Section 3 Isomorphic Binary Structures Instructor: Yifan Yang Fall 2006
Outline Isomorphic binary structure An illustrative example Definition Examples Structural properties Definition and examples Identity element
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 + 3 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1 + 3 1 1 2 1 1 1 2 1 1 2 1 2 2 1 1 + 3 1 ζ 2 1 1 ζ 2 ζ ζ 2 1 1 ζ ζ 2 1 1 ζ ζ 2 ζ ζ ζ 2 1 ζ 2 ζ 2 1 ζ 1 ζ ζ 2 1 1 ζ ζ 2 ζ ζ ζ 2 1 ζ 2 ζ 2 1 ζ 1 ζ ζ 2 1 1 ζ ζ 2 ζ ζ ζ 2 1
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.
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.
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.
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.
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.
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.)
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.
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.
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.
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.
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.
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.
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)
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.
Homework Do Problems 4, 6, 8, 16, 18, 26, 28, 30, 33 of Section 3.