Exercise 3, § 6

a) For each interval $I \subset \mathbf{R}$, let $G(I)$ denote the group of all homeomorphisms of $I$ onto itself. Show that if $I$ and $J$ are any two intervals of $\mathbf{R}$, each containing more than one point, then $G(I)$ and $G(J)$ are isomorphic. In $G(I)$, the set $F(I)$ of increasing homeomorphisms is a normal subgroup of index 2.

b) Let $G = G(\mathbf{R})$ and $F = F(\mathbf{R})$. For each $f \in G$ and each $x \in \mathbf{R}$, let $\sigma(x; f)$ denote $\operatorname{sgn}(f(x) - x)$. Let $T^+$ (resp. $T^-$) be the set of all $f \in F$ such that $\sigma(x; f)$ is constant and equal to 1 (resp. $-1$) on $\mathbf{R}$. Show that every element of $T^+$ (resp. $T^-$) is conjugate in $F$ to the translation $x \to x + 1$ (resp. $x \to x - 1$). [If $f \in T^+$, consider the sequence of points $f^n(o), n \in \mathbf{Z}$].

c) Let $T = T^+ \cup T^-$. Show that every $f \in F$ which does not belong to $T$ is the product in $F$ of two elements of $T_*$. [Write $f = f_1 f_2$ in $F$, where $f_1(x) = \sup (x, f(x))$ and $f_2(x) = \inf (x, f(x))$; if $g$ is the translation $x \to x + 1$, then $f_1 g$ and $g^{-1} f_2$ belong to $T$].

d) Two elements $f, g$ are conjugate in $F$ if and only if there exists $s \in F$ such that $\sigma(x; f) = \sigma(s(x); g)$ for all $x \in \mathbf{R}$. [Consider the components $I_n$ of the open set of all $x$ such that $\sigma(x; f) \neq 0$; observe that $F(I_n)$ and $F$ are isomorphic (by $a$) and use $b$]. In particular, if $[a, b]$ is an interval of $\mathbf{R}$ such that $\sigma(a; f) = \sigma(b; f) = 0$ for some f \in F, then the element $f'$ of $F$ which coincides with $f$ for $x \leq a$ or $x \geq b$ and is such that $f'(x) = x + \varepsilon(f(x) - x)$ for $a < x < b$, where $0 < \varepsilon < 1$, is conjugate to $f$ in $F$.

e) Let $f \in F$ and let $a, b, c, d$ be four real numbers such that $a < c < b < d$ and $f(x) - x = 0$ for $x = a, x = b, x = c, x = d$, $\sigma(x; f) = 1$ for $a < x < c$ and $b < x < d$. Let $a', b'$ be two numbers such that $a \leq a' < c$ and $b < b' \leq d$; let $J = [a, b], J' = [a', b']$, and let $f_1$ denote the restriction of $f$ to $J$; thus $f_1 \in F(J)$. Show that there is an increasing homeomorphism $s$ of $J$ onto $J'$ such that the element $f_2 = s f_1 s^{-1}$ of $F(J')$ is such that $f_2(x) > f^{-1}(x)$ whenever $a' < x < b'$ [use $d$].

f) Let $f \in F$ and let $[a, b]$ be an interval of $\mathbf{R}$ such that $f(x) - x = 0$ for $x = a$ and $x = b$, $\sigma(x; f) = -1$ for $x < a$ and $\sigma(x; f) = +1$ for $x > b$. Show that there exists $g \in F$, conjugate to $f$ in $F$, such that $g(x) > f(x)$ for all $x \in \mathbf{R}$ (same method).

g) Let $H^+$ (resp. $H^-$) denote the normal subgroup of $F$ consisting of all $f \in F$ such that $f(x) = x$ in a neighbourhood of $+\infty$ (resp. $-\infty$). Show that $H^+, H^-$ and $H = H^+ \cap H^-$ are the only normal subgroups of $F$ other than $F$ and $\{e\}$. [If $N$ is a normal subgroup of $F$, other than $F$, and if $f \in N$ does not belong to $H^+$, show that there is an element $g \in N$ such that the set of all $x \in \mathbf{R}$ for which $\sigma(x; g) = +1$ is an interval $]a, +\infty]$, and that we have $g(x) = x$ for $x \leq a$. To do this use the constructions of e) and f), as well as b) and c). To show that $N$ contains no subgroup, other than $H$ and $\{e\}$, which is normal in $F$, consider an element $f \in H$ distinct from $e$; let $a$ and $b$ be the greatest lower bound and least upper bound (both finite by hypothesis) of the set of $x \in \mathbf{R}$ such that $\sigma(x; f) \neq 0$. Observe then that $F([a, b])$ is isomorphic to $F$ and use the preceding result].

h) Show that the group $H$ is simple. (Same method as for proving that $H$ contains no non-trivial subgroups normal in $F$).