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Thread: Abelian groups

  1. #1
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    Abelian groups

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

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

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

    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 $\displaystyle 0<n<m$ I have if $\displaystyle g+...+g (m times) =0$ then $\displaystyle [g+...+g (m times)] + g+ ... +g (ntimes)= g+...+g (n times)$ 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 $\displaystyle m<n<0$.

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

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

    Consider the result of its euclidian division by $\displaystyle m:\ n=qm+r$ where $\displaystyle 0\leq r<m$
    We get $\displaystyle 0=ng=(qm+r)g=(qm)g+rg=q(mg)+rg=rg.$ But $\displaystyle 0\leq r<m...$ 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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