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# Fund Concpts Of Math MATH 300

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This 5 page Class Notes was uploaded by Reuben Hudson DDS on Friday October 30, 2015. The Class Notes belongs to MATH 300 at University of Massachusetts taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/232224/math-300-university-of-massachusetts in Mathematics (M) at University of Massachusetts.

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Date Created: 10/30/15

Math 300 Fall 2008 Note 7 Zhigang Han Umass at Amherst 1 Group Def 1 A group G 9 is a set G with a binary operation satisfying i Associativity Va bc E G a b gtk c a b 0 ii Identity element 3 an element 5 E G such that Va 6 Ge a a e a e is called an identity eleinent iii Inverse eleinent For each a E G 3 an element 1 E G such that a b b a e where e is an identity element We say I is an inverse of a Rink Because of the associativity it makes sense to write a b c Rink We often omit in a b to simply write ab Rink Note that the commutativity ab ba does not necessarily hold When it does the group G is called a commutative group or an Abelian group Rink Due to the uniqueness of identity and inverse below we will call 5 the identity element of the group G and call I the inverse of a and write I f1 Thus aa l a la 5 Prop 1 If G is a group then the following hold i Cancelation Laws For all a bc E G ab ac implies b c and ba ca implies b 0 ii Uniqueness of Identity G has a unique identity element iii Uniqueness of Inverse Every element in G has a unique inverse iv For all a b E G ab 1 19 Proof Let 1 be an inverse of a If ab ac then dab dac Since da e we get 51 50 Thus I 0 Similarly we can show if ba ca then b 6 ii Suppose 51 and 52 are both identity elements in G 51 is an identity implies 5152 52 52 is an identity implies 5152 51 These two equations imply that 51 52 so G has a unique identity element which is the element 5 given in the de nition of a group iii Suppose b1 and 2 are both inverses of a Then bla bga 5 By i we have 1 2 This proves the uniqueness of the inverse iv we need to show abb 1a 1 b la 1ab 5 This is true since abb 1a 1 ozbb lof1 aea l aa l e and b la 1ab bilailab bileb bill e Eg 1 Let G e be a set with only one element The binary operation is that e e 5 Then G 9 is a group 5 is the identity element and the inverse of e is 5 Eg 2 Let G be the set 1 71 with being the usual multiplication Then G 9 is a group 1 is the identity element The inverse of 1 is 1 and the inverse of 71 is 71 Eg 3 Z is a group 0 is the identity element and the inverse of n E Z is in Eg 4 Q is a group 0 is the identity element and the inverse of q E Q Eg 5 R is a group 0 is the identity element and the inverse of r E R Eg 6 Q 0 is a group 1 is the identity element and the inverse of q 0 om Eg 7 R 0 is a group 1 is the identity element and the inverse of r 0 Rm1 Eg 8 Zm is a group 0 is the identity element and the inverse of n E Zm is in m 7 Eg 9 lfp is prime then Zp0 is a group 1 is the identity element and using Fermat s little Theorem we know the inverse of a 7 0 E 217 is lap zl NonExample 10 Z 0 is not a group For instance 2 does not have an inverse NonExample 11 If m is composite then Zm 0 is not a group Some elements in Zm do not have an inverse Which ones Eg 12 Let X be a nonempty set Let G be the set of bijections from X to X and the binary operation 0 on G is the composite of functions Then G o is a group 1X is the identity element of G The inverse of f X a X is the inverse function f 1 X a X Rmk When X in the above example is a nite set say X 1 2 n a bijection from X to itself is called a permutation of X Then the set Sn of all permutations of X with the composite of permutations is a group called the symmetric group of X 2 Subgroup Def 2 A subset H C G is called a subgroup if H itself is a group with the same binary operation on G Prop 2 A subset H C G is a subgroup if and only if the following hold i closure lf 011 6 H then ab 6 H ii identity 5 E H where e is the identity element of G iii inverse If h E H then h 1 E H where h 1 is the inverse of h in G Proof This follows directly from the de nition How Eg 13 In Example 3 4 and 5 Q is a subgroup of R Z is a subgroup of R Z is a subgroup of Q ln Example 6 and 7 Q 0 is a subgroup of R 0 One can easily check this by Proposition 2 Eg 14 Denote by 22 the set of even numbers Show that 22 is a subgroup of Z Theorem If H and K are two subgroups of a group G then H O K is a subgroup of G Proof 1 lfab E H K then 011 6 H and ab E K Since H and K are subgroups ab 6 H and ab 6 K Hence ab 6 H O K 2 Since H and K are subgroups e is in both H and K Hence 5 is in H O K 3 If a 6 HO K then a E H and a E K Since H and K are subgroups7 f1 E H and f1 E K Hence7 f1 E H K By Proposition 27 H O K is a subgroup of G Eg 15 If G is a group7 de ne the center of G to be CG a E Gle 6 G7 am ma Show that CG is a subgroup of G Eg 16 Suppose H is a subgroup of the group G and a E G De ne aHa 1aha1lh e H Show that otHof1 is a also subgroup of G Eg 17 Suppose H is a subgroup of the group G De ne a relation on G such that aRb if and only of ab 1 6 H Show that R is an equivalence relation 3 Group homomorphism Def 3 Let G1 and G2 be two groups A function f G1 a G2 from G1 to G2 is called a group homomorphism7 if ab fafb for all 011 6 G1 More precisely7 if is the operation on G17 and o is the operation on G27 then f is a homomorphism if fa b fa o fb for all 071 6 G1 Eg 18 Let G1 and G2 be two groups De ne f G1 a G2 by fa 52 for all a 6 G1 Show that f is a group homomorphism Eg 19 Let H be a subgroup ofa group G De ne f H a G by fh h for all h E H Show that f is a group homomorphism Theorem If f G1 a G2 and g G2 a G3 are both group homomor phisms7 then 9 o f G1 a G3 is a group homomorphism Proof For all 071 6 G17 using the fact that f and g are group homomor phism we have 9 o ab 9fab 9fafb 9fagfb This proves g o f is a group homomorphism

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