Page 83 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Solution: Let us first verify that f is a homomorphism. Now, for any two elements

 a b  c d
 b a  and  d c  in G,
     

  a b  c d    a c b d 



r   b a    d c    f  (b d) a c   (a c) i(b d)
           
= (a + ib) + (c + id)
  a b     c d  
= f      f    
   b a     d c 
Therefore, f is a homomorphism.


   a b      a b    0 0  


0  
Now, Ker f = f    b a  a ib    b a  a 0,b    
0  
              0 0  
Therefore, by Theorem 4, f is 1-1.
Finally, since Im f = C. f is surjective
Therefore, f is an isomorphism.
We would like to make an important remark now.
Remark: If G and G are isomorphic groups, they must have the same algebraic structure and
2
1
satisfy the same algebraic properties. For example, any group isomorphic to a finite group must
be finite and of the same order. Thus, two isomorphic groups are algebraically indistinguishable
systems.
The following result is one of the consequences of isomorphic groups being algebraically alike
Theorem 6: If f : G  H is a group isomorphism and Y  G, then < x >  < f (x)> ,
Therefore.

(i) if s is of finite order, then o(x) = o(f(s)).
(ii) if x is of infinite order, so is f(x).
Proof: If we restrict f to any subgroup K of G, we have the function f | K ; K  f(K), Since f is
bijective, sc is its restriction f | k ; k  f(K) for any subgroup K of G. In particular, for any x  G,
< x >  f(< x >) = < f(x) >,
Now if x has finite order, then o(x) = o(< x >) = o(< f(x) >) = o(f(x)), proving (i)

To prove (ii) assume hat x is of infinite order. Then < x > is an infinite group.
Therefore, < f(x) > is an infinite group, and hence, f(x) is of infinite order. So, we have proved (ii).

Example: Show that (R*,.) is not isomorphic to (C*,.).
Solution: Suppose they are isomorphic, and f : C* — R* is an isomorphism. Then
o(i) = o(f(i)), by Theorem 6, Now o(i) = 4.  o(f(i)) = 4.

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