corollaries of basic theorem on ordered groups

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

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 $x^{-1}<1<x$ .

Q.E.D.

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.

Q.E.D.

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 $x=y$ .

Q.E.D.

Corollary 4 An ordered group cannot contain any elements offinite 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 aninteger 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 contradictsour hypothesis.

Q.E.D.

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

单词	CorollariesOfBasicTheoremOnOrderedGroups
释义	corollaries of basic theorem on ordered groups Corollary 1 Let $G$ be an ordered group. For all $x\\in G$ , either $x\\leq 1\\leq x^{-1}$ or $x^{-1}\\leq 1\\leq x$ . 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 $x^{-1}<1<x$ . Q.E.D. 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. Q.E.D. 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 $x=y$ . Q.E.D. Corollary 4 An ordered group cannot contain any elements offinite 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 aninteger 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 contradictsour hypothesis. Q.E.D. It is worth noting that, in the context of additive groups of rings,this result states that ordered rings have characteristic zero.
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