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Math Help - fun with "necessarily" abelian

  1. #1
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    fun with "necessarily" abelian

    I have this problem:

    Suppose that X is a group with x^2=1 for all x \in X. Show that X is necessarily abelian.

    Prove that if X is finite, then |X|=2^n for some n \geq 0 and X needs at least n generators.

    --------------

    For the first part I did:

    \forall x \in X , x(1)=x=(1)x, so: e_X=1

    and: x^2=e_X \, \forall x \in X \Rightarrow x = x^{-1} \, \forall x \in X

    \forall x,y \in X:

    xy=(xy)^{-1}=y^{-1}x^{-1}=yx

    So X is abelian. Is that the same as necessarily abelian?


    -------

    Not sure how to go about the second part...
    Last edited by MichaelMath; October 25th 2010 at 03:12 PM.
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  2. #2
    Senior Member Dinkydoe's Avatar
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    What you proved is that a group X with the given property must be Abelian. This means, there can not exist a group with the same property
    that is not Abelian. (that's what they mean with necessarily Abelian)

    suppose there's a prime p\neq 2 such that p divides |X| then there's an element of order p in X....so?

    If it X had less then n generators....then |X|=...?
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  3. #3
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    For the first part, could I just have said :

    \forall x,y \in X

    x^2=1 and y^2=1 and (xy)^2=1

    so: (xy)^2=x^2y^2

    and: xy=(xy)^{-1}=y^{-1}x^{-1}=yx
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  4. #4
    Senior Member Dinkydoe's Avatar
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    You could have, but the only important part is:

    (\forall x)(x=x^{-1})

    This observation is made by multiplying x^2=1 with x^{-1} on both sides...

    Hence (\forall x)(\forall y)(xy = (xy)^{-1}= y^{-1}x^{-1}=yx)
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  5. #5
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    Quote Originally Posted by Dinkydoe View Post
    suppose there's a prime p\neq 2 such that p divides |X| then there's an element of order p in X....so?

    If it X had less then n generators....then |X|=...?
    arghhh... I know it's easy but I'm still not getting this stuff.
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  6. #6
    Senior Member Dinkydoe's Avatar
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    If p\neq 2 divides |X| then there's an element of order p in X....wich is a contradiction (why?)

    So p can not divide |X| and consequently, only 2 divides |X| so |X|=2^n

    If X has n-k generators then |X|=2^{n-k}... (the generators must also be of order 2...)
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Dinkydoe View Post
    If p\neq 2 divides |X| then there's an element of order p in X....wich is a contradiction (why?)

    So p can not divide |X| and consequently, only 2 divides |X| so |X|=2^n

    If X has n-k generators then |X|=2^{n-k}... (the generators must also be of order 2...)
    This is perfect, but you're using a fairly advanced theorem (that there exists an element of order p for a prime p dividing |G|).

    Another way to see it is that G can be considered as a vector space over the field with two elements. Switching to additive notation, the above identities read as 2(x+y)=2x+2y... since every vector space has a basis, we have an isomorphism G \cong (Z_2)^n for some n, hence |G|=2^n.
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