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Math Help - proof that subgroup has finite index

  1. #1
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    proof that subgroup has finite index

    Q) If H is of finite index in G prove that there exists a subgroup N of G contained in H, and of finite index in G such that  aNa^{-1} = N .


    In a previous problem, I proved that if H is a subgroup of G, then  N = \bigcap xHx^{-1} is a subgroup of G such that  aNa^{-1} = N . So I'm assuming that this is the same subgroup that is being talked about in the above question. How do I prove that it has finite index?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sashikanth View Post
    Q) If H is of finite index in G prove that there exists a subgroup N of G contained in H, and of finite index in G such that  aNa^{-1} = N .


    In a previous problem, I proved that if H is a subgroup of G, then  N = \bigcap xHx^{-1} is a subgroup of G such that  aNa^{-1} = N . So I'm assuming that this is the same subgroup that is being talked about in the above question. How do I prove that it has finite index?
    Because H has finite index we have that there are only finitely many of these xHx^{-1}, and each has finite index.

    It then suffices to prove that the intersection of finitely many cosets (or whatever these thing are called) of finite index is a coset of finite index. Use induction to prove this. (Although be warned, I wouldn't call this proof "easy"...)
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    Quote Originally Posted by Swlabr View Post
    Because H has finite index we have that there are only finitely many of these xHx^{-1}, and each has finite index.
    What is the proof that each of the xHx^{-1} has finite index? I'm sorry if its a trivial question, but I can't seem to hit the nail on the head!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sashikanth View Post
    What is the proof that each of the xHx^{-1} has finite index? I'm sorry if its a trivial question, but I can't seem to hit the nail on the head!
    Why isn't the mapping \lambda:H\mapsto xHx^{-1} a surjection?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Why isn't the mapping \lambda:H\mapsto xHx^{-1} a surjection?
    It is surjective. Let me try and answer my question now. The mapping \lambda:H\mapsto xHx^{-1} is onto, I think it is actually a bijection. So the number of elements in H equal the number of elements in xHx^{-1}. Since the size of the coset equals the order of the subgroup, the size of the cosets of xHx^{-1} equal the size of the cosets of H. Since cosets span the group, their index has to be equal.

    What if the order of the subgroup H is not finite?
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  6. #6
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    Quote Originally Posted by sashikanth View Post
    It is surjective. Let me try and answer my question now. The mapping \lambda:H\mapsto xHx^{-1} is onto, I think it is actually a bijection. So the number of elements in H equal the number of elements in xHx^{-1}. Since the size of the coset equals the order of the subgroup, the size of the cosets of xHx^{-1} equal the size of the cosets of H. Since cosets span the group, their index has to be equal.

    What if the order of the subgroup H is not finite?
    If N=\bigcap_{x \in G}xHx^{-1}, then it is the kernel of h:G \rightarrow A(M), where A(M) is the group of permutations of M and M is the set of all left cosets H in G. Here, h is the induced homomorphism of the group G acting on left cosets of H in G (see here).
    Since G/N is isomorphic to the subgroup of A(M), the index of N in G should be finite.
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  7. #7
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    Thank you for your reply. I have just started studying groups, I'm on section 2.5 of Topics in Algebra. I dont yet understand isomorphisms. I'll check this solution after I've covered the same.
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