Page 81 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes (b) Im f  , since f(e ) E Im f.
l
Now, let x , y  Im f. Then x , y  G such that f(x ) = x and f(y ) = y .
2
1
1
2
1
2
2
1
1
-1
-1
-1
 x y = f(x ) f(y ) = f(x y )  Imf.
1
1
2 2
1 1
 Im f  G .
2
Using this result, we can immediately see that the set of purely imaginary numbers is a normal
subgroup of C.
Consider  : (R, +)  (C*,.) (x) = cos x + i sin x. We have seen that(x + y) = (x)(y), that is,  is a
group homomorphism. Now (x) = 1 iff x = 2n for some n  Z . Thus, by Theorem 3, Ker =
(2 n | n  Z) is a normal subgroup of (R. +). Note that this is cyclic, and 2n is a generator.
Similarly, Im  is a subgroup of C*. This consists of all the complex numbers with absolute value
1, i.e., the complex numbers on the circle with radius 1 unit and centre (0, 0).
You may have noticed that sometimes the kernel of a homomorphism is {e} and sometimes it is
a large subgroup. Does the size of the kernel indicate anything? We will prove that a
homomorphism is 1 – 1 iff its kernel is {e}.
Theorem 4: Let f : G  G be a group homomorphism. Then f is injective iff Ker f = {e }, where
1
1
2
e is the identity element of the group G .
1
1
Proof: Firstly, assume that f is injective. Let x  Ker f. Then f(x) = e , i.e., f(x) = f(e ). But f is 1 – 1.
2
1
 x = e .
1
Thus, Kerf = {e }.
1
Conversely, suppose Ker f = {e ]. Let x, y  G such that
1
1
f(x) = f(y). Then f(xy ) = f(x) f(y )
-1
-1
= f(x) [f(y)] = e .
-1
2
-1
-1
 xy  Ker f = {e1}.  xy = e and x = y.
1
This shows that f is injective.
So, by using Theorem 4, we can immediately say that any inclusion i : B  G is 1-1, since
Ker i = {e}.
Let us consider another example.
Example: Consider the group T of translations of R . We define a map  : (R + )  (T, o) by
2
2
4 (a, b) = f . Show that  is an onto homomorphism, which is also 1-1.
a, b
Solution: For (a, b), (c, d) in R*, we have seen that
f a+c,h+d = f o f c.d
a,b
 ((a, b) + (c, d)) = (a, b) o (c, d).
Thus, , is a homomorphism of groups.
Now, any element of T is -f(a, b). Therefore,  is surjective. We now show that  is also injective.
Let (R, b)  Ker . Then $(a, b) = f
0, 0
i.e., f = f 0,0
a, b
 f (0, 0) = f (0, 0),
0, 0
a,b
i.e., (a, b) = (0, 0)
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