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Math Help - Abelian Subgroup

  1. #1
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    Abelian Subgroup

    I have the following question which i do not know how to continue.

    Question: Let H be a subgroup of G. G and H has order 2n and n respectively. Suppose exactly half of the elements in G has order 2 and there is no elements of order 2 in H, prove that H is abelian.

    Attempts: I started by considering the homomorphism f:H \rightarrow G defined by f(h) = gh{g^{-1}}. I know that since f is an homomorphism, order of h = order of gh{g^{-1}}. Also, since no elements in H has order 2, their inverse are not themselves. I attempt to prove by contradicting by assuming that H is not abelian. However, i cant proceed on after that. Is the approach correct or should i consider something else? Thank You!
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  2. #2
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    Re: Abelian Subgroup

    suppose g is an element of G not in H (any such g is of order 2). since gH ≠ H; for any h in H, gh is not in H, hence has order 2. thus:

    ghgh = e, whence
    ghg = h^-1, so that
    ghg^-1 = ghg = h^-1, for all h in H.

    now suppose h,k are any two elements of H.

    hk = (gh^-1g)(gk^-1g) = g(h^-1)(gg)(k^-1)g = g(h^-1k^-1)g = g(kh)^-1g = kh,

    so H is abelian.
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