OrderedSemigroups

1 Groups

In this chapter, we prove Holder’s theorem for groups. We follow the proof in “Groups, Orders, and Dynamics” by Deroin, Navas, and Rivas.

We choose an element of the ordered group \(f\) and map it to \(1\) in the real numbers. Then for any other element \(g\) of the ordered group, we use \(f\) to construct a sequence of rational approximations of \(g\). We map \(g\) to the real number that is the limit of this sequence of approximations. We then prove that this map is injective and order preserving.

1.1 Definitions

We begin with basic definitions of ordered groups.

Definition 1
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A left ordered group \(G\) is a group and a partial order such that for all \(x, y, z \in G\), if \(x \le y\), then \(z * x \le z * y\).

Definition 2
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A left linear ordered group \(G\) is a left ordered group that is also a linear order.

Definition 3
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An Archimedean group is a left ordered group such that for any \(g, h \in G\) where \(g \ne 1\), there exists an integer \(z\) such that \(h {\lt} g^z\).

1.2 Approximation

In this section we assume that \(G\) is a left linear ordered group that is Archimedean. Furthermore, we assume we have an element \(f\in G\) such that \(1 {\lt} f\).

Theorem 4
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For any \(g \in G\) and \(p \in \mathbb {N}\), there exists an integer \(q \in \mathbb {Z}\) such that

\[ f^q \le g^p {\lt} f^{q+1} \]
Proof

Since \(G\) is an Archimedean group, we can construct exponents \(l\) and \(u\) such that \(f^l {\lt} g^p {\lt} f^u\). Therefore, there must exist some integer \(q\) which satisfies what we want.

Definition 5
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We define a function \(q \colon G \to \mathbb {N} \to \mathbb {R}\) using Theorem 4 such that for any \(g\in G\) and \(n \in \mathbb {N}\),

\[ f^{q_g(n)} \le g^n {\lt} f^{q_g(n)+1} \]
Theorem 6
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For any sequence \(a_n\) of real numbers, if there exists \(C \in \mathbb {R}\) such that for all \(m,n\in \mathbb {N}\) we have that

\[ |a_{m+n} - a_m - a_n| \le C \]

then sequence \(\frac{a_n}{n}\) converges.

Proof

Not included here as the ideas are separate from this project.

Theorem 7
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For any \(g \in G\) and \(a,b \in \mathbb {N}\), we have that

\[ f^{q_g(a) + q_g(b)} \le g^{a+b} {\lt} f^{q_g(a) + q_g(b) + 2} \]
Proof

We know the following two things by the definition of \(q\)

\begin{align*} f^{q_g(a)}\le & g^a {\lt} f^{q_g(a)+1}\\ f^{q_g(b)}\le & g^b {\lt} f^{q_g(b)+1}\\ \end{align*}

And so it follows that

\[ f^{q_g(a)+q_g(b)}\le g^{a+b} {\lt} f^{q_g(a)+q_g(b) + 2} \]
Theorem 8
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For any \(g \in G\), the sequence \(\frac{q_g(n)}{n}\) converges.

Proof

From Theorem 7, we have that

\[ q_g(a) + q_g(b) \le q_g(a+b) \le q_g(a)+q_g(b) + 1 \]

and so

\[ |q_g(a+b) - q_g(a) - q_g(b)| \le 1 \]

Therefore, by Theorem 6 with \(C = 1\), we have that the sequence \(\frac{q_g(n)}{n}\) converges.

1.3 Map

We make the same assumptions as in the previous section. So we assume that \(G\) is a left linear ordered group that is Archimedean, \(f\in G\), and \(1 {\lt} f\).

We now define the map from the \(G\) to \(\mathbb {R}\) and prove its properties.

Definition 9
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We define a map \(\phi \colon G \to \mathbb {R}\) by mapping \(g\) to the real number that \(\frac{q_g(n)}{n}\) converges to as we know from Theorem 8.

Theorem 10
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For all \(g_1,g_2 \in G\) and \(p\in \mathbb {N}\),

\[ q_{g_1}(p) + q_{g_2}(p) \le q_{g_1g_2}(p) \le q_{g_1}(p) + q_{g_2}(p) + 1 \]
Proof

Let \(q_1 = q_{g_1}(p)\) and \(q_2 = q_{g_2}(p)\). Then we know that

\begin{align*} f^{q_1} \le & g_1^p {\lt} f^{q_1+1}\\ f^{q_2} \le & g_2^p {\lt} f^{q_2+1}\\ \end{align*}

And so we also have the following two facts

\begin{align*} f^{q_1+q_2}& \le g_1^p g_2^p\\ g_2^p g_1^p& {\lt} f^{q_1 + q_2 + 2} \end{align*}

We look at the case where \(g_1g_2 \le g_2g_1\). Then \(g_1^p g_2^p \le (g_1g_2)^p \le g_2^p g_1^p\). And so combined with the previous facts, we have that

\[ f^{q_1+q_2}\le g_1^p g_2^p \le (g_1g_2)^p \le g_2^p g_1^p {\lt} f^{q_1+q_2+2} \]

Therefore,

\[ q_1+q_2 \le q_{g_1g_2}(p) \le q_1+q_2+1 \]

And so we are done. The case where \(g_2g_1 \le g_1g_2\) follows similarly.

Theorem 11
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The map \(\phi \) is a homomorphism.

Proof

Let \(g_1, g_2 \in G\). Then from Theorem 10 we have that

\[ q_{g_1}(p) + q_{g_2}(p) \le q_{g_1g_2}(p) \le q_{g_1}(p) + q_{g_2}(p) + 1 \]

And so since \(\lim _{p\to \infty }\frac{q_{g_1}(p) + q_{g_2}(p)}{p} = \lim _{p\to \infty }\frac{q_{g_1}(p) + q_{g_2}(p) + 1}{p}\), we have that

\[ \lim _{p\to \infty }\frac{q_{g_1}(p) + q_{g_2}(p)}{p} = \lim _{p\to \infty }\frac{q_{g_1g_2(p)}}{p} \]

Therefore, by the definition of \(\phi \), we have shown that \(\phi (g_1)+\phi (g_2) = \phi (g_1g_2)\).

Theorem 12
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For all \(g,h \in G\), if \(g \le h\) then \(\phi (g) \le \phi (h)\).

Proof

First, notice that since \(g\le h\), then for all \(p \in \mathbb {N}\), \(q_g(p) \le q_h(p)\). Then from the definition of \(\phi \), it follows that \(\phi (g) \le \phi (h)\).

Theorem 13
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We have that \(\phi (f) = 1\) where \(f\) is our fixed positive element.

Proof

We have that for all \(n\in \mathbb {n}\) that \(f^n \le f^n {\lt} f^{n+1}\) and so \(q_f(n) = n\). Therefore, \(\phi (f) = 1\).

Theorem 14
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The map \(\phi \) is injective.

Proof

Since from Theorem 11 we have that \(\phi \) is a homomorphism, it suffices to show that for any \(g\in G\), if \(\phi (g) = 0\), then \(g = 1\).

Assume for the sake of contradiction that there exists \(g \in G\) such that \(\phi (g) = 0\) but \(g\) is not equal to \(1\). Then since \(G\) is Archimedean, there exists an integer \(z\) such that \(g^z {\gt} f\). Therefore, since by Theorem 13 we have that \(\phi (f) = 1\),

\begin{align*} 0 & = \phi (g) = \phi (g)^z = \phi (g^z)\\ & {\gt} \phi (f) = 1 \end{align*}

Contradiction.

Theorem 15
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For all \(g,h \in G\), we have that \(g \le h\) if and only if \(\phi (g) \le \phi (h)\).

Proof

(\(\Rightarrow \)) This is Theorem 12.

(\(\Leftarrow \)) We have that \(\phi (g) \le \phi (h)\). Assume for the sake of contradiction that \(h {\lt} g\). Then by Theorem 12, we know that \(\phi (h) \le \phi (g)\). Therefore, \(\phi (g) = \phi (h)\). And so by Theorem 14, we know that \(\phi \) is injective and so \(g = h\). Contradiction.

1.4 Holder’s Theorem

Theorem 16
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If \(G\) is a left linear ordered group that is Archimedean, then \(G\) is isomorphic to a subgroup of \(\mathbb {R}\).

Proof

First, we look at the case where there exists a positive element \(f\) in \(G\). Given such an element, we have an order preserving, injective homomorphism \(\phi \). And so \(G\) is isomorphic to the image of \(\phi \) which is a subgroup of \(\mathbb {R}\).

If there does not exist a positive element in \(G\), then \(G\) is trivial and is isomorphic to the trivial subgroup of \(\mathbb {R}\).