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Math Help - Abelian groups

  1. #1
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    Abelian groups

    Let G be an abelian group and let g,h \in G.

    If |g| = m is finite then prove that, for n \in \mathbb {Z}, ng=0 iff m|n.

    Now the obvious cases are if n=0, ng=0 and m|n trivially. Same for n=m.

    The rest all seems rather straightforward, however I'm having trouble with finding a structure for my proof, especially since n is an integer, not just a natural number.

    For the case where 0<n<m I have if g+...+g (m times) =0 then [g+...+g (m times)] + g+ ... +g (ntimes)= g+...+g (n times) which, by definition of the order of an element cannot be equal to zero. I find it hard to prove this for the similar case of m<n<0.

    Am I making this too difficult or what is it that I'm missing?
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  2. #2
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    Hi

    Since g and -g have the same order m, proving that for any integer n,\ m|n\Rightarrow ng=0 is quite easy. (just need to write what m|n means and use the associativity of +)

    Now let n be an integer such that ng=0. We can assume n\geq 0 because ng=0 iff (-n)(-g)=0. (so if n<0, we can work with -g and -n>0)

    Consider the result of its euclidian division by m:\ n=qm+r where 0\leq r<m
    We get 0=ng=(qm+r)g=(qm)g+rg=q(mg)+rg=rg. But 0\leq r<m... Can you conclude?
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  3. #3
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    Got it, thanks for structuring that for me!!
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  4. #4
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    Re: Abelian groups

    Let G be a group. If x^(-1)=x, for all x E G, then prove that G is abelian
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  5. #5
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    Re: Abelian groups

    suppose that x-1 = x, for all x in G.

    then, for any x,y in G:

    (xy)-1 = xy (since xy is likewise in G, by closure)
    y-1x-1 = xy (since (xy)-1 = y-1x-1, in any group)
    yx = xy (since y-1 = y, and x-1 = x, since x,y are in G)

    thus G is abelian.
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