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#1
Dec1809, 08:29 AM

P: 108

1. The problem statement, all variables and given/known data
Let A,B be normal subgroups of a group G. G=AB. Prove that: G/AnB is isomorphic to G/A*G/B Have no idea how to start...Maybe the second isom. theorem can help us... TNX! 2. Relevant equations 3. The attempt at a solution 


#2
Dec1809, 12:08 PM

P: 352

Use the internal characterization of direct products of groups: if [tex]G[/tex] has two normal subgroups [tex]H, K[/tex] such that [tex]HK = G[/tex] and [tex]H \cap K = 1[/tex], then [tex]G \cong H \times K[/tex].
Also, the third isomorphism theorem may help you (if [tex]K \subset H[/tex] are both normal subgroups of [tex]G[/tex], then [tex]G/H \cong (G/K)/(H/K)[/tex]). 


#3
Dec1809, 12:38 PM

P: 108

Sry but I rly can't figure out the Latex text (I see it in black, and it's realy not clear)...
If I understand what you're saying, then we don't have the right conditions to use "internal characterization of direct products of groups"... A,B are normal subgroups of G and AB=G but who said AnB={1}? The isomorphism you've put afterwards is relevant only when G=A*B and it isn't the case/// Am I wrong? TNx 


#4
Dec1809, 03:09 PM

P: 108

Abstract AlgebraIsomorphism
I've managed to prove it...TNX a lot anyway...



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