Possible Duplicate:
Prove that for any element $b$, $|b|$ divides $|a|$ (order of $b$ divides order of $a$).
Finite abelian group generated by elements of maximal order
Let $G$ be a finite abelian group. Let $a\in G$ be an element of maximal order. Prove $|b|$ divides $|a|$ for all $b\in G$
I have been given a general outline of a proof but I am stuck on a few parts: I suppose that $g$ is an element of maximal order in $G$ and $h$ is a non identity element of $G$.
I am stuck at a few parts namely,
showing $|gh|\gt |g|$ and from there assuming $|h|$ doesn't divide $|g|$ proving that there exist integers i,j such that $ |g^i h^j |> |g|$
Any help would be appreciated,
Thanks,
This was the basic outline (Note that if g and h are elements of a group and $gh=hg$, it is possible to show that $(gh)^n = g^n h^n$ for any integer n. If you want to use this fact, please include a short proof.)
Suppose that g is an element of maximal order. Let h be a nonidentity element of G.
Step 1: Assume that $gcd(|g|,|h|)=1$. Prove that $|gh| > |g|$.
Step 2: Assume that $|h|$ does not divide $|g|$. Prove that there exist integers i and j such that $|g^i h^j |> |g|$.
