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Math Help - Proof:Finite group with even order has at least one element x with x = x^{-1}

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    Senior Member x3bnm's Avatar
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    Proof:Finite group with even order has at least one element x with x = x^{-1}

    Let G be a finite group, and let S be the set of all the elements of G

    The set S can be divided up into pairs so that element is paired off with its own inverse. Prove that:

    If the order of G is even, there is at least one element x in G such that x \neq e (neutral element) and x = x^{-1}(inverse element)


    Is it possible to kindly tell me how I can prove this?
    Last edited by x3bnm; November 8th 2012 at 05:35 PM.
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    Re: Proof:Finite group with even order has at least one element x with x = x^{-1}

    consider the set {g,g-1}. this set has two elements if g ≠ g-1, and one element if g = g-1.

    so the set of all elements of order > 2 (where g ≠ g-1) has an even number of elements, for any finite group G, call it 2k.

    the set of all elements of order 1 (which is one of the elements g for which g = g-1, since e = e-1) has just one element, e.

    therefore, the number of elements of order 2 (that is elements g with g ≠ e, and g = g-1) is:

    |G| - 2k - 1.

    if |G| is an even number, then |G| = 2m, and the number of elements of order 2 is:

    2m - 2k - 1 = 2(m - k) - 1, which is an odd number.

    in particular, since 0 is an even number, this number cannot be 0, so we have at least ONE element of order 2.
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    Senior Member x3bnm's Avatar
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    Re: Proof:Finite group with even order has at least one element x with x = x^{-1}

    Quote Originally Posted by Deveno View Post
    consider the set {g,g-1}. this set has two elements if g ≠ g-1, and one element if g = g-1.

    so the set of all elements of order > 2 (where g ≠ g-1) has an even number of elements, for any finite group G, call it 2k.

    the set of all elements of order 1 (which is one of the elements g for which g = g-1, since e = e-1) has just one element, e.

    therefore, the number of elements of order 2 (that is elements g with g ≠ e, and g = g-1) is:

    |G| - 2k - 1.

    if |G| is an even number, then |G| = 2m, and the number of elements of order 2 is:

    2m - 2k - 1 = 2(m - k) - 1, which is an odd number.

    in particular, since 0 is an even number, this number cannot be 0, so we have at least ONE element of order 2.
    Thank you Deveno. That's all I wanted to know.
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