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Math Help - Abstract Algebra-Isomorphism

  1. #1
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    Abstract 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!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    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!
    Define \phi:G\rightarrow G\slash A\times G\slash B\,\,\,by\,\,\,\phi(g):=(gA,gB) and apply the first isomorphism theorem.

    Hint to show the above map is onto: use that every g\in G can be written as g=ab\,,\,\,a\in A\,,\,\,b\in B.

    Tonio
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  3. #3
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    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!
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  4. #4
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    Quote Originally Posted by WannaBe View Post
    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!

    Yes, that's right....but please do note that showing \phi(G)=G\slash A\times G\slash B (i.e., that \phi is onto) is NOT trivial: you MUST prove this.

    Tonio
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  5. #5
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    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
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  6. #6
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    Quote Originally Posted by WannaBe View Post
    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

    Yeppers!

    Tonio
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  7. #7
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    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!
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