1. SECTION 3 ISOMORPHIC BINARY STRUCTURES Definition Let  S,  and  S’,  ’  be binary algebraic structures. An isomorphism of S with S’ is a one-to-one function  mapping S onto S’ such that  (x  y)=  (x)  ’  (y) for all x, y  S. homomorphism property If such a map  exists, than S and S’ are isomorphic binary structures, which we denote by, omitting the  and  ’ from the notation.

2. How to show that binary structures are isomorphic To show that two binary structures  S,  and  S’,  ’  are isomorphic: Step 1 Define the function  that gives the isomorphism of S with S’. That is, we have to describe what  (s) is to be for every s  S. Step 2 Show that  is a one-to-one function. That is, suppose that  (x) =  (y) in S’ and deduce from this that x=y in S. Step 3 Show that  is onto S’. That is, suppose that s’  S’ is given and show that there does exist s  S such that  (s)=s’. Step 4 Show that  (x  y)=  (x)  ’  (y) for all x, y  S. This is just a question of computation. Compute both sides of the equation and see whether they are the same.

3. Example Let 2Z = { 2n|n  Z }, so that 2Z ia the set of all even integers, positive, negative, and zero. We claim that  Z,  +  is isomorphic to  2Z,  +  Step 1 Define  : Z  2Z by  (n)=2n for n  Z. Step 2 If  (m)=  (n), then 2m=2n so m=n. Thus  is one to one. Step 3 If n  2Z, then n is even so n=2m for m=n/2  Z. Hence  (m)= 2(n\2)=n so  is onto 2Z. Step 4 Let m, n  Z. The equation  (m+ n)= 2(m+n)=2m+2n=  (m)+  (n) then shows that  is an isomorphism.

4. Example Show that the binary structure  R,  +  with operation the usual addition is isomorphic to the structure  R +,   where  is the usual multiplication. Step 1 Define  : R  R + by  (x)=e x for x  R. Step 2 If  (x)=  (y), then e x = e y Taking the natural logarithm, x=y. So  is one to one. Step 3 If r  R +, then ln (r)  R and  (ln r)=e ln(r) =r. Thus  is onto R +. Step 4 For x, y  R. We have  (x+y)= e x+y = e x  e y =  (x)   (y). Thus we see that  is an isomorphism.

5. How to show that binary structures are not isomorphic To show that two binary structures  S,  and  S’,  ’  are not isomorphic, we need to show there is no one-to-one function  from S onto S’ with the property  (x  y)=  (x)  ’  (y) for all x, y  S. If there is no one-to-one function  from S onto S’, then two are not isomorphic. This is the case precisely when S and S’ do not have the same cardinality. Recall: |Z|=|Z + |=|Q|=|Q + |=  0 but |R|=|R + |>  0. Example: The binary structure  Q,  +  and  R,  +  are not isomorphic because |Q|=  0, but |R|>  0.

6. A structure property A structure property of a binary structure is one that must be shared by any isomorphic structure. It is not concerned with names or some other nonstructural characteristics of the elements. In the event that there are one-to-one mappings of S onto S’, we usually show that  S,  is not isomorphic to  S’,  ’  (if this is the case) by showing that one has some structural property that the other does not possess.

7. Examples The sets Z and Z + both have cardinality  0, and there are lots of one- to-one functions mapping Z onto Z +. However, the binary structure  Z,   and  Z +,  ,where  is the usual multiplication, are not isomorphic. In  Z,  , there are two elements x such that x  x=x, namely, 0 and 1. However, in  Z +,  , there is only the single element 1 such that x  x=x.

8. Example Show that the binary structures  Q,  +  and  Z,  +  under the usual addition are not isomorphic. (|Q|=|Z|=  0, so there are lots of one-to- one functions mapping Q onto Z.) The equation x + x=c has a solution x for all c  Q, but this is not the case in Z. For example, the equation x + x = 3 has no solution in Z. Here we have exhibited a structural property that distinguishes these two structures.

9. Example The binary structures  C,   and  R,   under the usual multiplication are not isomorphic. (It can be shown that C and R have the same cardinality.) The equation x  x=c has a solution x for all c  C, but x  x= -1 has no solution in R.

10. Structural and nonstructural properties We list a few examples of possible structural properties and nonstructural properties of a binary structure  S,  : Possible Structural Properties Possible Nonstructural Properties 1.The set has 4 elements a. The number 4 is an element 2.The operation is commutative b. The operation is called “addition” 3.x  x = x for all x  S. c. The elements of S are matrices 4.The equation a  x = b has a d. S is a subset of C. solution x in S for all a, b  S.

11. Identity Element 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. Theorem (Uniqueness of Identity Element) A binary structure  S,  has at most one identity element. That is, if there is an identity element, it is unique. Proof: Suppose that both e and e’ are elements of S serving as identity elements. Regarding e as an identity element, we must have e  e’=e’. However, regarding e’ as an identity element, we must have e  e’=e. We thus obtain e=e’, showing that an identity element must be unique.

12. Theorem The following theorem shows that having an identity element for  is indeed a structural property of a structural  S, . Theorem Suppose  S,  has an identity element e for . If  : S  S’ is an isomorphism of  S, , then  (e) is an identity element for the binary operation  ’ on S’.

13. Proof Let s’  S. We must show that  (e)  ’ s’=s’  ’  (e) =s’. Because  is an isomorphism, it is a one-to-one map of S onto S’. In particular, there exists s  S such that  (s) =s’. Now e is an identity element for  so that we know that e  s=s  e=s. Because  is a function, we then obtain  (e  s)=  (s  e)=  (s). Using Definition 3.7 of an isomorphism, we can rewrite this as  (e)  ’  (s)=  (s)  ’  ( e)=  (s). Remembering that we chose s  S such that  (s)= s’, we obtain  (e)  ’ s’=s’  ’  (e) =s’. This completes the proof of Theorem.