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Math Help - Proving an abelian group properties

  1. #1
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    Proving an abelian group properties

    Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

    just a bit stuck on this question, i know lagrange's theorem is involved and have applied it but im just not sure how to finish off the proof, can anyone show me?

    thx
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TrueTears View Post
    Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

    just a bit stuck on this question, i know lagrange's theorem is involved and have applied it but im just not sure how to finish off the proof, can anyone show me?

    thx
    This only holds if n is coprime to 2.

    Personally, I would note that there is an element of order 2 in there (why?) and one can quotient out the subgroup generated by it (why?).

    An element has order 2 is and only if it is in the kernel of this homomorphism and non-trivial (why?). This proves the result.
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  3. #3
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    a somewhat elementary way, is to observe that we have 2n-1 non-identity elements. by pairing each such element with its inverse,

    we wind up with an odd number of elements such that x^-1 = x. so we must have at least one element of order 2 (because 0 is not an odd number).

    can we have two? well if we did, say, x and y, since G is abelian {e,x,y,xy} is a subgroup of order 4. but 4 does not divide 2n.
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