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isomorphism 2nd theorem


 

7 years ago

Answers : (1)

Aman Bansal
592 Points
							

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(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 .

 

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Thanks

Aman Bansal

Askiitian Expert

7 years ago
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