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Let $G$ be a commutative group, and let $g$ be an element of $G$ be an element of maximal finite order, then $|h|\leq |g|$. Prove that in fact if $h$ is finite order in $G$, then $|h|$ divides $|g|$.

This is what I have:

Proof by contradiction If $|h|$ is finite but doesn't divide $|g|$, then there is a prime integer $p$ such that $|g|=rp^m$, $|h|=sp^n$, with $r$ and $s$ relatively prime to $p$ and $m < n$.

(This is where I do not know where to go.)

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Your conclusion on $r,s,p,m$ is wrong. –  Olivier Bégassat Feb 26 '13 at 22:10
    
How? Please explain? –  Username Unknown Feb 26 '13 at 22:13
    
Why would $r$ and $s$ be relatively prime to $m$? –  Olivier Bégassat Feb 26 '13 at 22:20
    
Sorry there was an error in the question. I fixed it –  Username Unknown Feb 26 '13 at 22:23
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1 Answer 1

up vote 3 down vote accepted

Suppose the $|h|$ doesn't divide $|g|$, then $|h|$ contains a prime factor with multiplicity higher than in $|g|$. This tells us that $gh$ has order greater than that of $g$, contradicting the hypothesis.

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That's it!?!?!?!? That was simple. Is this a rigourous proof? –  Username Unknown Feb 26 '13 at 22:57
    
I mean, you might want to flesh out the details, but I'm pretty sure the reasoning is sound. –  orlandpm Feb 26 '13 at 23:11
    
okie dokie thank you –  Username Unknown Feb 26 '13 at 23:12
    
Since the statement in question is in general false for non-commutative groups, a correct argument needs to use the commutativity. Where did you do so in your answer? –  Pete L. Clark Jan 2 at 6:44
    
Commutativity tells us that $|gh| = \hbox{lcm}(|g|,|h|)$? –  orlandpm Jan 4 at 2:18
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