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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 S S ' , omitting the and ’ from the notation.
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.
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.
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)=ex for x R .
Step 2 If (x)= (y), then ex = ey Taking the natural logarithm, x=y. So 
is one to one.
Step 3 If r  R+, then ln (r)  R and  (ln r)=eln(r)=r. Thus  is onto R+.
Step 4 For x, y R. We have (x+y)= ex+y= ex  ey = (x)  (y). Thus we
see that  is an isomorphism.
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.
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.
Examples
The sets Z and Z+ both have cardinality 0, and there are lots of oneto-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.
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-toone 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.
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.
Structural and nonstructural properties
We list a few examples of possible structural properties and
nonstructural properties of a binary structure S, :
Possible Structural Properties
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 x in S for all a, b S.
Possible Nonstructural Properties
a. The number 4 is an element
b. The operation is called “addition”
c. The elements of S are matrices
d. S is a subset of C.
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.
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’.
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.