ab'= a'b,cd' = c'd那么ab'dd' = a'bddcd'bb' = c'dbb'(ad +bc) b'd'= (a'd'+b'c') bdad +bcTa'd' +b'c'b'd'bd另一方面,ab'cd'=a'bc'd(ac)(b'd') =(ac)(bd)["] [岁]
那么 另一方面, ab a b cd c d = = , ab dd a bdd = cd bb c dbb = (ad bc b d a d b c bd + = + ) ( ) ad bc a d b c bd b d + + = ab cd a bc d = (ac b d ac bd )( ) = ( )( ) ac a c bd b d =
两类加法相乘的结果与类的代表无关9.对于加法来说作成一个加群:[][a][a][](1)[([[](2)adf+bcf+bdebdf(:1-[D-[,]-[]-[;]adf +bcf +bdebdf
两类加法相乘的结果与类的代表无关。 Q0 对于加法来说作成一个加群: a c c a b d d b + = + a c e a cf de b d f b df + + + = + adf bcf bde bdf + + = (1) (2) a c e ad bc e b d f bd f adf bcf bde bdf + + + = + + + =