⟩⇒H⊆⟨g^d ⟩ ○ Therefore H=⟨g^d ⟩ • Statement (2) ○ If G is finite, then for all positive integers a dividing n ○ ∃! subgroup H≤G of order a ○ Moreover, this subgroup is ⟨g^d ⟩, where d=n/a • Proof (2) ○ Let a be a positive divisor of n=|G| ○ Let d≔n/a⇒n/d=a ○ Existence § |⟨g^d ⟩|=n/((d,n) )=n/d=a by Proposition 19 § This proves existence ○ Uniqueness § Without loss of generality, assume a 1 § Suppose H≤G and |H|=a § We must show H=⟨g^d ⟩ § By (1), H=⟨g^b ⟩, where b is the smallest positive integer s.t. g^b∈H § We have n/d=a=|H|=|⟨g^b ⟩|=n/((n,b) ) by Proposition 19 § Thus d=(n,b) i.e. d|b § Thus g^b∈⟨g^d ⟩⇒⟨g^b ⟩≤⟨g^d ⟩ § Since |⟨g^b ⟩|=|⟨g^d ⟩|, ⟨g^b ⟩=⟨g^d ⟩ § i.e. H=⟨g^d ⟩ Subgroups Generated by Subsets of a Group (Section 2.4) • Lemma: If {H_i }_(i∈I) is a family of subgroups of G, then ⋂136_(i∈I)▒〖H_i≤G〗 ○ Let H≔⋂136_(i∈I)▒H_i ○ H≠∅ because 1∈H_i, ∀i∈I ○ Let h1,h2∈H, then h1,h2∈H_i, ∀i∈I ○ ⇒h1 h2^(−1)∈H_i,∀i∈I ○ ⇒h1 h2^(−1)∈H • Definition ○ Let G be a group and A⊆G ○ The subgroup generated by A is ○ the intersection of every subgroup of G containing A ○ ⟨A⟩≔⋂8_█(H≤G@A⊆H)▒H • Example ○ If A=∅, then ⟨A⟩={1} ○ If A={1}, then ⟨A⟩={1}$

Math 541 - 2/21

$Cyclic Group • Definition ○ A group G is cyclic if ∃g∈G s.t. ⟨g⟩=G • Note ○ A finite group G of order n is cyclic iff ∃g∈G s.t. |g|=n • Example 1: Z is cyclic ○ Z=⟨1⟩ ○ Z=⟨−1⟩ • Example 2: Z\/nZ is cyclic ○ If (a,n)=1, Z\/nZ=⟨a ̅ ⟩ ○ (We will prove this later) • Example 3: S_3 is not cyclic ○ Note: If (a_1,…,a_t )∈S_n is a t-cycle, then |(a_1,…,a_t )|=t ○ S_3={(1),(1 2),(1 3),(2 3),(1 2 3),(1 3 2)} ○ Every element in S_3 have order 1,2, or 3 ○ So S_3 cannot be cyclic Proposition 18 • Let G be a cyclic group • If |G|=∞, then G≅Z ○ Choose g∈G s.t. G=⟨g⟩ ○ Define a map f:Z→G by n↦g^n ○ Check homomorphism § If n_1,n_2∈Z § then f(n_1+n_2 )=g^(n_1+n_2 )=g^(n_1 ) g^(n_2 )=f(n_1 )f(n_2 ) § Thus, f is a homomorphism ○ Check surjectivity § Surjectivity is clear ○ Check injectivity § Suppose f(n_1 )=f(n_2 ) § Then g^(n_1 )=g^(n_2 ) § Without loss of generality, assume n_1≥n_2 § Then g^(n_1−n_2 )=1 § Since |g|=∞ § n_1−n_2=0 § i.e. n_1=n_2 § Thus f is injective • If |G|=n ∞, then G≅Z\/nZ ○ Choose g∈G s.t. G=⟨g⟩ ○ Define a map f:Z\/nZ→G by a ̅↦g^a ○ Check well-definedness § We need to check that f is well-defined. § That is we must show that if a ̅=b ̅ in Z\/nZ, then f(a ̅ )=f(b ̅ ) § Let a,b∈Z, suppose a ̅=b ̅ in Z\/nZ § Choose q∈Z s,t, nq=a−b § f(a ̅ )=g^a=g^(nq+b)=g^nq g^b=g^b=f(b ̅ ) § Thus, f is well-defined ○ Check homomorphism § f(a ̅+b ̅ )=g^(a+b)=g^a g^b=f(a ̅ )f(b ̅ ) § Thus, f is a homomorphism ○ Check surjectivity § Surjectivity is clear ○ Check injectivity § If f(a ̅ )=f(b ̅ ) § g^a=g^b § g^(a−b)=1 § ├ |g|┤|├ (a−b)┤ § n|(a−b) § a ̅=b ̅ § Thus f is injective Least Common Multiple • Definition ○ Let a,b∈Z where one of a,b is nonzero. ○ A least common multiple of a and b is a positive integer m s.t. § a|m and b|m § If a|m′ and b|m′, then m|m^′ ○ We denote the lcm of a and b by [a,b] ○ Define [0,0]≔0 • Uniqueness ○ Similar to the proof of uniqueness of gcd • Existence: If a,b∈Z, and one of a,b is nonzero, then ab/((a,b) ) is the lcm of a,b ○ ab/((a,b) ) is a multiple of a and b ○ Suppose m^′∈Z s.t. a|m′ and b|m′ ○ We must show ├ ab/((a,b) )┤|├ m′┤ ○ Choose q,q^′∈Z s,t, aq=m′ and bq^′=m′ ○ Choose x,y∈Z s.t. ax+by=(a,b) ○ m^′ (a,b)=m^′ (ax+by)=m^′ ax+m^′ by=bq^′ ax+aqby=ab(q^′ x+qy) ○ Thus ab|(m^′ (a,b)) ○ i.e. ab/((a,b) )=m^′ Proposition 19 • Statement ○ If G=⟨g⟩ is cyclic, and |G|=n ∞, then |g^a |=n/((a,n) ) • Proof ○ Let a∈Z ○ If a=0, this is clear ○ So, assume a≠0 ○ (g^a )^(n/((a,n) ))=g^(an/((a,n) ))=g^[a,n] =g^kn for some integer k ○ Thus, (g^a )^(n/((a,n) ))=(g^n )^k=1 since n=|g| ○ Suppose t∈Z( 0) and (g^a )^t=1 ○ By HW3, g^at=1⇒n|at ○ Thus, at is a common multiple of n and a ○ ├ [a,n]┤|├ at┤ ○ ⇒├ an/((a,n) )┤|├ at┤ ○ ⇒├ n/((a,n) )┤|├ t┤ ○ In particular, n/((a,b) )≤t Theorem 20 • Let G=⟨g⟩ be a cyclic group • Statement (1) ○ Every subgroup of G is cyclic ○ More precisely, if H≤G, then either H={1} or ○ H=⟨g^d ⟩, where d is the smallest positive integer s.t. g^d∈H • Statement (2) ○ If G is finite, then for all positive integers a dividing n ○ ∃! subgroup H≤G of order a. ○ Moreover, this subgroup is ⟨g^d ⟩, where d=n/a$

Math 541 - 2/19

$Regular n\-gon • A regular n\-gon is a polygon with all sides and angles equal Symmetry • Definition ○ A symmetry of a regular n-gon is a way of § picking up a copy of it § moving it around in 3d § setting it back down ○ so that it exactly covers the original • Examples ○ Rotations ○ Reflection Dihedral Groups (Section 1.2) • Definition ○ D_2n≔{symmetries of the n-gon} is called n-th dihedral groups • Note ○ |D_2n |=2n (proof on page 24) ○ There are n rotations and n reflections ○ Symmetries of n-gons are determined by ○ the permutations of the vertices they induce • Example: n=3 ○ Rotations § 120°:(1 2 3) § 240°:(1 3 2) § 360°:(1) ○ Reflections § (2 3) § (1 3) § (1 2) ○ D_6≅{(1),(2 3),(1 3),(1 2),(1 3 2),(1 2 3)}=S_3 • Example: n=4 ○ Rotations § 90°:(1 2 3 4) § 180°:(1 3)(2 4) § 270°:(1 4 3 2) § 360°:(1) ○ Reflections § (2 4) § (1 3) § (1 4)(2 3) § (1 2)(3 4) ○ D_8≅{(1),(1 2 3 4),(1 3)(2 4),(1 4 3 2),(1 3),(2 4),(1 4)(2 3),(1 2)(3 4)}≤S_4 • Fact ○ In general D_2n is isomorphic to a subgroup of S_n ○ Every finite group is isomorphic to a subgroup of a symmetric group Proposition 17 (The Subgroup Criterion) • Statement ○ A subset H of a group G is a subgroup iff ○ H≠∅ ○ ∀x,y∈H, xy^(−1)∈H • Recall the original definition ○ A subset H of a group G is a subgroup iff ○ H≠∅ ○ ∀h,h′∈H, hh′∈H ○ ∀h∈H, h(−1)∈H • Proof ○ (⟹) Clear ○ (⟸) We must check that H is closed under multiplication and inversion ○ Let x∈H ○ 1⋅x^(−1)∈H ○ Thus x^(−1)∈H ○ Let y∈H, then y^(−1)∈H ○ So x(y^(−1) )^(−1)∈H ○ xy∈H Examples of Subgroups • Example 1 ○ Z≤Q≤R≤ℂ • Example 2 ○ Definition § Fix n∈Z( 0) § SL_n (R≔{A∈GL_n (R│det⁡A=1} is called the special linear group ○ Claim § SL_n (R≤GL_n (R ○ Proof § SL_n (R≠∅, since I_n∈SL_n (R § Let A,B∈SL_n (R § det⁡(AB^(−1) )=det⁡A⋅det⁡〖B^(−1) 〗=det⁡A/det⁡B =1/1=1∎ • Example 3 ○ Definition § If G is a group § Z(G)≔{a∈G│ag=ga, ∀g∈G} is called the center or G ○ Claim § Z(G)≤G ○ Proof § Z(G)≠∅, since 1∈Z(G) § Let a,b∈Z(G) § If g∈G, abg=agb=gab § so Z(G) is closed under multiplication § Also a^(−1) g=(g^(−1) a)^(−1)=(ag^(−1) )^(−1)=ga^(−1) § so Z(G) is closed under inversion Cyclic Groups • Definition ○ A group G is cyclic if ∃g∈G s.t. ⟨g⟩=G ○ In this case, we say G is generated by g • Note ○ If G is finite of order n, then ○ G is cyclic iff ∃g∈G s.t. |g|=n$

Math 541 - 2/16

$Proposition 16 • Let f:G→H be an isomorphism • G is abelian if and only if H is abelian ○ (⟹) Suppose G is abelian ○ Let h,h′∈H ○ Choose g,g^′∈G s.t. f(g)=h,f(g′)=h′ ○ Then hh′=f(g)f(g^′ )=f(gg^′ )=f(g^′ g)=f(g^′ )f(g)=h′ h ○ (⟸) Apply the same argument with f^(−1):H→G • ∀g∈G, |g|=|f(g)| ○ Proof: f(1_G )=1_H § Let g∈G, then § f(g)=f(1_G⋅g)=f(1_G )⋅f(g) § By Cancellation Law, f(1_G )=1_H ○ Proof: When |g| ∞ § Let n≔|g|, then § 1_H=f(1_G )=f(g^n )=f(g)^n § (This last equality follows from an induction argument) § Therefore, |f(g)|≤n § Now, apply this same argument with f replaced by f^(−1) § So we can conclude that |f(g)|=n ○ Proof: When |g|=∞ § If |f(g)| ∞ § The above argument shows |g| ∞ § This is impossible § Thus, |f(g)|=∞ ○ Note § This result also holds if we only assume f is injective Order and Homomorphism • G,H are groups, and |G|=|H|, is it the case that G≅H? No • Counterexample 1 ○ Z and Q ○ |Z=|Q, but Z≇Q • Fact: Any homomorphism f:Z→Q is not surjective ○ Let f:Z→Q be a homomorphism ○ If f(a)=0,∀a∈Z § Obviously f is not surjective ○ Assume otherwise § By induction, f(a)=f ⏟((1+1+…+1) )┬(n copies)=a⋅f(1) § By assumption, f(1)≠0; otherwise f=0 § We know that f(1)/2∈Q § But ∄a∈Z s.t. f(1)/2=af(1) § i.e. f(1)/2∉im(f) § Thus f is not surjective • Counterexample 2 ○ Z\/6Z and S_3 ○ |Z6Z=|S_3 |, but Z\/6Z≇S_3 ○ Because Z\/6Z is abelian, but S_3 is not ○ Also |1 ̅ |=6 in Z\/6Z, but S_3 have no element of order 6 Orders of elements in S_n • Let σ∈S_n • If σ=σ_1⋯σ_m, where σ_1⋯σ_m are disjoint cycles • Then |σ|=lcm(|σ_1 |,…,|σ_m |) • Also, if τ is a t-cycle, then |τ|=t Subgroup • Definition ○ Let G be a group, and let H⊆G ○ H is a subgroup if § H≠∅ (nonempty) § If h,h′∈H, then hh′∈H (closure under the operation) § If h∈H, then h(−1)∈H (closure under inverse) ○ We will write H≤G • Note ○ Subgroups of a group are also groups • Example 1 ○ If G is a group, then G≤G and {1}≤G • Example 2 ○ If m,n∈Z( 0), and n≥m, then S_m≤S_n • Example 3 ○ Let G be a group, and let g∈G ○ Then ⟨g⟩≔{g^n│n∈Z≤G ○ ⟨g⟩ is called the cyclic subsgroup generated by g ○ ⟨g⟩≠∅, since g∈⟨g⟩ ○ Let g^i,g^j∈⟨g⟩, then g^i g^j=g^ij∈⟨g⟩ ○ If g^i∈⟨g⟩, then (g^i )^(−1)=g^(−i)∈⟨g⟩ Regular n\-gon • A regular n\-gon is a polygon with all sides and angles equal$

Math 541 - 2/14

$Homomorphism • Definition ○ Let G,H be groups ○ A function f:G→H is a homomorphism if ○ f(g_1 g_2 )=f(g_1 )f(g_2 ), ∀g_1,g_2∈G ○ One says f "respects", or "preserves" the group operation • Trivial Examples ○ Let G be a group ○ The identity map of G→G is a homomorphism § f(g_1 )f(g_2 )=1⋅1=1=f(g_1 g_2 ) ○ The map G→G given by g↦1 is a homomorphism § This only works if we send every element of G to 1 § If x∈G∖{1}, and f:G→G is given by g↦x,∀g § f(g_1 g_2 )=f(g_1 )(g_2 ) § i.e. x=x^2 § Thus x=1 § This is impossible since x∈G∖{1} • Example 1 ○ Let f:R→R∗ be given by f(x)=e^x ○ f is a homomorphism ○ f(x_1+x_2 )=e^(x_1+x_2 )=e^(x_1 ) e^(x_2 )=f(x_1 )f(x_2 ) • Example 2 ○ Let G be a group, and let x∈G ○ The map f:G→G, g↦xgx^(−1) is a homomorphism ○ f(g_1 g_2 )=xg_1 g_2 x^(−1)=xg_(1 ) x^(−1) xg_2 x^(−1)=f(g_1 )f(g_2 ) ○ This homomorphism is called conjugation by x • Example 3 ○ Let n∈Z ○ Is f:Z→Z, x↦x+n a homomorphism? ○ Answer: Only when n=0 ○ f(0+0)=f(0) ○ i.e. n+n=n ○ Thus n=0 • Example 4 ○ Let n∈{0,1,2,3,…} ○ Is α:Z→Z, x↦x^n a homomorphism? ○ Answer: Only when n=1 ○ When x=0 § α(x)=1 § 1 is not the identity (0 is) § So this doesn’t work ○ For n≥2 § α(x_1+x_2 )=α(x_1 )+α(x_2 ) § (x_1+x_2 )^n=x_1^n+x_2^n § This is not always true § For instance, when x_1=x_2=1, 2^n≠2 for n≥2 • Example 5 ○ Let n∈Z ○ Is β:Z→Z, x↦nx a homomorphism? ○ Answer: Yes ○ β(x_1+x_2 )=n(x_1+x_2 )=nx_1+nx_2=β(x_1 )+β(x_2 ) • Example 6 ○ The previous examples is a special case of the following: ○ Let G be a group, and n∈Z ○ Define β:G→G, g↦g^n ○ Then β is homomorphism ∀n∈Z iff G is abelian ○ Proof: homomorphism ⇒ abelian § Say n=−1 § Let g_1,g_2∈G § β(g_1,g_2 )=β(g_1 )β(g_2 ) § (g_1 g_2 )^(−1)=g_1^(−1) g_2^(−1) § g_2^(−1) g_1^(−1)=g_1^(−1) g_2^(−1) § (g_2^(−1) g_1^(−1) )^(−1)=(g_1^(−1) g_2^(−1) )^(−1) § (g_1^(−1) )^(−1) (g_2^(−1) )^(−1)=(g_2^(−1) )^(−1) (g_1^(−1) )^(−1) § g_1 g_2=g_2 g_1 § Thus G is abelian ○ Proof: abelian ⇒ homomorphism § (Homework) Isomorphism • Definition ○ Let G,H be groups ○ A homomorphism α:G→H is a isomorphism if ○ there is a homomorphism β:H→G s.t. ○ αβ=id_H and βα=id_G ○ In this case, we say G and H are isomorphic • Fact ○ L:G→H is an isomorphism iff it is a bijective homomorphism ○ Proof: isomorphism ⇒ bijective homomorphism § Clear ○ Proof: bijective homomorphism ⇒ isomorphism § Need to show α^(−1) is a homormorphism § (Homework) • Example ○ R(0)≔{r∈Rr0} is a group under multiplication ○ Define f:R→R(0) where f(x)=e^x ○ Then f is a homomorphism ○ Moreover, f is an isomorphism ○ The inverse of f is ln • Observation ○ If G,H are isomorphic groups, then |G|=|H| Proposition 16 • Statement ○ If f:G→H is an isomorphism, then ○ G is abelian iff H is abelian ○ ∀g∈G, |g|=|f(g)| ○ Note |g|=inf⁡{n∈Z(0)│g^n=1}$

Shawn Zhong

Shawn Zhong

Math 541

Math 541 - 2/23

Math 541 - 2/21

Math 541 - 2/19

Math 541 - 2/16

Math 541 - 2/14

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