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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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