Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Foundations of Pure Mathematics: Lecture 28 Dr. Charles Eaton [email protected] December 6th, 2016 Dr. Charles Eaton Lecture 28 December 6th, 2016 1/6 Identity elements Let ∗ be a binary operation on a set S. We say e ∈ S is an identity element for S (with respect to ∗) if ∀ a ∈ S, e ∗ a = a ∗ e = a. If there is an identity element, then it’s unique: Proposition Let ∗ be a binary operation on a set S. Let e, f ∈ S be identity elements for S with respect to ∗. Then e = f . Proof. Since e is an identity element and f ∈ S, we have e ∗ f = f . Since f is an identity element and e ∈ S, we have e ∗ f = e. Hence e = e ∗ f = f . Dr. Charles Eaton Lecture 28 December 6th, 2016 2/6 Examples (i) 0 is the identity element for S = R when ∗ = +. (ii) 1 is the identity element for S = R when ∗ = ×. (iii) Let S = Q \ {0} and a ∗ b = ba . Then there is no identity element. Suppose that e ∈ Q \ {0} is an identity element. Then e ∗ 1 = 1 ∗ e = 1. Now e ∗ 1 = e1 = e, so e = 1. But 1 ∗ 2 = 12 6= 2, so 1 cannot be an identity element. (iv) S = {a, b, c, d}, and define a binary operation ∗ on S by the multiplication table a b c d a c a a d b d c b a c a b c d d d b d a In this case c is an (the) identity element. Dr. Charles Eaton Lecture 28 December 6th, 2016 3/6 Groups Let G be a non-empty set and ∗ a binary operation on G . We call (G , ∗) a group if the following hold: (G1) ∗ is associative (G2) G has an identity element e with respect to ∗ (G3) ∀ g ∈ G , ∃ h ∈ G such that g ∗ h = h ∗ g = e. [We sometimes say G forms a group under ∗.] Remark: In (G3), the element h is called an inverse to g . It is unique, since suppose h1 , h2 ∈ G with g ∗ h1 = h1 ∗ g = e, g ∗ h2 = h2 ∗ g = e. Then h1 ∗ (g ∗ h2 ) = h1 ∗ e = h1 and h1 ∗ (g ∗ h2 ) = (h1 ∗ g ) ∗ h2 = e ∗ h2 = h2 , since ∗ is associative. Hence h1 = h2 . Hence we may refer to the (unique) inverse to g , and write g −1 . Dr. Charles Eaton Lecture 28 December 6th, 2016 4/6 Examples (i) G = Z, ∗ = +. The identity element is 0, and ∗ is associative. Let a ∈ Z. Then (−a) + a = a + (−a) = 0, so −a is the inverse of a (i.e., a−1 = −a). Hence (Z, +) forms a group. (ii) Let G = R \ {0}, with ∗ = ×. 1 is the identity element, since ∀ a ∈ R, a × 1 = 1 × a = a. Multiplication is associative. Let a ∈ R \ {0}. Then ( 1a )a = a( 1a ) = 1. So a−1 = 1a . Hence (R \ {0}, ×) forms a group. (iii) Let G = {−1, 1, −i, i} ⊆ C, with ∗ multiplication of complex numbers. On board Dr. Charles Eaton Lecture 28 December 6th, 2016 5/6 Cancellation in groups Let (G , ∗) be a group. Let g , h1 , h2 ∈ G . Suppose that g ∗ h1 = g ∗ h2 . Then g −1 ∗ (g ∗ h1 ) = g −1 ∗ (g ∗ h2 ). Since ∗ is associative, this means (g −1 ∗ g ) ∗ h1 = (g −1 ∗ g ) ∗ h2 . g −1 ∗ g = e, so h1 = e ∗ h1 = e ∗ h2 = h2 . We have shown that g ∗ h1 = g ∗ h2 ⇒ h1 = h2 . Hence in any given row of the multiplication table for (G , ∗) we cannot have any repetitions (i.e., each row must contain every element of G , in some order). Similarly h1 ∗ g = h2 ∗ g ⇒ h1 = h2 , so each column of the multiplication table must contain every element of G , with no repetition. Dr. Charles Eaton Lecture 28 December 6th, 2016 6/6