Page 92 - DMTH403_ABSTRACT_ALGEBRA

Unit 7: Homomorphism Theorem

Proof: We must first verify that the quotient groups H/(H K) and (HK)/K are well defined. Notes
You know that H K  H. You know that HK  G. Again, you know that K  HK. Thus, the
given quotient groups are meaningful.

Now, what we want to do is to find an onto homomorphism f : H — (HK)/K with kernel H  K.
Then we can apply the Fundamental Theorem of Homomorphism and get the result. We define
f : H  (HK)/K : f(h) = hK.
Now, for x, y  H,
f(xy) = xyK = (xK) (yK) = f(x) f(y).
Therefore, f is a homomorphism.

We will show that Im f = (HK)/K. Now, take any element hK  Im f. Since h  H, h  HK
 hK  (HK)/K.  Im f  (HK)/K. On the other hand, any element of (HK)/K is
hkK = hK, since k  K.

 hkK  Im f. (HK)/K  Im f.
 Im f = (HK)/K.
Finally, Ker f = { h  H | f(h) = K } = { h  H hK = K }
= { h  H | h  K }
= H  K .

Thus, on applying the Fundamental Theorem, we get H / (H  K)  (HK) / K
We would like to make a remark here.
Remark: If H and K are subgroups of (G.+ ), then Theorem 3 says that

(H + K) / K  H/H  K.
Theorem 4: Let H and K be normal subgroups of a group G such that K  H. Then (G/K)/(H/K)
 G/H.

Proof: We will define a homomorphism from G/K onto G/H, whose kernel will turn out to be
H/K.
Consider f : G/K  G/H : f(Kx) = Hx. f is well-defined because Kx = Ky tor x, y  G

 xy  K  H  xy  H  Hx = Hy (Kx) = f(Ky)
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7.2 Automorphisms

Let us start discussing the concept of automorphism
Let G be a group. Consider
Aut G = { f : G  G | f is an isomorphism }.
You have already seen that the identity map I  Aut G. You know that Aut G is closed under the
G
binary operation of composition. Iff E Aut G, then f  Aut G. We summarise this discussion in
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the following theorem.
An isomorphism from a group (G,*) to itself is called an Automorphisms of this group. Thus it
is a bijection f : G  G such that

f(u) * f(v) = f(u * v).

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