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Math Help - Complement of a subgroup

  1. #1
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    Complement of a subgroup

    Let G be a finite group. Suppose that every element of order 2 of G has a complement in G, then G has no element of order 4.

    Proof. Let x be an element of G of order 4. By hypothesis, G=\langle x^{2} \rangle K and \langle x^{2} \rangle \cap K=1 for some subgroup K of G. Since [G:K]=2, then K is normal in G. Clearly, G=\langle x \rangle K and \langle x\rangle \cap K=1, but |G|=|\langle x^{2} \rangle||K|<|\langle x \rangle ||K|=|G|, a contradiction. Therefore G has no element of order 4.

    Is above true? Thanks in advance.
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  2. #2
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    Re: Complement of a subgroup

    Quote Originally Posted by moont14263 View Post
    Let G be a finite group. Suppose that every element of order 2 of G has a complement in G, then G has no element of order 4.

    Proof. Let x be an element of G of order 4. By hypothesis, G=\langle x^{2} \rangle K and \langle x^{2} \rangle \cap K=1 for some subgroup K of G. Since [G:K]=2, then K is normal in G. Clearly, G=\langle x \rangle K and \langle x\rangle \cap K=1
    to me, reading this, this is NOT so clear. how do we KNOW x is not in K?

    of course, if it is, then x2 is in K as well (by closure, since x*x must lie in K), contradicting <x2>∩K = {1}.

    , but |G|=|\langle x^{2} \rangle||K|<|\langle x \rangle ||K|=|G|, a contradiction. Therefore G has no element of order 4.

    Is above true? Thanks in advance.
    rather than argue that G = <x>K (which still requires further proof), i would use the fact that x is in G, and G = <x^2>K, to assert that x = x2k, for some k in K (we ruled out x = k above).

    thus x-1 = x3 = (x2)x = (x2)(x2k) = k, so x = k-1 which is in K, contradiction.
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  3. #3
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    Re: Complement of a subgroup

    Your proof is clearer, thank you very much.
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