Group Homework 2 Abstract Algebra Names: Score: Select any 10 of the following problems. Provide full justification for your answers. 1. For a set S, and X, Y ? S, we define X + Y = (X ? Y ) \ (X n Y...

1. For a set S, with , ,X Y S we define    \X Y X Y X Y    and .X Y X Y   We
wish to determine whether  P S is a ring or a field with these two operations.

In order for  P S to be a ring, it must be an abelian group with respect to addition, a
associative with respect to multiplication, and distributive with respect to addition and
multiplication.
It is useful to define X Y in terms of complementary sets. For a subset X of S, we let
\ .X S X Then we have \X Y X Y  for all subsets X and Y of S. Thus we have

   
   
   
\
.
X Y
X Y X Y
X Y X Y
X Y X Y

  
   
   
Now we check whether  P S satisfies the axioms of a ring. First we check whether it is
commutative with respect to addition. We have
,
X Y
X Y
Y X
Y X

 
 
 
whence  P S is commutative with respect to addition.
Next we check whether  P S is associative with respect to addition. We have
 
   
         
        
       
   
       
       
       .
X Y Z
X Y X Y Z
X Y X Y Z X Y X Y Z
X Y Z X Y Z X Y X Y Z
X Y Z X Y Z X Y X Y Z
X Y Z X Y Z
X X Z X Y Z X Y Z Y Y Z
X Y Z X Y Z Z X Y Z X Y Z Z
X Y Z X Y Z X Y Z X Y Z
 
      
               
          
            
     
           
             
           
We also have
 
 
       
       
       
  .
X Y Z
Y Z X
Y Z X Y Z X Y Z X Y Z X
X Y Z X Y Z X Y Z X Y Z
X Y Z X Y Z X Y Z X Y Z
X Y Z
 
  
           
           
           
  
Thus we see that  P S is associative with respect to addition.

Next we check whether  P S has an additive identity element, 0. We claim that 0.
To see this, note that
   
 
0
.
X
X X
X X S
X S
X

   
  
 

Thus we see that  P S has an additive identity element. Thus,  P S is an abelian group
with respect to addition.
Next we check whether  P S is associative with respect to multiplication. We have
 
 
 
 .
X Y Z
X Y Z
X Y Z
X Y Z
 
  
  
  
Thus we see that  P S is associative with respect to multiplication.

Finally we must check whether  P S satisfies the two distributive properties. We have
 
 
   
     
   
       
   
   .
X Y Z
X Y Z
X Y Z Y Z
X Y X Z Y Z
X Y Y Z X Z Y Z
X Y Y X Y Z X Y Z X Z Z
X Y Z X Y Z
X Y Z X Y Z
 
  
    
       
             
           
      
     
We also have
   
       
     
     
       
   
   
   
 .
X Y X Z
X Y X Z
X Y X Z X Y X Z
X Y X Z X Y X Z
X Y X Z X Y Z
X X Y X X Z X Y Y X Y Z
X Y Z X Y Z
X Y Z X Y Z
X Y Z X Y Z
X Y Z
  
   
            
         
        
           
     
      
     
  
Thus we see that  P S satisfies the first distributive property.

We also have
,
X Y
X Y
Y X
Y X

 
 
 
whence  P S is commutative with respect to multiplication.

Finally we have
 
 
,
X Y Z
Y Z X
Y X Z X
X Y X Z
 
  
   
   
whence  P S satisfies the second distributive property as well. Thus we see that  P S
is a ring.
Now in order for  P S to be a field, it must also have a multiplicative identity element,
1, and every nonempty subset X of S must have a multiplicative inverse. First we check
whether  P S has a multiplicative identity element.

We claim that 1.S  To see this, note that for every subset X of S, we have
1
.
X
S X
X

 

Next we must check whether every nonempty element of  P S has a multiplicative
inverse. Let X be a nontrivial nonempty subset of S (which only exists if # 2.S  ) We
look for a subset Y of S such that 1.X Y  But then we would have ,X Y S  which is
impossible since X is a nontrivial subset of S. Thus we see that  P S is not a field unless
S has cardinality 0 or 1.
2. We are given the sets

 
2
2 : , and : , .
a
R a b a b S a
a
  
      
  
We wish to show that R is a su
ing of R , S is a su
ing of  2 ,M and that R and S are
isomorphic.
To show that R is a su
ing of R , we must show that R is closed under addition and
multiplication (the other ring axioms follow from those of ). Let a, b, c, and d be
integers. Then we have
2 and 2 .a b R c d R      
We also have
   
   
2 2
2
,
a b c d
a c b d
R
     
   

whence R is closed under addition.
Finally we have
  
   
2 2
2 2
,
a b c d
ac bd ad bc
R
   
   

whence R is also closed under multiplication and hence is a su
ing of R .
To show that S is a su
ing of  2 ,M we must show that S is also closed under addition
and multiplication. Let a, b, c, and d be integers. Then we have

2 2
and .
a b c d
A S B S
a d c
   
      
   
Now we have

 
2 2
2
,
a b c d
A B
a d c
a c b d
d a c
S
   
     
   
   
  
  

whence S is closed under addition. We also have

 
2 2
2 2
2
,
a b c d
AB
a d c
ac bd ad bc
ad bc ac bd
S
   
    
   
   
  
  

whence S is also closed under multiplication and hence is a su
ing of  2 .M
Finally we must show that R and S are isomorphic. Consider the map : R S  defined
y
 
2
2 .
a
a
a

 
   
 
Now let 2 and 2.a b c d     Then we have
 
    
    
 
   
   
2 2
2
2
2 2
2 2
.
a b c d
a c b d
a c b d
d a c
a b c d
a d c
a b c d
  


 
   

   
   
   
  
  
   
    
   
   
 
We also have
 
    
    
 
   
   
2 2
2 2
2 2
2
2 2
2 2
.
a b c d
ac bd ad bc
ac bd ad bc
ad bc ac bd
a b c d
a d c
a b c d
 


 
   
  
   
   
  
  
   
    
   
  

Thus we see that  is a ring homomorphism.
To show that R and S are isomorphism, it suffices to show that  is one-to-one and onto
and hence is a ring isomorphism. Suppose that    2 2 .a b c d    Then we have


2 2
,
a b c d
a d c
 ...