Abelian groups

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November 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 $0<n<m$ 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?
November 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 $0\leq r<m$
We get $0=ng=(qm+r)g=(qm)g+rg=q(mg)+rg=rg.$ But $0\leq r<m...$ Can you conclude?
November 28th 2009, 12:56 PM
nmatthies1

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