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    How to show that any finite group of even order contains an element "a" not equal to 1 such that a*a = 1
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    Quote Originally Posted by thomas_donald View Post
    How to show that any finite group of even order contains an element "a" not equal to 1 such that a*a = 1
    The simplest approach is to use Cauchy's theorem on finite groups: http://en.wikipedia.org/wiki/Cauchy'..._(group_theory)
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    I am new to Abstract Algebra. Could you give me some more hints.
    Thanks a lot for the help.
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    Quote Originally Posted by thomas_donald View Post
    I am new to Abstract Algebra. Could you give me some more hints.
    Thanks a lot for the help.
    Cauchy theorem is the most directly way to explain it. But there is still another way..
    a^2=e\Leftrightarrow a=a^{-1}
    suppose no element fit this except e.
    you may consider all the pairs (a,a^{-1})and "join" them by a line.
    Then for all a\in G, there is exactly one line connected to it.
    if a=a^{-1}is only possible for e,then all the other lines will occupy two elements, and the line (e,e)occupys one element. Thus |G|must be an odd number!!!
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  5. #5
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    Thanks,
    Great!
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  6. #6
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    Quote Originally Posted by thomas_donald View Post
    How to show that any finite group of even order contains an element "a" not equal to 1 such that a*a = 1
    I don't think you need Cauchy's theorem.

    Suppose this is not true. Since the inverse of 1 is itself, that leaves an odd number of elements such that each is NOT its own inverse. Pair each element with its inverse. What happens? (Of course, you need the fact that inverses are unique- if a*b= 1 and a*c= 1, then b= c.)
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    Thanks a lot for your help.
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