Section 3 Isomorphic Binary Structures

Transcription

1 Section 3 Isomorphic Binary Structures Instructor: Yifan Yang Fall 2006

2 Outline Isomorphic binary structure An illustrative example Definition Examples Structural properties Definition and examples Identity element

3 An illustrative example Let ζ = e 2πi/3 be a 3rd root of unity. Consider the following two binary structures Z 3, + 3 and U 3,. The tables of Z 3, + 3 and U 3, are given by ζ ζ 2 ζ ζ ζ ζ ζ ζ 2 ζ ζ ζ 2 1 ζ 2 ζ 2 1 ζ 1 ζ ζ ζ ζ 2 ζ ζ ζ 2 1 ζ 2 ζ 2 1 ζ 1 ζ ζ ζ ζ 2 ζ ζ ζ 2 1

4 An example This shows that Z 3, + 3 and U 3, have the same algebraic structure. (For example, we have (first element) (second element) = (second element) (third element) (second element) = (first element) and so on.) In mathematics, we say these two binary structures are isomorphic.

5 Isomorphic binary structures Definition Let S, and S, be two binary algebraic structures. An isomorphism of S with S is a one-to-one function mapping S onto S such that φ(x y) = φ(x) φ(y) for all x, y S (homomorphism property). If such a map φ exists, then S, and S, are isomorphic binary structures, which we denote by S S, deleting and from the notation.

6 Examples of isomorphic binary structures Example The binary structure R, + is isomorphic to R +,. Proof. We claim that φ : R R + defined by φ(x) = e x satisfies 1. one-to-one, 2. onto, 3. φ(x + y) = φ(x) φ(y) for all x, y R. Proof of Claim 1. If φ(x) = φ(y), then e x = e y and x = y. Proof of Claim 2. For r R +, let x = ln r. Then φ(x) = e ln r = r. Proof of Claim 3. Easy.

7 Examples of isomorphic binary structures Example The binary structures Z, + and 2Z, + are isomorphic. Proof. We claim that φ : Z 2Z defined by φ(n) = 2n satisfies 1. one-to-one, 2. onto, 3. φ(m + n) = φ(m) + φ(n) for all m, n Z. Proof of Claim 1. If φ(m) = φ(n), then 2m = 2n and m = n. Proof of Claim 2. For n 2Z, we have n = 2m for some integer m. Then φ(m) = 2m = n. Proof of Claim 3. Easy.

8 Remarks 1. Given two isomorphic binary structures S, and S,, there may be more than one isomorphisms between them. For the first example, any positive real number a 1 will define an isomorphism φ a : x a x between R, + and R +,. For the second example, the function ψ(n) = 2n is also an isomorphism. 2. The second example also shows that an infinite set can be isomorphic to a proper subset with the induced binary operation.

9 Examples of non-isomorphic binary structures 1. The binary structures Q, + and R, + can not be isomorphic because Q and R have different cardinalities. 2. The binary structures Q, + and Z, + are not isomorphic even though they have the same cardinality. To see this, observe that the equation x + x = c has a solution for every c in Q, but this is not the case in Z. (Suppose that φ : Q Z is an isomorphism. Let r Q be the element such that φ(r) = 1. Then φ(r/2) + φ(r/2) = φ(r) = 1. Thus, φ(r/2) = 1/2, but it is not in Z.)

10 Examples of non-isomorphic binary structures 1. The binary structures Z, and 2Z, are not isomorphic, even though Z, + and 2Z, + are isomorphic. This is because in Z, there is an element e = 1 such that e n = n for all n Z, but there is no such element e in 2Z satisfying e n = n for all n in 2Z. Suppose that φ is an isomorphism between S, and S,. If e is an element in S such that e s = s for all s S, then for all s S we have s = φ(s) for some s S and then φ(e) s = φ(e) φ(s) = φ(e s) = φ(s) = s. That is, the element e = φ(e) satisfies e s = s for all s S. 2. The binary structures R, and C, are not isomorphic. This is because in C, the equation x x = c has solutions for all c C, but in R, the equation x x = c have no solutions in R when c < 0.

11 In-class exercises Determine whether the given map φ is an isomorphism of the first binary structure with the second. 1. Z, + with Z, +, where φ(n) = n for n Z. 2. Z, + with Z, +, where φ(n) = 2n for n Z. 3. Q, + with Q, +, where φ(r) = r/2 for r Q.

12 Structural properties Definition A structural property of a binary structure is one that must be shared by any isomorphic structure. Example The following properties are structural. 1. The set has 4 elements. 2. The operation is commutative. 3. x x = x for all x S. 4. The equation a x = b has a solution in S for all a, b S. 5. The equation x x = s has a solution in S for all s S. 6. There is an element e in S such that e s = s for all s S.

13 Examples of non-structural properties Example The following properties are non-structural. 1. The set S is a subset of C. 2. The number 4 is an element. 3. The operation is called addition. 4. The elements of S are matrices. In-class exercises 1. Give a few more structural properties. 2. Prove that they are indeed structural properties.

14 Identity element Definition (3.12) Let S, be a binary structure. An element e of S is an identity element for if e s = s e = s for all s S. Example 1. In Z, +, the element 0 is an identity element. 2. In Z,, the element 1 is an identity element. 3. In Z n,, the element 1 is an identity element. 4. In M 2 (R), + (the set of all 2 2 matrices with entries in R), the zero matrix is an identity element.

15 Identity element Theorem (3.13) A binary structure S, has at most one identity element. That is, if there is an identity element, then it is unique. Proof. Suppose that e and e are both identity elements. Consider e e. On the one hand, since e is an identity element, we have e e = e. On the other hand, because e is an identity element, we have We conclude that e = e. e e = e.

16 Identity element Theorem (3.14) Suppose that S, has an identity element e for. If φ : S S is an isomorphism of S, with S,, then φ(e) is an identity element for. proof We need to show that for all s S. φ(e) s = s φ(e)

17 Identity element Proof of Theorem 3.14 (continued) Since φ is an isomorphism, φ is onto. Thus, there exists s S such that φ(s) = s. Then φ(e) s = φ(e) φ(s). Now, by the assumption that φ is an isomorphism again, we have φ(e) φ(s) = φ(e s). It follows that φ(e) s = φ(e s) = φ(s) = s. By the same token, we can also show that s φ(e) = s. We conclude that φ(e) is an identity element.

18 Homework Do Problems 4, 6, 8, 16, 18, 26, 28, 30, 33 of Section 3.