isomorphism 2nd theorem

asad ullah, 11 years ago

Grade:

1 Answers

Aman Bansal

592 Points

11 years ago

Daer Asad,

First, we shall prove that HK is a subgroup of G : Since eH and eK , clearly e=e2HK . Take h1h2Hk1k2K . Clearlyh1k1h2k2HK . Further,

h1k1h2k2=h1(h2h2−1)k1h2k2=h1h2(h2−1k1h2)k2

Since K is a normal subgroup of G and h2G , then h2−1k1h2K . Therefore h1h2(h2−1k1h2)k2HK , so HK is closed under multiplication.

Also, (hk)−1HK for hH , kK , since

(hk)−1=k−1h−1=h−1hk−1h−1

and hk−1h−1K since K is a normal subgroup of G . So HK is closed under inverses, and is thus a subgroup of G .

Since HK is a subgroup of G , the normality of K in HK follows immediately from the normality of K in G .

Clearly HK is a subgroup of G , since it is the intersection of two subgroups of G .

Finally, define :HHKK by (h)=hK . We claim that is a surjective homomorphism from H to HKK . Let h0k0K be some element ofHKK ; since k0K , then h0k0K=h0K , and (h0)=h0K . Now

ker(

(h)=K

hK=K

and if hK=K , then we must have hK . So

ker(

Thus, since (H)=HKK and ker=HK , by the First Isomorphism Theorem we see that HK is normal in H and that there is a canonical isomorphism between H(HK) and HKK .

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