301 Abstract Algebra
2:00 - 2:50 MWF
Hentzel 432 Carver
hentzel@iastate.edu
Web Page
http://www.math.iastate.edu/hentzel/class.301

Text: A First Course in Abstract Algebra, sixth Edition 

by John B. Fraleigh

Course: Chapter 1,2,3,4,5 

The material we are covering today is on Page  52-61

Section 1.3 beginning with "Definitions and Examples."   


Main Idea: There is nothing that cannot be undone. 

Key Words: Group, Associative, Identity, Inverse, Modulo,

Goal: Introduce the defining properties of groups.


Definition:  A group is a set G together with a binary operation *

satisfying:

1.   Associative:           (a*b)*c = a*(b*c)   for all a,b,c in G.

2.   Identity element:    e*a = a*e = a  for all a in G.

3.   Inverse:   Given a in G, there exists an element a' in G such

                that a*a' = a'*a = e.

If in addition the product is commutative, that is,

a*b=b*a for all a,b in G, we call G an Abelian Group. 


Lemma:  (a)  The identity is unique.

        (b)  The inverse in unique.

        (c)  (x')' = x.  The inverse of the inverse of x is x. 


Proof:   If e and f are identities, then  look at e*f.

Since e is an identity it must be f.  Since f is an identity

it must be e.  Therefore   e = e*f = f.

        Suppose that x' and x" were both inverses for x.


Then          x'(x x")  = x' e = x'
               ||
              (x'x)x"   = e x" = x".

So x' = x".

       If x' is the inverse of x, then x'x = x x' = e.

       But then by the definition of inverse, x is also

the inverse of x'.     
 
Lemma:  The value of the product is independent of the association.

Example:   Show ((ab)c)d = a(b(cd)) directly.

((ab)c)d = (ab)(cd) = a(b(cd)) using the associative law twice.

Proof that any the value of the product is independent of the

association.  We use induction.  We know that for the case 3, 

the association does not matter since we explicitly know that

(ab)c = a(bc) and these are the only ways to associate a product

of degree three.

Now assume that the product is independent of association for

all degrees less than n.  We shall show that the product

must be independent of association for degree n as well.  This

will prove the result for all degrees. 

              ================
Suppose that  x1 x2 x3 ... xn   is a product associated in 

               -------------- 
some way, and  x1 x2 x3 ... xn is another association. 

     There has to be an "outer" product.  I mean by this the 

last product computed when the expression is evaluated.  

Indicate this by "outer" product in each product by parentheses. 

=============  ===========           ------------  -----------
(x1 x2 ... xi)(xi+1 ... xn)         (x1 x2 ... xj)(xj+1 ... xn)

With out loss of generality, i <= j.   By induction we can

reassociate the second term on the left so that its "outer product"

occurs at j.   _                     _
              |                       |
============= | =========  ===========|         ------------  -----------
(x1 x2 ... xi)|(xi+1.. xj)(xj+1... xn)|    =   (x1 x2 ... xj)(xj+1 ... xn)
              |_                     _|

Now we can reassociate the above three terms by the ordinary associative

law.
 _                       _
|                         |            
| ============  ========= | ==========     ------------  -----------
|(x1 x2 ... xi)(xi+1.. xj)|(xj+1... xn) = (x1 x2 ... xj)(xj+1 ... xn)
|_                       _|            

Now we can reassociate the first factor on the left to

agree with the first factor on the right by induction.  

Wc can also reassociate the second factor on the left to

agree with the second factor on the right by induction.

we have proved that when the product of degree n is evaluated, it

is independent of association.


Examples of Groups:

Z+     Integers under addition.

Q+     Rationals under addition.

C+     Complex numbers under addition.

Q# *   Non zero rationals under multiplication

C# *   Non zero complexes under multiplication

S      The symmetric group on n elements.
 n

Symmetries of a Triangle, Square, pentagon, hexagon, etc.

Zn +   The integers modulo n under addition.

 
Zp *   Non zero elements of Zp when p is a prime under

       multiplication.
  

Assignment:

1.  Reassociate  (((ab)c)d)e to  a(b(c(de))) using only 

the associative law.

            2
2.  Solve  x  + x  + 1 = 0   in Z    and Z     and Z
                                 7        11        19


3.  Define a product on the Rationals by

aob = a + b + ab.

(a)  Show the product is commutative.

(b)  Show the product has an identity.  (what is the identity

element).

(c)  Show that the product is associative. 

(d)  Show that every element except one of them has an inverse.  

Which element does not have the inverse?