1. ## 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?

2. 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?

3. Got it, thanks for structuring that for me!!

4. ## Re: Abelian groups

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

5. ## 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.