Introduction to Group Theory

Transcription

1 Introduction to Group Theory Let G be n rbitrry set of elements, typiclly denoted s, b, c,, tht is, let G = {, b, c, }. A binry opertion in G is rule tht ssocites with ech ordered pir (,b) of elements of G unique element c of the sme set G. A binry opertion is therefore function ψ : G G G;(, b) ψ (, b) In group theory, it is customry to only consider one binry opertion t time nd to write ψ(,b) = b nd cll b the product of nd b. The binry opertion is then clled multipliction. Behind this mthemticl nottion re some hidden requirements. First, for every pir of elements, b G there must exists product b. Second, this product is uniquely identified. However, the definition does not sy tht every element of G hs to be product of two elements of G. There is no requirement tht the product of nd b is the sme s the product of b nd, however, for some importnt binry opertions, this is lwys the cse. Generl binry opertions re very generl structures. Usully, we investigte multiplictions tht hve specil properties. One of the most importnt property is ssocitivity: Definition 1: (Associtivity) Let G be set nd let ψ : G G G;(, b) ψ (, b) = b be multipliction defined between elements of G. Then ψ is clled ssocitive if for ll, b, c G the following property holds: ( b c) = ( b) c

2 In the definition, we used prentheses to explin the order in which we multiply. On the left side, we multiply with the product of b nd c, wheres on the right side, we multiply the product of nd b with c. We sy tht two elements nd b of G commute (with respect to given binry opertion), if b = b. Definition 2: (Commuttivity) Let G be set nd let ψ : G G G;(, b) ψ (, b) = b be multipliction defined between elements of G. Then ψ is clled commuttive if for ll, b G the following property holds: b = b. Thus, product is commuttive if ll elements in G commute with ech other. We now ssume tht we hve set G with n ssocite product. We cn then define the notion of powers recursively. As usul, we cll the set of nturl numbers N = {1, 2, 3, 4, }. Sometimes we do wnt to include 0 s nturl number, then we write N 0 = {0, 1, 2, 3, 4, }. Definition 3: (Powers) Let G be set nd let ψ : G G G;(, b) ψ (, b) = b be multipliction defined between elements of G. We define the powers of n element G recursively by 1 = n+ 1 n for n = N

3 According to this definition 1 =, 2 =, 3 = ( ), etc. According to the ssocite lw, 3 = ( ) = = ( ) = 2. We cn generlize this observtion: Proposition 1: Let G be set nd let ψ : G G G;(, b) ψ (, b) = b be multipliction defined between elements of G. Then the following rules re vlid for ll elements G nd ll nturl numbers m, n N: () m n m = + n (b) ( ) = m n mn If, b G re two rbitrry elements tht commute, then in ddition (c) ( b ) n = n b n For your first homework, you hve to prove these three properties using induction. A proof by induction is typiclly proof of the vlidity of some sttement S(n) tht contins nturl number n N s prmeter. The proof proceeds in two steps, the induction bse nd the induction step. The first step, the induction bse, sks you to prove directly tht S(1) is true. For the second step, you prove tht if S(n) is true (the induction ssumption), then lso S(n+1) is true. The second step is typiclly more difficult. As n exmple, we prove Proposition 1, (). Proof (Proposition 1.): We proceed by induction on n. Number m stys fixed. Induction Bse: We hve to show tht + m m 1 =. However, this follows directly from the definition of powers. Induction Step: We ssume tht for given m

4 = m n m+ n We then need to show tht = m n+ 1 m+ n+ 1 Proof of the Induction Step: m n+ 1 m n = ( ) [Definition of powers] m n = ( ) [Associtivity of product] m+ n = = ( m+ n) + 1 [Induction Assumption] [Definition of power] m+ n+ 1 = [Addition of numericl numbers is ssocitive] qed. Associtivity is nice feture, but there is not whole lot tht we cn sy bout ssocitive products without further properties. Definition 4: (Identity Element) Let G be set nd let ψ : G G G;(, b) ψ (, b) = b be multipliction defined between elements of G. An element e G is clled n identity element if for ll elements G the following identity holds: e = e =. Proposition 2: If G is set with multipliction nd if there exists n identity element, then this identity element is uniquely determined. Proof: Assume tht there re two different identity elements, e, f G. Then e f = e becuse f is n identity element nd e f = f becuse e is n identity element. Tking both equtions together we hve e = e f = f. This proves uniqueness. Definition 5: (Inverse Element) Let G be set with multipliction (written s usul). Assume tht e G is n identity element. Then n inverse element of n element G is n element such tht = = e.

5 Proposition 3: Let G be set with multipliction (written s usul) nd with n identity element e G. Assume tht G hs n inverse element. Then this inverse element is uniquely determined. Becuse of Proposition 3, we denote the inverse of G with -1. This nottion would be troublesome if we were to look t two different products defined over G but such is rrely the cse. We cn extend the definition of powers of to negtive powers, i.e. extend the definition to the set Z of ll entire numbers { 2, 1, 0, 1, 2, } if the inverse of exists: Definition 6 (Powers): Let G be set with multipliction (written s usul) nd with n identity element e G. Assume tht G hs n inverse element. We then define recursively 0 1 = e = ' = ' ( n+ 1) n Proposition 4: The identities of Proposition 2 lso pply to negtive powers. To be more precise, ssume tht G be set with multipliction (written s usul) nd with n identity element e G nd ssume tht G hs n inverse element -1. Then for ll m, n Z the following identities re vlid: () m n m n = + (b) ( ) = m n mn If, b G re two rbitrry elements tht commute, then in ddition (c) ( b) n = n b n We lso hve the following reltionship between powers nd tking the inverse. If hs n inverse, then so does n nd the inverse is (d) n 1 n ( ) =.

6 We leve the proofs to the reder s n exercise.