Let $G$ be a finite abelian group. Let $a\in G$ be an element of maximal order. Prove $|b|$ divides $|a|$ for all $b\in G$. A different proof [duplicate]

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 $|g^i h^j |> |g|$.

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marked as duplicate by amWhy, YACP, user7530, Asaf Karagila, MicahJan 30 '13 at 1:39

This question was marked as an exact duplicate of an existing question.

These steps sound, to me, very specific to the outline we cannot see. For instance, it is impossible to show $|gh|>|g|$ in general, so that means there is some assumption in the outline that we do not know. I think it is best just to copy the outline into your question. – peoplepower Jan 30 '13 at 0:58
@peoplepower hopefully that helps, this was the outline i was given. – bobdylan Jan 30 '13 at 3:10
I take it you want this question to be reopened because you feel it differs significantly from the other one. You can ask for this by posting an answer to meta.math.stackexchange.com/questions/6424/… being sure to explain why you believe your question is not a duplicate. – Gerry Myerson Jan 30 '13 at 3:35