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Week 8 Factor groups (From Homomorphisms.) Let : G G be a homo. and H Ker . Then a G, aH Ha . Let G H be the set of left cosets of H in G and define aH bH abH . It is well-defined. Hb bH ! Note: aH a and bH b . G H is read as “G over H”, “G modulo H” or “G mod H”. (*) G H is a group. Theorem Let : G G be a group homo. with kernel H . Then the cosets of H form a factor group, G H , where aH bH abH . Also, the map : G H G defined by aH a is an isomorphism. Both coset multiplication and are well-defined, independent of the choices a and b from the cosets. Proof. aH bH abH ab a b aH bH . a, aH , s.t. aH a . (Onto) aH e a e a H aH H . e.g. : Z Z n , m r if m qn r, 0 r n . Ker nZ Cosets: nZ , 1 nZ , …, n 1 nZ . (The residue classes modulo n) ■ The group of cosets Z nZ . : Z nZ Z n (Isomorphism). Note: We also use Z n to denote the factor group Z nZ . G H is read as factor group of G modulo H Theorem. (From Normal subgroups) Let H G . Then left coset multiplication is well-defined by aH bH abH if and only if H G . Proof. ( ) a H , consider aH and let x aH . Since a 1 a 1 H xa1 aH a 1 H H x Ha . Similarly, y Ha , y h1a , y 1 a 1h1 1 , y 1 a 1 H y 1a a 1 H aH H a 1 y H y aH . Hence H G . ( ) x aH and y bH , we claim: xH yH abH . z xH yH , z xh1 yh2 ah3 h1bh4 h2 ah1' bh2' abh3' abH . z abH , z abh ah1h11bh ah1bh2' h aH bH . Corollary ■ Let H G . Then the cosets of H in G from a group G H under the operation aH bH abH . Proof. It is easy to check. Defn. G H is the factor group (or quotient group) of G by H. e.g. nZ Z Z nZ is a group. e.g. Let c 5.8 c x | x and x 0, c c . c , is a group. c 5.8 7.5 c 1.2 6.3c 0.5 . c 3c, 2c, c, 0, c, 2c, 3c c c cZ x 0, c, x c is a coset of c in . given by x xH is a homo with Ker H . Theorem H G :G G Theorem (The fundamental homo. thm.) H Let : G G be a group homo. with Ker ( ) H . Then : G H G given by gH g g G is an isomorphism. If : G G H is a homo. given by g gH , then g g g G or . Proof. K e r G . G H is a group. is a homo. is a homo. Ker H . is 1-1. The others are easy to check. ■ (1) ghg 1 H g G and h H . H G (2) gHg 1 H g G. (3) gH Hg g G. (*) In order to check H G , it suffices to show that g G gHg 1 H . (Why?) Since h H , h g g 1hg g 1 g g 1hg 1 g 1 gh1 g 1 gHg 1 . 1 H gHg 1 whenever gHg 1 H . Note: g G , gHg 1 is also a subgroup of G and | H | | gHg 1 | . (**) A normal subgroup plays an important role in finding a factor group. Theorem (Intersection of normal subgroups) Let H 1 and H 2 be two normal subgroups of G . Then H1 H 2 G . Proof. g G , g H1 H 2 g 1 H1 and g H 1 H 2 g 1 H 2 . Hence g H1 H 2 g 1 H1 H 2 . ■ Theorem Let S be a subset of a group G then there exists a normal subgroup of smallest order which contains S. Proof. Let K H i , H i S and H i G . ■ iI Defn. (Automorphism) An isomorphism : G G of a group G with itself is an automorphism of G . The automorphism i g x gxg 1 x G is the inner automorphism of G by “ g ”. gxg 1 is known as the conjugate of x . x gxg 1 conjugation Defn. (Conjugate subgroup) A subgroup K of G is a conjugate subgroup of H if K i g H for some g G , i.e. K gHg 1 . Defn. (Commutator subgroup) Let G be a group. An element of G that can be expressed in the form aba 1b 1 for some a, b G is a commutator of G . The commutator subgroup C of G is the smallest normal subgroup which contains all commutators of G . Ex. G C is abelian. Theorem Proof. C is normal in G . It suffices to show that for every g , a , b in G , g aba 1b 1 g 1 is also in 1 1 1 1 1 C .(?) By g aba b ga bga gb g = ga bga b 1bgb 1 g 1 1 = ga bga 1 b 1 bgb 1 g 1 C , we conclude the proof. ■ Ex. Solution aba 1b 1C C aC bC a 1C b 1C C aC bC bC aC . Remark Defn. A subgroup H of G is said to be a characteristic subgroup of G if for all automorphisms of G , H H . ■ Theorem Proof. If H is a characteristic subgroup of G , then H G . Since the mapping defined by i g x gxg 1 is an automorphiam and i g x gxg 1 H for each x H . Hence gHg 1 H for each g G . This implies ■ that H G . Theorem of G . Proof. The commutator subgroup C of G is also a characteristic subgroup This is a direct conclusion of aba 1b 1 a b a 1 b 1 a b a 1 b 1 . ( C C .) ■