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We have the theorem that if a partially ordered group is directed and integrally closed then it must already be commutative.

Now we have a partial order on real functions, which is: $f$ is finally greater than $g$, in symbols $f>_\infty g$, i.e. there is an $x_0$ such that $f(x)>g(x)$ for all $x>x_0$. An analytic function different from identity function is not comparable with the identity function iff there are infinitely many fixpoints (if these fixpoints are bounded then the function must be already the identity function, otherwise the function is neither finally greater nor finally smaller than the identity)

My question is now the following: We generate a group $G$ of functions - which are strictly increasing and bijective on the positive real axis - with the two generators, say, $f(x)=x^2$ and $g(x)=x^2+x$ and the operation of composition. Every element of this group has the form $f^{i_{1}}\circ g^{k_{1}}\circ f^{i_{2}}\circ g^{k_2}\circ \dots \circ f^{i_n}\circ g^{k_n}$ for some integers $k_1,\dots,k_n$ and $i_1,\dots,i_n$, where $f^k$ denotes the k-time composition of $f$ and negative $k$ the $|k|$ time composition of the inverse.

The above mentioned order $\lt_\infty$ is linear on the monoid generated by $f$ and $g$. But if $\lt_\infty$ is linear on the generated group then it is directed and integrally closed/archimedian which would imply commutativity which $G$ not have. Hence there must be non-comparable functions and hence there must be a function not comparable with the identity function, which further implies that there must be a function in $G$ with infinitely many fixpoints.

Find the error in the above argumentation or find a function in $G$ with infinitely many fixpoints.

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up vote 2 down vote accepted

What is the proof that the order is archimedean on your group? $f=x^2$, $g=x^2+x$, so $\alpha := f^{-1}\circ g \sim x+1/2+O(x^{-1})$ and more generally $\alpha^n \sim x+n/2+O(x^{-1})$. Thus $\alpha^n \lt_\infty f$ for all $n$ but $\alpha \gt_\infty x$.

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Yeah, you are right. (And what a quick answer! I like this way of communication.) So I guess we have a linear non-Archimedian order. Does anyone have a proof for this or similar families of functions? –  bo198214 Mar 31 '11 at 9:42
    
Yes, this order is linear and non-archimedean. Goes back at least to Hardy, "Orders of Infinity". –  Gerald Edgar Mar 31 '11 at 16:49
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