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Math Help - Group Theory and Order of Elements

  1. #1
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    Group Theory and Order of Elements

    Hello. I have a question about the order of elements in a group:

    let G be a group of order 2n. Show that the number of elements in G with order 2 is odd.

    Here is what I have done so far:

    Let H be the subset of G consisting of the identity and all the elements of G with order 2. Then H is a subgroup of G and so, by Lagrange's Theorem, the order of H divides the order of G.

    My strategy was to use this to arrive at the order of H is even, completing the proof. But I am stuck at this point.
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  2. #2
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    Re: Group Theory and Order of Elements

    Hint: Prove that |\{ a \in G | a \neq a^{-1} \}| is even. Check the parity of what remains, and notice that  e \notin \{ a \in G | a \neq a^{-1} \}
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  3. #3
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    Re: Group Theory and Order of Elements

    Oh my goodness. This makes it very clear. Thank you for the hint Isomorphism.
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    Re: Group Theory and Order of Elements

    Hey, I'm trying to work this exact problem right now. I do not understand the conclusion of post #2. Why is the parity of the set of elements x such that x^2 is not e, necessarily even? This is where I'm stuck.
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  5. #5
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    Re: Group Theory and Order of Elements

    Don't inverse-pairs occur in pairs if an element is NOT its own inverse?
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    Re: Group Theory and Order of Elements

    Quote Originally Posted by Deveno View Post
    Don't inverse-pairs occur in pairs if an element is NOT its own inverse?
    Huh?
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  7. #7
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    Re: Group Theory and Order of Elements

    Ok. Imagine this: you start with a finite group, G. First you take out all the elements that equal their own inverses (this will include the identity element).

    Now, for any element g that is left, there is some OTHER element g^{-1} (which is, by assumption, NOT g) also in the leftovers. So take them out as a pair. Keep doing this.

    One of two things HAS to happen:

    1) We wind up with a single element left over when we're done.

    2) Everything comes out in pairs, so we wind up with NO elements left over.

    Convince yourself that if #1 happens, that remaining element must be its own inverse, a contradiction, so #2 must occur, the number of elements who are NOT their own inverses must be EVEN.
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