Let A,B be normal sub-groups 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!

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- December 18th 2009, 12:12 PMWannaBeAbstract Algebra-Isomorphism
Let A,B be normal sub-groups 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! - December 18th 2009, 12:16 PMtonio
- December 18th 2009, 01:06 PMWannaBe
Well... So after we define this map and show that G is homomorphic to G/AXG/B , by the first iso. theorem, we know that G/AXG/B is isomorphic to G/kerPhi

The only thing that is missing is to show toat kerPhi=AnB...

We can show that this way: (I'll use the symbol p for phi)

If k is in kerp then p(k)=(A,B), which means that: (kA, kB) = (A,B) hence:

kA=A , kB=B which means k is in A and k is in B which means k is in AnB and we're done...

Am I right?

TNX a lot for the answer and for your verification! - December 18th 2009, 03:46 PMtonio
- December 19th 2009, 01:12 AMWannaBe
Ofcourse it's not trivial...But according to the hint you gave me:

Each g in G can be written as ab when a is in A and b is in B...

So we can write: p(g)=p(a)*p(b)=(A,aB)*(bA,B)...So if we have (kA,kB) in G/AXG/B, we only need to prove that (kA,kB)=(A,aB)*(bA,B) i.e (kA,kB)=(A*bA, aB*B)=(abA, abB)...

But it's pretty much obvious, because if kA is in G/A then k is in G hence k=ab for some a in A and b in B... Am I right?

TNX again - December 19th 2009, 03:55 AMtonio
- December 19th 2009, 05:33 AMWannaBe
Tnx a lot man! I realy need your help in a homework assigment I need to send tommorow...I'll be delighted if you'll be able to look at my post in Differential Equations forum...I'm not sure what I need to do...

TNX a lot!