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I've been playing around with the problem for a while now but haven't managed to make any progress. I would really appreciate a nudge in the right direction -- but please no full solutions.

Attempt:

Assume $|h|$ is finite but does not divide $|g|$. Then there is a prime integer $p$ such that $|g|=p^m r$ and $|h|=p^n s$, with $\gcd(r,s)=1$ and $m<n$. Note that $$\left| g^{p^m} \right| = \frac{|g|}{\gcd\left( p^m, |g| \right)}=r$$ $$\left| h^{s} \right| = \frac{|h|}{\gcd\left( s, |h| \right)}=p^n$$ thus $\gcd(\left| g^{p^m} \right|, \left| h^{s} \right|)=1$, so by a previous exercise we have $$\left| g^{p^m} h^{s} \right|=\left| g^{p^m} \right|\left| h^{s} \right| = p^n r .$$

Andrew Tawfeek
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  • You're very, very close. Take a long look at your last line. There is an immediate contradiction there. – Ben G. Jun 03 '18 at 00:58
  • @BenG. Since $m<n$ we have $p^m<p^n$, in order for $g$ to be maximal order it would require $r>s$... So $$\left| g^{p^m} h^{s} \right|=p^n r>|h|=p^n s$$ and thus... hmm, I can't immediately see what I'm missing.. – Andrew Tawfeek Jun 03 '18 at 01:05
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    Compare $|g^{p^m}h^s|$ to $|g|$. – Ben G. Jun 03 '18 at 01:21
  • @BenG. Oh! $$\left| g^{p^m} h^{s} \right| =p^n r > p^m r = |g|$$ which contradicts $g$ being a maximal order element! I feel beyond silly for being blind to that for so long. Thank you for the help! – Andrew Tawfeek Jun 03 '18 at 01:31
  • @BenG. If you're willing to rewrite your comment as an answer, I would be more than delighted to marked it as accepted. – Andrew Tawfeek Jun 03 '18 at 01:32

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Compare $|g^{p^m}h^s|$ to $|g|$.

Ben G.
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