Ordering of $\mathbb{Z_n}$

Question

How can one show that there is no ordering on $\mathbb{Z}_n$?

At first the answer seemed trivial to me as there is no inclusion of the sets, meaning they are all pairwise disjoint elements. Hence Ordering Cannot exists. However we are asked to specifically show the following properties do not hold

i.) if $a<b$ and $b<c$ then $a<c$

ii.) if $a<b$ then $a+c<b+c$

How would one show these specifically do not hold since there is no ordering I can come up with at all? Or is it enough to say that there is no ordering without proving these specific examples?

Answer 1

Of course you need to prove it rigorously. If there is a total order $<$ over $\mathbb Z_n$, then either $1>0$ or $1<0$. What happens if you add the inequality to itself $n$ times?

Edit: note that the proof uses the fact that if $a<b$ and $c<d$ then $a+c<b+d$. This is because $a+c<b+c$ (by property $\mathrm{ii}$), and $b+c<b+d$ (again by property $\mathrm{ii}$). Then by property $\mathrm{i}$, we get $a+c<b+c$ and $b+c<b+d$ implies $a+c<b+d$.

Answer 2

More generally,

If $G$ is a finite group, then the unique partial order on $G$ compatible with the product is the equality.

Proof. Suppose that $a \leqslant b$ for some $a, b \in G$. Setting $c = ba^{-1}$, one gets $1 = aa^{-1} \leqslant ba^{-1} = c$. It follows that, for every $n > 0$, $c^n \leqslant c^{n+1}$. Thus $$ 1 \leqslant c \leqslant c^2 \leqslant \dotsm \leqslant c^{|G|} = 1 $$ whence $c = 1$ and $a = b$.

Answer 3

To answer my own question thanks to the help of all of you, I have come up with the following reasoning:

If you define any total ordering on the set $\mathbb{Z}_n$, there must exist a unique maximal element. Since two elements in the set produce this maximal element, say $a+b=m$. One can say that though $a<m$ this does not imply $a+b<m+b$ since we defined $a+b=m$ and $m$ is the maximal element hence $m+b$ cannot be “greater” than it so to say.