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Math Help - Tough Problem from Rotman's Group Theory, 3rd Ed

  1. #1
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    Tough Problem from Rotman's Group Theory, 3rd Ed

    Given a group G of order (2^m)k, k odd. If G contains an element of order 2^m, then (show that) the set of elements of G having odd order is a (normal) subgroup of G.

    This is from Rotman, 3rd ed, p.46. He gives the following HINT: Consider G as permutations via Cayley's Th. and show that it contains an odd permutation.

    I have gotten to the following facts:
    The postulated element of order 2^m is an odd permutation, since it is regular and each cycle has order 2^m (so each cycle is odd) and no letter is fixed (so there are k cycles, k odd). Thus the group contains both odd permutions and even ones. Since these are exchanged by applying the odd permutation, the even and odd are equinumerous. The odd permuations are of even order, so the odd ordered elements, if any, are a proper subset. This subset is invariant, even characteristic, but it is not clear that it is closed under group operation (ie, a subgroup) since the group may not be abelian...

    Thanks,
    Gary

    PS: This comes early in the book, before Sylow and other powerful results. Therefore, I assume there is an "elementary" argument....
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  2. #2
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    Quote Originally Posted by signaldoc View Post
    Given a group G of order (2^m)k, k odd. If G contains an element of order 2^m, then (show that) the set of elements of G having odd order is a (normal) subgroup of G.

    This is from Rotman, 3rd ed, p.46. He gives the following HINT: Consider G as permutations via Cayley's Th. and show that it contains an odd permutation.

    I have gotten to the following facts:
    The postulated element of order 2^m is an odd permutation, since it is regular and each cycle has order 2^m (so each cycle is odd) and no letter is fixed (so there are k cycles, k odd). Thus the group contains both odd permutions and even ones. Since these are exchanged by applying the odd permutation, the even and odd are equinumerous. The odd permuations are of even order, so the odd ordered elements, if any, are a proper subset. This subset is invariant, even characteristic, but it is not clear that it is closed under group operation (ie, a subgroup) since the group may not be abelian...

    Thanks,
    Gary

    PS: This comes early in the book, before Sylow and other powerful results. Therefore, I assume there is an "elementary" argument....
    prove it by induction over m: if m = 0, there's nothing to prove. now let G be a group of order n=2^mk, with k odd and m \geq 1. let g \in G be such that o(g)=2^m and H be the set of elements

    of odd order in G. let G_1=G \cap A_n and g_1=g^2. we have o(g_1)=2^{m-1}. looking at G as a subgroup of S_n (Cayley's theorem), we'll see that the length of each cycle in the decomposition of an

    element x \in G is equal to o(x). so H \subseteq G_1 and g_1 \in G_1. also [G:G_1]=2 because \forall x,y \in G-A_n: \ xy^{-1} \in G_1. hence |G_1|=2^{m-1}k and the proof is complete by induction.
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  3. #3
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    Many thanks to NonCommAlg

    Nicely done... My problem may have been in (obstinately) not believing the result in the first place.
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