# Abelian groups

• Nov 28th 2009, 11:18 AM
nmatthies1
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 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.

Am I making this too difficult or what is it that I'm missing?
• Nov 28th 2009, 12:34 PM
clic-clac
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
We get $0=ng=(qm+r)g=(qm)g+rg=q(mg)+rg=rg.$ But $0\leq r Can you conclude?
• Nov 28th 2009, 12:56 PM
nmatthies1
Got it, thanks for structuring that for me!!
• Oct 28th 2012, 11:25 PM
tejashree24
Re: Abelian groups
Let G be a group. If x^(-1)=x, for all x E G, then prove that G is abelian
• Oct 28th 2012, 11:32 PM
Deveno
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.