《Discrete Mathematics II教学-华南理工》Lecture 2 Groups I.pptVIP

Theorem Let (A, *) and (B, *) be semigroups, monoids or groups. Then the inverse f-1:B?A of any isomorphism f:A ? B is itself an isomorphism. Proof： The inverse f-1:B?A of an isomorphism f:A?B is itself a bijective function whose inverse is the function f:A ? B. It remains to show that f-1:B?A is a homomorphism. Let u and v be elements of B, and let x = f-1(u) and y = f-1(v). Then u = f(x) and v = f(y), and therefore f(x* y) = f(x)* f(y) = u *v and therefore f-1(u*v) = x *y = f-1(u) *f-1(v); showing that the function f-1:B ?A is a homomorphism from (B; *) to (A;* ), as required. Isomorphism Definition: Let (A,* ) and (B, *) be semigroups, monoids or groups. If there exists an isomorphism from (A,* ) to (B, *) then (A,* ) and (B,* ) are said to be isomorphic. Example The integers are isomorphic to the subgroup of Q* consisting of elements of the form 2n. Define a map f : Z ? Q* by f (n) = 2n. Then Q(m + n) = 2m+n = 2m2n = Q(m)Q(n) By definition the map f is onto the subset {2n : n ? Z} of Q*. To show that the map is injective, assume that m ? n. Suppose that m n and assume that f(m) = f(n). Then 2m = 2n or 2m-n = 1, which is impossible since m - n 0. Groups(I) South China University of Technology Groups A group (G,·) is a nonempty set G together with a binary operation · on G such that the following conditions hold: (i) Closure: For all a,b in G, the element a · b is a uniquely defined element of G. (ii) Associativity: For all a,b,c in G, we have a · (b · c) = (a · b) · c. (iii) Identity: There exists an identity element e in G such that e · a = a ? ? and ? ? a · e = a for all a in G. (iv) Inverses: For each a in G there exists an inverse element a-1 in G such that a · a-1 = e ? ? and ? ? a-1 · a = e. ?We will usually write ab for the product a · b. Example The set Q× of nonzero rational numbers, the set R× of nonzero real numbers, and the set C× of nonzero complex numbers form groups under ordinary multiplication. The integers Z form a g