Page 91 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Solution: Define f : C  R : f(a + ib) = b. Then f is a homomorphism, Ker f = R and Im f = R.
Therefore, on applying Theorem 1 we see that C/R  R.

 1, if n is even
Example: Consider f : Z  ({1, - 1),.) : f(n)  
  1, if n is odd.
At the beginning, you saw that f is a homomorphism. Obtain Ker f and Im f. What does Theorem
1 say in this case?
Solution: Let Z and Z denote the set of even and odd integers, respectively. Then
e o
Ker f = {n  Z | f(n) = 1 } = Z,
Im f = {f(n) | n  Z ) = { l , – 1}
Thus, by Theorem 1, Z/Z,  { 1, - 1 }.

This also tells us that o(Z/Z ) = 2. The two cosets of Z in Z are Z and *.
e
e
e
 { Z , Z }  { 1, -1 }.
o
e
Example: Show that GL (R)/SL (R)  R*, where SL (R) = {A  GL (R) | det (A) = 1 }.,
2
2
2
2
Solution: We know that the function
f : GL (R)  R* : f(A) = der(A) is a homomorphism. Now, Ker f = SL (R).
2
2
  1 0  
Also, Im f = R*, since any r  R* can be written as det   0 1 .  
   
Thus, using Theorem 1, GL (R)/SL (R)  R*.
2
2
Now-we will use the Fundamental Theorem of Homomorphism to prove a very important
result which classifies all cyclic groups.
Theorem 2: Any cyclic group is isomorphic to (Z, +) or (Z , +).
n
Proof: Let G = < x > be a cyclic group. Define
f : Z  G : f(n)=x .
n
f is a homomorphism because
f(n + m) = x n+m = x . x = f(n) f(m).
n
m
Also note that Im f = G.
Now, we have two possibilities for Ker I  Ker f = {0) or Ker f  {0}.
Case 1 (Ker f = {0)): In this case f is 1-1. Therefore, f is an isomorphism. Therefore, by Theorem 7
of unit 6, f-1 is an isomorphism. That is, G  (Z, +).
Case 2 (Ker f # {0)): Since Ker f  Z, we know that Ker f = nZ, for some n  N. Therefore, by the
Fundamental Theorem of Homomorphism, Z/nZ  G.
 G = Z/nZ = (Z , +).
n
Over here note that since < x > = Z , o(x) = n. So, a finite cyclic group is isomorphic to Z , where
n
n
n is the order of the group.
Theorem 3: If H and K are subgroups of a group G, with K normal in G, then H/(H K) 
(HK)/K.

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