Multiplicative Group of Z\nZ • Let n∈Z( 0) fixed, Proposition 9 implies that there is a well-defined function ○ Z\nZ×Z\nZ→Z\nZ ○ (a ̅,b ̅ )→(ab) ̅ • Check group property ○ Identity: 1 ̅⋅a ̅=(1⋅a) ̅=1 ̅ ○ This operation is associative ○ And 1 ̅ is a reasonable candidate for an identity, but there is no inverse ○ Example: Z\4Z § 2 ̅⋅0 ̅=0 ̅ § 2 ̅⋅1 ̅=2 ̅ § 2 ̅⋅2 ̅=0 ̅ § 2 ̅⋅3 ̅=2 ̅ • Definition ○ Define (ZnZ^×≔{a ̅∈ZnZ(a,n)=1} ○ By HW 2 #2, a ̅∈(ZnZ^× iff ○ ∃c ̅∈Z\nZ s.t. (ac) ̅=1 ̅ Proposition 11: (ZnZ^× • Statement ○ (ZnZ^× is a group with opeation given by multiplication • Proof ○ Closure: If a ̅,b ̅∈(ZnZ^×, then (ab) ̅∈(ZnZ^× as well ○ Associativity: Clear, from associativity of multiplication of integers ○ Identity: 1 ̅ ○ Inverses: Built in HW 2 #2 List of Groups Set Operation Z,Q,R,ℂ + Q∗,R∗,ℂ^∗ ⋅ GL_n (R, n 0 Matrix multiplication Z\nZ, n 0 + 〖Z/nZ^∗, n 0 ⋅ Proposition 12: Properties of Group • Let G be a group, then G has the following properties • The identity of G is unique ○ i.e. If ∃〖1,1〗^′∈G s.t. ∀g∈G,1g=g1=g and 1^′ g=g1^′=g, then 1 =1^′ ○ Proof: 1=1⋅1′=1′∎ • Each g∈G has a unique inverse ○ i.e. If g∈G and ∃h,h′∈G s.t. hg=gh=1 and h′ g=gh′=1 ○ Let g∈G, and suppose h,h′∈G are inverses of g ○ h=h⋅1=h(gh′ )=(h�) h′=1⋅h′=h′ • (g^(−1) )^(−1)=g, ∀g∈G ○ Let g∈G, then gg(−1)=1=g(−1) g ○ Since the inverse is unique, g=(g^(−1) )^(−1) • The Generalized Associative Law ○ i.e. If g_1,…,g_n∈G, then g_1…g_n is independent of how it is bracketed ○ First show the result is true for n=1,2,3 ○ Assume for any k n any bracketing of a product of k elements ○ b_1 b_2⋯b_k can be reduced to an expression of the form b_1 (b_2 (b_3⋯b_k )) ○ Then any bracketing of the product a_1 a_2⋯a_n must break into ○ 2 sub-products, say (a_1 a_2⋯a_k )(a_(k+1) a_(k+2)⋯a_n ) ○ where each sub-product is bracketed in some fashion ○ Apply the induction assumption to each of these two sub-products ○ Reduce the result to the form a_1 (a_2 (a_3…a_n )) to complete the induction. • (gh^(−1)=h(−1) g^(−1), ∀g,h∈G ○ By the generalized associative law ○ (gh(h(−1) g^(−1) )=g(h^(−1) ) g(−1)=gg(−1)=1 ○ (h(−1) g^(−1) )(gh=h(gg^(−1) ) h(−1)=hh(−1)=1 • Notation ○ We will apply the Generalized Associative Law without mentioning it ○ In particular, if G is a group and n∈Z( 0), we will write § g^n=⏟(g…g)┬(n copies) § g(−n)=⏟(g(−1)…g^(−1) )┬(n copies) § g^0=1 Proposition 13: Cancellation Law • Statement ○ Let G be a group, and let a,b,u,v∈G ○ If au=av, then u=v ○ If ua=va, then u=v • Proof ○ au=av⇒a^(−1) au=a^(−1) av⇒u=v ○ ua=va⇒uaa(−1)=vaa(−1)⇒u=v • Warning ○ ua=av⇏u=v ○ This holds in abelian groups, but not in general Corollary 14 • Let G be a group, and let g,h∈G • If gh=g,then h=1 ○ gh=g ○ ⇒gh=g1 ○ ⇒h=1 • If gh=1,then h=g^(−1) ○ gh=1 ○ ⇒gh=gg^(−1) ○ ⇒h=g^(−1)
