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Math Help - Group theory

  1. #1
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    Group theory

    I am trying to do all the exercises in Rotman (3rd ed.). It's an integral part of his exposition to do many of them, but this one is optional. I don't get it... (problem 2.83 in the book).

    G=H x K, N is normal in G. Prove: Either N is abelian OR N intersects one of the factors non-trivially.

    Suppose N intersects H in the identity AND N intersects K in the identity (i.e., trivially in both cases). The proposition is then that N must be abelian.

    I think we have the following fact: Every element of N must be of the form n=hk with neither h,k the identity unless both are. From this, we can deduce that every first component, h, has the same order, m, as every second component, k.

    h^m=1
    k^m=1

    But does this make N abelian? No.

    Note: Presumably, Sylow theory is not used to solve this, since it precedes that theory.

    Any help will be greatly appreciated, and apologies in advance if I'm being dense.
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  2. #2
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    Suppose that N intersects each of H and K trivially. Let n = h_0k_0\in N, and let h_1\in H. Then n^{-1} = h_0^{-1}k_0^{-1}\in N. Also, h_1^{-1}h_0h_1k_0\in N (by normality). Therefore the product h_0^{-1}k_0^{-1}h_1^{-1}h_0h_1k_0 = h_0^{-1}h_1^{-1}h_0h_1 is in N. But N intersects H trivially, so h_0^{-1}h_1^{-1}h_0h_1 is the identity. Thus h_0 is in the centre of H. Similarly k_0 is in the centre of K. That's enough to show that N is abelian.
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  3. #3
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    Quote Originally Posted by signaldoc View Post
    I am trying to do all the exercises in Rotman (3rd ed.). It's an integral part of his exposition to do many of them, but this one is optional. I don't get it... (problem 2.83 in the book).

    G=H x K, N is normal in G. Prove: Either N is abelian OR N intersects one of the factors non-trivially.

    Suppose N intersects H in the identity AND N intersects K in the identity (i.e., trivially in both cases). The proposition is then that N must be abelian.

    I think we have the following fact: Every element of N must be of the form n=hk with neither h,k the identity unless both are. From this, we can deduce that every first component, h, has the same order, m, as every second component, k.

    h^m=1
    k^m=1

    But does this make N abelian? No.

    Note: Presumably, Sylow theory is not used to solve this, since it precedes that theory.

    Any help will be greatly appreciated, and apologies in advance if I'm being dense.

    Ok, what you did was fine: take n_1\,,\,n_2\in N\Longrightarrow\,n_1=h_1k_1\,,\,\,n_2=h_2k_2\,,\,  \,with\,\,\,h_i\in H\,\,and\,\,k_i\in K

    Now use that both H,K are normal subgroups to show that:

    [h,n]=[k,n]=1\,\,\forall h\in H ,\,\,n\in N

    Finally, use the basic relation between commutators [xy,z]=[x,z]^y[y,z] , so that finally: [n_1,n_2]=[h_1k_1,n_2]\Longleftrightarrow N is abelian.

    Tonio
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  4. #4
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    Very quick....you must do this in your sleep. I guess that's why I need to do the problems... Thanks!
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  5. #5
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    Thanks... you guys are good!

    Quote Originally Posted by tonio View Post
    Ok, what you did was fine: take n_1\,,\,n_2\in N\Longrightarrow\,n_1=h_1k_1\,,\,\,n_2=h_2k_2\,,\,  \,with\,\,\,h_i\in H\,\,and\,\,k_i\in K

    Now use that both H,K are normal subgroups to show that:

    [h,n]=[k,n]=1\,\,\forall h\in H ,\,\,n\in N

    Finally, use the basic relation between commutators [xy,z]=[x,z]^y[y,z] , so that finally: [n_1,n_2]=[h_1k_1,n_2]\Longleftrightarrow N is abelian.

    Tonio
    Much thanks!
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