# isomorphism 2nd theorem

Aman Bansal
592 Points
10 years ago

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

h1k1h2k2=h1(h2h21)k1h2k2=h1h2(h21k1h2)k2

Since K is a normal subgroup of G and h2G , then h21k1h2K . Therefore h1h2(h21k1h2)k2HK , so HK is closed under multiplication.

Also, (hk)1HK for hH , kK , since

(hk)1=k1h1=h1hk1h1

and hk1h1K 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()=hH(h)=K=hHhK=K

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

ker()=hHhK=HK

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 .

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal