Math 541 - 2/28

Coset • If G is a group, H≤G, and g∈G • Define gH≔{ghhH} • Similarly, define Hg≔{h�│hH} • gH is caleld a left coset, and Hg is called a right coset • An element of a coset is called a representative of the coset Proposition 22: Properties of Coset • Let G be a group and H≤G, then • For g_1,g_2∈G,g_1 H=g_2 H⟺g_2^(−1) g_1∈H ○ (⟹) Choose h∈H s.t. g_1=g_2 h (since g_1=g_1⋅1∈g_1 H=g_2 H) ○ Therefore g_2^(−1) g_1=h∈H ○ (⟸) Choose h∈H s.t. g_1=g_2 h ○ ∀h′∈H, g_1 h′=g_2 ⏟(h^′ )┬(∈H)∈g_2 H⇒g_1 H⊆g_2 H ○ ∀h′∈H,g_2 h′=g_1 ⏟(h(−1) h′ )┬(∈H)∈g_1 H⇒g_2 H⊆g_1 H ○ Therefore g_1 H=g_2 H • The relation ~ on G given by g_1~g_2 iff g_1∈g_2 H is a equivlance relation ○ Reflexive § If g∈G, then g=g⋅1∈gH § So g~g ○ Symmetric § If g_1,g_2∈G, and g_1~g_2 i.e. g_1∈g_2 H, then § g_1=g_2 h for some h∈H § Thus g_1 h(−1)=g_2 § So g_2∈g_1 H, which means g_2~g_1 ○ Transitive § Suppose g_1~g_2 and g_2~g_3 § This means g_1∈g_2 H and g_2∈g_3 H § Choose h1,h2∈H s.t. g_1=g_2 h, and g_2=g_3 h § Then g_1=g_3 h2 h1∈g_3 H § So g_1~g_2 • In particular, left/right cosets are either equal or disjoint ○ Suppose g_1,g_2∈G, and z∈g_1 H∩g_2 H ○ Suppose x∈g_1 H, then x_{g_1}z~g_2 ○ So x∈g_2 H ○ This implies that g_1 H⊆g_2 H ○ To get g_2 H⊆g_1 H, exchange the roles of g_1 and g_2 ○ Therefore g_1 H=g_2 H • Example 1 ○ Let G be a group, H≤G. If h∈H, then hH=H ○ Let h′∈H, then h′=h(h(−1) h′ )∈hH ○ Thus H⊆hH ○ By closure under the operation, hH⊆H ○ Therefore hH=H • Example 2 ○ Let G=Z\6Z, and H= unique subgroup of Z\6Z of order 2 ○ H={0 ̅,3 ̅ }≤Z\6Z ○ Left cosets of H in G § 0 ̅+{0 ̅,3 ̅ }={0 ̅,3 ̅ } § 1 ̅+{0 ̅,3 ̅ }={1 ̅,4 ̅ } § 2 ̅+{0 ̅,3 ̅ }={2 ̅,5 ̅ } § 3 ̅+{0 ̅,3 ̅ }={0 ̅,3 ̅ } § 4 ̅+{0 ̅,3 ̅ }={1 ̅,4 ̅ } § 5 ̅+{0 ̅,3 ̅ }={2 ̅,5 ̅ } ○ Note § |G|=6, |H|=2, and H has 3 distinct cosets (2⋅3=6) § If G is a finite group, and H≤G, then ├ |H|┤|├ |G|┤, and § H has |G|/|H| distinct left (or right) cosets in G Normal Subgroup • Definition ○ Let G be a group, let N≤G ○ N is a normal subgroup if gng^(−1)∈N, ∀n∈N,∀g∈G ○ In other words, N is closed under conjugation ○ If N≤G is normal, we write N⊴G • Example 1 ○ If G is abelian, every subgroup of G is normal ○ Suppose H≤G ○ Let h∈H and g∈G ○ Thus ghg^(−1)=hgg(−1)=h∈H • Example 2 ○ Let G=S_3, H=⟨(1 2)⟩ ○ Suppose g=(1 2 3), h=(1 2) ○ ghg^(−1)=(1 2 3)(1 2) (1 2 3)^(−1)=(1 2 3)(1 2)(1 3 2)=(2 3)∉H ○ Therefore H⋬G • Example 3 ○ ⟨(1 2 3)⟩ in S_3 is normal • Example 4 ○ Let G=GL_n (R ○ Let P,A∈G, then PAP^(−1) is change of basis matrix ○ Note: In GL_n (R, conjugation amounts to changing basis Proposition 23: Properties of Normal Subgroup • Statement ○ Let N be a subgroup of a group G ○ N⊴G iff gN=Ng,∀g∈G • Proof (⟹) ○ Suppose N⊴G ○ Let g∈G, n∈N ○ gn=gn(g^(−1) g)=⏟(gng^(−1) )┬(∈N) g∈Ng⇒gN⊆Ng ○ ng=(gg^(−1) )ng=g ⏟(g^(−1) ng)┬(∈N)∈gN⇒Ng⊆gN ○ Therefore gN=Ng • Proof (⟸) ○ Suppose gN=Ng,∀g∈G ○ Let g∈G,n∈N ○ We must show gng^(−1)∈N ○ Choose n^′∈N s.t. gn=n^′ g ○ Then gng⁽⁻¹⁾⁼ⁿ′∈N ○ Therefore N⊴G • Example ○ Let f:G→H be a homomorphism, then ker⁡f⊴G ○ ker⁡f is a subgroup of G § ker⁡f≠∅, since f(1_G )=1_H § If k_1,k_2∈ker⁡f § f(k_1 k_2^(−1) )=f(k_1 )f(k_2 )^(−1)=1_H § Thus k_1 k_2^(−1)∈ker⁡f § Therefore ker⁡f≤G ○ ker⁡f is normal § Let g∈G,k∈ker⁡f § f(gkg^(−1) )=f(g)f(k)f(g)^{(−1)=f(g)f(g)}(−1)=1_H § ⇒gkg^(−1)∈ker⁡f∎