Page 90 - DMTH403_ABSTRACT_ALGEBRA

Unit 7: Homomorphism Theorem

Therefore, y is a well-defined function, Notes
Now, let us check that  is a homomorphism. For Hx, Hy  G /H,
1
(Hx)(Hy)) = (Hxy)
= f(xy)
= f(x) f(y), since f is a homomorphism.
= (H) (HY)

Therefore,  is a group homomorphism.
Next, let us see whether  is bijective or not.
Now, (Hx) = (Hy) for Hx, HY in G /H
1
 f(x) = f(y)

 f(x) [f(y)l = e 2
-1
 f(xy ) = e 2
1
 xy  Ker f = H.
-1
 Hx = Hy

Thus, , is 1-1.
Also, any element of Im f is f(x) = (Hx), where x  G .
1
 Im  = Im f.
So, we have proved that  is bijective, and hence, an isomorphism. Thus, G1/Ker f = Im f.

Now, if f is surjective, Im f = G . Thus, in this case G /Ker f  G .
2
1
2
The situation in Theorem 1 can be shown in the following diagram.

Here, p is the natural homomorphism.
The diagram says that if you first apply p, and then , to the elements of G , it is the same as
1
applying f to them. That is,
 p = f.


Also, note that Theorem 1 says that two elements of G have the same image under f iff they
1
belong to the same coset of Ker f.
Let us look at a few examples.
One of the simplest situations we can consider is I : G  G. On applying Theorem 1 here, we see
G
that G/{e}  G. We will be using this identification of G/{e) and G quite often.

Example: Prove that C/R  R.

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