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Can we define a compatible ordering in $\Bbb{Z}_m$?

The ordering is compatible with the operation if:

$x,y,z \in G$ with $x\leq y$ then $x\cdot z\leq y\cdot z$

I have the belief that we can not find this ordering, but how can I prove it?

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2 Answers

Hint: Consider whether $x<e$ or $x>e$ for arbitrary $x$; where $e$ is the identity element.

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Thank you for your help. –  art Mar 30 '13 at 16:50
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In case you ask about $(\mathbb{Z}/m,+)$: Try to prove that ordered groups are trivial or infinite.

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It appears that OP is interested in the multiplicative monoid of $\Bbb{Z}_{m}$, where $m$ looks arbitrary. –  Andreas Caranti Mar 23 '13 at 18:21
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