
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 $\displaystyle mn$.
Now the obvious cases are if $\displaystyle n=0$, $\displaystyle ng=0$ and $\displaystyle mn$ 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?

Hi
Since $\displaystyle g$ and $\displaystyle g$ have the same order $\displaystyle m$, proving that for any integer $\displaystyle n,\ mn\Rightarrow ng=0$ is quite easy. (just need to write what $\displaystyle mn$ 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?

Got it, thanks for structuring that for me!!

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

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^{1}x^{1} = xy (since (xy)^{1} = y^{1}x^{1}, in any group)
yx = xy (since y^{1} = y, and x^{1} = x, since x,y are in G)
thus G is abelian.