Page 79 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes In Examples 1 and 2 we observed that the homomorphisms carried the identity to the identity
and the inverse to the inverse. In fact, these observations can be proved for any group
homomorphism.
Theorem 1: Let f ; (G , * )  (G , * ) be a group homomorphism.
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1
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Then
(a) f(e ) = e , where e is the identity of G and e is the identity of G .
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l
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(b) f(x ) = [f(x)] for all x in G .
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Proof: (a) Let x  G . Then we have e * 1x = x. Hence,
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f(x) = f(e * x) = f(e ) * f(x), since f is a homomorphism. But
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f(x) = e * f(x) in G .
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Thus, f(e ) * f(x)= e * f(x).
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So, by the right cancellation law in G , f(e ) =e .
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(b) Now, for any x  G , f(x) * f(x ) = f(x * x ) = f(e ) = e .
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Similarly, f(x ) * f(x) = e .
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Hence, f(x ) = [f(x)]  x E G .
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Note that the converse of Theorem 1 is false. That is, if f : G  G is a function such that f(e ) = e 2
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and [f(x)] = f(x ) = f(x )  x  G , then f need not be a homomorphism. For example, consider
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f : Z  Z : f(0) = 0 and ,

 n 1  n  0
f(n)   0

 n 1  n 
Since f(l + 1)  f(1) + f(l), f is not a homomorphism. But f(e ) = e and f(n) = - f(- n)  n  Z.
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Let us look at a few more examples of homomorphisms now. We can get one important class of
homomorphisms from quotient groups.
Example: Let H  G. Consider the map p : G  G/H : p(x) = Hx. Show that p is a
homomorphism. Also show that p is onto. What is Ker p?
Solution: For x, y  G, p(xy) = Hxy = Hx Hy = p(x) p(y). Therefore, p is a homomorphism.
Now, Im p = { p(x) ( x  G} = ( Hx | x  G } = G/H. Therefore, p is onto.
Ker p = { x  G | p(x) = H ). (Remember, H is the identity of G/H.)
= ( x  G | H x = H }
= { x  G | x H ), by Theorem 1 of Unit 4.
= H.
In this example you can see that Ker p  G. You can also check that Theorem 1 is true here.
Example: Let H be a subgroup of a group G. Show that the map i : H  G, i(h) = h is a
homomorphism. This function is called the inclusion map.
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