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Math Help - group intersection question

  1. #1
    nhk
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    group intersection question

    Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. I have no idea where to go on this besides letting A= the intersection of all subgroups of order n.
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  2. #2
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    Quote Originally Posted by nhk View Post
    Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. I have no idea where to go on this besides letting A= the intersection of all subgroups of order n.
    This is the sketch of the proof.

    Let H be a subgroup of group G of order n and M be the intersection of all subgroups of G of order n. It follows immediately that M is a subgroup of G. We remain to show that M is normal in G. We see that all subgroups of order n in G are either conjugate subgroups of H or subgroups of order n in G that are not conjugate to H but are conjugate to other subgroups of order n in G.

    Let K be the intersection of all conjugates of H in G. Since the intersection of all conjugates of H is normal in G (link 1)(link 2) and M \subset K, we see that gMg^{-1}=M for every g \in G. Thus M is a normal subgroup of G.

    Edit : The last line holds because M is the intersection of all conjugate subgroups of subgroups of order n in G, where M = \bigcap_{i \in I}\bigcap_{g \in G}gH_ig^{-1} and each H_i for i \in I denotes a subgroup of a group G of order n.
    Last edited by aliceinwonderland; April 8th 2010 at 05:54 PM.
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