Let \(a\in S\) and \(b\in S\). Since the \(S\) is a linear order we have one of the following cases.

The first case is that \(b * a {\gt} b\). Then for all \(x\in S\) we have that \(b * a * x {\gt} b * x\) and so \(a * x {\gt} x\). Therefore, \(a\) is positive.

The second case is that \(b * a {\lt} b\). Then for all \(x\in S\) we have that \(b * a * x {\lt} b * x\) and so \(a * x {\lt} x\). Therefore, \(a\) is negative.

The last case is that \(b * a = b\). Then for all \(x\in S\) we have that \(b * a * x = b * x\) and so \(a * x = x\). Therefore, \(a\) is zero.

Since \(S\) is non-Archimedean, there exists \(a,b\in S\) such that \(a\) and \(b\) have the sign and are not mutually Archimedean.

First, we look at the case where \(a\) and \(b\) are positive. Then since they are not mutually Archimedean, without loss of generality, for all \(n\in \mathbb {N}^+\), \(a^n {\lt} b\).

Then we either have that \(a * b \le b * a\) or that \(b*a\le a * b\). In the first case, we have that

which means that, since \(a^n {\lt} b\),

And so \(b\) and \(a*b\) form an anomalous pair.

The other cases follow similarly.

Let \(a\) and \(b\) be elements of an ordered semigroup \(S\). If either \(a\) or \(b\) is one, then they commute.

We begin with the case that \(a\) and \(b\) are positive. If \(a * b {\lt} b * a\), then for all \(n\in \mathbb {N}^+\), we have that

And so \(a*b\) and \(b*a\) form an anomalous pair.

The same for the case that \(b * a {\lt} a * b\). Therefore, we must have that \(a*b = b*a\).

The case where \(a\) and \(b\) are negative follows similarly.

We now look at the case where \(a\) is negative and \(b\) is positive. If the element \(1\) exists and \(a*b = 1\), then \(a * b * a = a\) and so \(b*a = 1\). Therefore, \(a\) and \(b\) commute.

If \(a * b\) is positive, then

We already showed that positive elements commute and so

Now we look at the case where \(a*b {\lt} b*a\). Then we have that

Which is a contradiction. We get a similar contradiction in the case that \(b*a {\lt} a*b\). Therefore, \(a*b = b*a\).

The case where \(a*b\) is negative follows similarly. The case where \(b\) is negative and \(a\) is positive is symmetric.

(\(\Rightarrow \)) If \(a\) and \(b\) are positive and \(a {\lt} b\), then we have that \(a^n {\lt} b^n\) for all \(n\). Therefore, there must exist an \(n\) such that \(a^{n+1} \le b^n\) as otherwise \(a\) and \(b\) would form an anomalous pair. Thus, the theorem is satisfied with \(c = a\).

Similarly if \(a\) and \(b\) are negatiave.

(\(\Leftarrow \)) We look first at the case where \(a\) and \(b\) are positive and \(a {\lt} b\). Then we have \(c\in S\) Archimedean with respect to \(a\) and \(m\in \mathbb {N}^+\) such that \(a^m * c \le b^m\).

Therefore, for any \(n \in \mathbb {N}^+\), either

or

holds.

Since \(c\) is Archimedean with respect to \(a\), there exists an \(N\) such that for all \(n \ge N\), \(a {\lt} c^n\). Thus, for any \(n \ge N\), we get from the previous relations that

and so \(a\) and \(b\) do not form an anomalous pair.

Similalry if \(a\) and \(b\) are negative.

We do casework on whether or not \(S\) has an element that is one.

If \(S\) has an element that is one then it is already a monoid. Furthermore, since it has no anomalous pairs, by Theorem 32, it is commutative. And so we are done.

If \(S\) does not have an element that is one, then we let \(M\) be \(S\) with one added. We define one to be less than every positive element and greater than every negative element. By Theorem 32, we know that \(S\) is commutative and so \(M\) is too. Furthermore, it is clear that \(M\) has no anomalous pairs still. Then \(S\) is isomorphic to the subsemigroup of \(M\) that is all its elements except for the added one.

We look at the positive case. If \(x\) is positive then we are done as \(x*x\) is positive.

Next we look at the case where \(x\) is negative. Assume for the sake of contradiction that for all \(y \in S\), if \(y\) is positive then \(x*y\) is negative. Let \(z\) be a positive element we assumed existed in \(S\). Then for all positive \(n \in \mathbb {N}\), we have that \(x * z^{n+2}\) is negative. Recall that from Theorem 32 we have commutativity. From there we have an anomalous pair:

Contradiction. And similarly for the negative case.

Let \(\frac{a}{b}\in G\) where \(a,b\in M\). Since \(M\) has a positive element, by Theorem 35, we have \(a_2, b_2\in M\) such that \(a*a_2\) and \(b*b_2\) are positive. Let \(a' = a*a_2\) and \(b' = b*b_2\). Then \(\frac{a}{b} = \frac{a'}{b'}\) and \(a'\) and \(b'\) are positive.

And similarly for the negative case.

If \(M\) is trivial then we are done. Otherwise, it has a positive element of a negative element.

We look at the case where \(M\) has a positive element \(t\). We now want to show that the Grothendieck group \(G\) of \(M\) is Archimedean. It suffices to show that for any positive \(g, h \in G\), there exists an integer \(n\) such that \(g^n {\gt} h\). Since \(g, h \in G\), there exist \(g_1,g_2,h_1,h_2 \in M\) such that \(g = \frac{g_1}{g_2}\) and \(h = \frac{h_1}{h_2}\), and by Theorem 36, we can assume that \(g_1\), \(g_2\), \(h_1\), and \(h_2\) are positive.

Then since \(M\) does not have anomalous pairs, we have by Theorem 32 that \(M\) is Archimedean. Since \(g_1\) and \(h_1\) are positive and \(M\) is Archimedean, there exists a positive \(N \in \mathbb {N}\) such that for all \(n \ge N\), \(g_2^n {\gt} h_1\). So in particular, we have that \(g_2^N {\gt} h_1\).

Notice that since \(\frac{g_1}{g_2}\) positive, we have that \(g_2 {\lt} g_1\). And therefore, since \(M\) does not have anomalous pairs, there exists a positive natural number \(N_1\) such that \(g_2^{N_1+N} {\lt} g_1^{N_1}\).

We now claim that \(g^{N_1} {\gt} h\). This is equivalent to showing that \(g_2^{N_1}*h_1 {\lt} h_2*g_1^{N_1}\). And we have that

And so we are done.

The final case where \(M\) has a negative element follows similarly.

By Theorem 34, we have that \(S\) is isomorphic to a subsemigroup of a linear ordered cancel commutative monoid \(M\) that does not have anomalous pairs.

We let \(G\) be the Grothendieck group of \(M\). Then by Theorem 37 we know that \(G\) is Archimedean. Since \(G\) is the Grothendieck group of \(M\), \(M\) is isomorphic to a submonoid of \(G\). Thus, we have that \(S\) is isomorphic to a subsemigroup of \(G\).

By Theorem 38, there exists a linear ordered commutative Archimedean group \(G\) such that \(S\) is isomorphic to a subsemigroup of \(G\). By Theorem 16, \(G\) is isomorphic to a subgroup of \(\mathbb {R}\). Therefore, \(S\) is isomorphic to a subsemigroup of \(\mathbb {R}\).