Corollary 1 Let $G$ be an ordered group. For all $x \in G$ , either $x \le 1 \le x^{-1}$ or $x^{-1} \le 1 \le x$ .

Proof: By conclusion 1, either $x < 1$ or $x = 1$ or $1 < x$ . If $x < 1$ , then, by conclusion 5, $1^{-1} < x^{-1}$ , so $x < 1 < x^{-1}$ . If $x = 1$ , the conclusion is trivial. If $1 < x$ , then, by conclusion 5, $x^{-1} < 1^{-1}$ , so $x^{-1} < 1 < x$ .

Corollary 2 Let $G$ be an ordered group and $n$ a strictly positive integer. Then, for all $x, y \in G$ , we have $x < y$ if and only if $x^n < y^n$ .

Proof: We shall first prove that $x < y$ implies $x^n < y^n$ by induction. If $n = 1$ , this is a simple tautology. Assume the conclusion is true for a certain value of $n$ . Then, conclusion 4 allows us to multiply the inequalities $x < y$ and $x^n < y^n$ to obtain $x^{n+1} < y^{n+1}$ .

As for the proof that $x^n < y^n$ implies $x < y$ , we shall prove the contrapositive statement. Assume that $x < y$ is false. By conclusion 1, it follows that either $x = y$ or $x > y$ . If $x = y$ , then $x^n = y^n$ so, by conclusion 1 $x^n < y^n$ is false. If $x > y$ then, by what we have already shown, $x^n > y^n$ so $x^n < y^n$ is also false in this case for the same reason.

Corollary 3 Let $G$ be an ordered group and $n$ a strictly positive integer. Then, for all $x, y \in G$ , we have $x = y$ if and only if $x^n = y^n$ .

Proof: It is trivial that, if $x = y$ , then $x^n = y^n$ . Assume that $x^n = y^n$ . By conclusion 1 of the main theorem, it is the case that either $x < y$ or $x = y$ or $y < x$ . If $x < y$ then, by the preceding corollary, $x^n < y^n$ , which is not possible. Likewise, if $y < x$ , then we would have $y^n < x^n$ , which is also impossible. The only remaining possibility is $x = y$ .

Corollary 4 An ordered group cannot contain any elements of finite order.

Let $x$ be an element of an ordered group distinct from the identity. By definition, if $x$ were of finite order, there would exist an integer such that $x^n = 1$ . Since $1 = 1^n$ , we would have $x^n = 1^n$ but, by Corollary 3, this would imply $x=1$ , which contradicts our hypothesis.

It is worth noting that, in the context of additive groups of rings, this result states that ordered rings have characteristic zero.