1 3 assume g 1 h g h for all g g and all h h so the

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( 1 3 ) Assume g 1 h g H for all g G and all h H . So the conjugacy class of h is in H . Suppose H is not a union of conjugacy classes in G . Then there exists z H not in a conjugacy class. Now g 1 z g H , so the conjugacy class of z is in H . Contradiction. ( 2 4 ) g 1 H g = H g g 1 H g = g H H g = g H . Definition: Quotient Group The quotient group G / H is the set of cosets { gH }∀ g G with the operation x H y H = x y H where is the group operation from G . Note: This only works if H is a normal subgroup in G . Example Let H ={ e ,r ,r 2 }⊂ D 3 be normal. Then D 3 / H ={ eH ,sH } . 8 of 17
MAT301H1: Groups and Symmetries Fact G / H = G H (since L g is 1:1). Properties of Quotient Groups G / G ={ g G g G } { e } . If G is abelian, then G / H is abelian. Proof: x H y H = x y H = y x H = y H x H since x y = y x . If G is cyclic, then G / H is cyclic. Proof: G ={ g = 1, ,n } . The cosets in G / H are { g H } , so g H generates G / H whenever g generates G . If G is finitely generated, then G / H is finitely generated. Proof: If g 1, ,g n generate G , then g 1 H , , g n H generate G / H . Claim x H y H = x y H if and only if H is normal. Proof: ( ) x H y H = x y H x 1 x H y H = x 1 x y H H y H = y H , so h i yh j = y h k (fixing h i and h j we get h k ), so h i y = y h k h j 1 . Let h = h k h j 1 which can be anything in H . Then h i y = y h H y = y H , so H is normal. ( ) Clear from definition. Definition: Kernel, Image Let f :G G' . Then Ker f ={ g G f g = e } , and Im f ={ g ' G ' g G such that f g = g ' } . Theorem: 1 st Isomorphism Theorem Let f :G G' be a homomorphism. Then Ker f ≙ G and G / Ker f ≅ Im f . Proof: Want: Ker f ≙ G . Let x Ker f . Then f g 1 x g = f g 1 f x f g = f g 1 e f g = e , so g 1 x g Ker f . Now, f z f e z = f e f z  f e = e , so Ker f ≠∅ . Hence Ker f ≙ G . Want: G / Ker f ≅ Im f . Let K = Ker f . Let :G / Ker f  Im f where x K = f x . Then is a homomorphism since  a K b K = ab K = f a b = f a f b = a K  b K . is onto Im f since is onto Im  by definition of image, and Im = Im f by construction. is 1:1 since  a K = b K ⇒ f a = f b ⇒ f a f b 1 = e f a f b 1 = e f ab 1 = e a b 1 K , so a b 1 K = K ab 1 K b K = K b K a K = b K . Hence :G / Ker f  Im f is an isomorphism and so G / Ker f ≅ Im f . Corollary Let f :G G' be a homomorphism.

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