## ord$(g)|\max\{\text{ord}(g)|g\in G\}$ for all $g\in G$.

Let $$G$$ be a finite abelian group and $$n:=\max\{\text{ord}(g)|g\in G\}$$. Now I have to proof that ord$$(g)|n$$ for all $$g\in G$$.

My idea was:

Let $$g\in G$$ with ord$$(g)=m. Then because of the euclidean divsion in $$\mathbb{Z}$$ one can write $$n=km+r$$ for $$k,r\in\mathbb{Z}$$ and $$r. So what I have to show now is $$g^n = g^m = e$$. From that follows $$m|n$$ by the definition of ord, right?

So this gives me $$g^n=g^{km+r}=g^{km}g^r=(g^m)^kg^r=e^kg^r=g^r$$

But from here on I dont know where to go next.