Page 80 - DMTH403_ABSTRACT_ALGEBRA

Unit 6: Group Isomorphism

Solution: Since i(h h ) = h h = i(h )i(h )  h , h  H. i is a group homomorphism. Notes
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Let us briefly look at the inclusion map in the context of symmetric groups. Consider two
natural numbers m and n, where m I n.
Then, we can consider S  S , where any   Sm, written as
n
m
 1 2 .... m 
  , is considered to be the same as
  (1)  (2) ....  (m) 

 1 2 .... m m 1 .... n 
   S , i.e., (k) = k for m + 1  k  n.
n

  (1)  (2) ....  (m) m 1 .... n 
Then we can define an inclusion map i : S  S . r
m
 1 2 3 4 
For example, under i : S  S , (1 2) goes to   .
4
3
 2 1 4 4 
We will now prove some results about homomorphisms. Henceforth, for convenience, we shall
drop the notation for the binary operation, and write a * b as ab.
Now let us look at the composition of two homomorphisms. Is it a homomorphism? Let us see.’
Theorem 2: If f : G  G and g : G  G are two group homomorphisms, then the composite
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map g . f : G  G is also a group homomorphism.
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Proof: Let x. y  G . Then

g o f(xy) = g(f(x : y))
= g(f(x)f(y), since f is a homomorphism.
= g(f(x)) g(f(g)), since g is a homomorphism.
= g o f(x).g o f(y).
Thus g, f is a homomorphism.
Theorem 3: Let f : G  G be a group homomorphism. Then
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(a) Ker f is a normal subgroup of G .
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(b) Im f is a subgroup of G .
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Proof: (a) Since [(e ) = e , e  Ker f.  Ker f  .
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Now, if x, y  Ker f, then f(x) = e and f(y) = e .
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 f(xy ) = f(x) f(y ) = f(x) [f(y)] = e . 2
-1
-1
-1
xy  Ker f.
-1
Therefore, by Theorem 1 of Unit 3, Ker f  G . Now, for any y  G and x E Ker f,
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1
f(y xy) = f(y ) f(x)f(y)
-1
-1
= [f(y)] e f(y), since f(x) = e and by Theorem 1
-1
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= e .
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 Ker f  G .
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