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The ambient group

Let

$$I \;=\; \{\, x \in \mathbb{R} \;:\; 0 \le x \le 1 \,\} \;=\; [0,1],$$

carried with the order it inherits from $\mathbb{R}$, and let

$$G \;=\; \operatorname{Aut}_{\le}(I)$$

be the set of all order isomorphisms of $I$: bijections $f : I \to I$ such that $x \le y$ if and only if $f(x) \le f(y)$ (equivalently, strictly increasing bijections of $[0,1]$). No continuity, smoothness or piecewise-linearity is built into $G$; it is the full group of order-preserving bijections of the interval.

$G$ is a group under composition, with the convention that in a product the right-hand factor acts first:

$$(f \cdot g)(x) \;=\; f\bigl(g(x)\bigr) \qquad\text{for all } x \in I .$$

Its identity element $1$ is the identity map of $I$, the inverse $f^{-1}$ is the inverse bijection, and for a natural number $c$ the power $f^{c}$ is the $c$-fold composite of $f$ with itself, with $f^{0} = 1$.

All equalities and inequalities of elements below are equalities and inequalities in $G$ — that is, of order isomorphisms; two elements of $G$ are equal exactly when their underlying functions $[0,1] \to [0,1]$ agree, so this is the same as equality of maps.

Two distinguished elements of $G$

Let $A \in G$ be the strictly increasing, continuous, piecewise-linear bijection of $[0,1]$ given by

$$A(x) \;=\; \begin{cases} x/2, & 0 \le x \le 1/2,\\[2pt] x - 1/4, & 1/2 \le x \le 3/4,\\[2pt] 2x - 1, & 3/4 \le x \le 1, \end{cases}$$

whose breakpoint data is $A(0) = 0$, $A(1/2) = 1/4$, $A(3/4) = 1/2$, $A(1) = 1$ (the three formulas agree at the overlaps, so $A$ is well defined and continuous).

Let $B \in G$ be the strictly increasing, continuous, piecewise-linear bijection of $[0,1]$ given by

$$B(x) \;=\; \begin{cases} x, & 0 \le x \le 1/2,\\[2pt] x/2 + 1/4, & 1/2 \le x \le 3/4,\\[2pt] x - 1/8, & 3/4 \le x \le 7/8,\\[2pt] 2x - 1, & 7/8 \le x \le 1, \end{cases}$$

whose breakpoint data is $B(0)=0$, $B(1/2) = 1/2$, $B(3/4) = 5/8$, $B(7/8) = 3/4$, $B(1) = 1$. So $B$ is the identity on $[0,1/2]$ and on $[1/2,1]$ is a rescaled copy of $A$.

The family $X_0, X_1, X_2, \dots$

Define elements $X_i \in G$, one for each natural number $i \ge 0$, by

$$X_0 \;=\; A, \qquad X_{n+1} \;=\; \bigl(A^{\,n}\bigr)^{-1} \, B \, A^{\,n} \quad (n \ge 0),$$

the product again being composition with the right-hand factor acting first; so $X_{n+1}$ is the map $x \mapsto A^{-n}\bigl(B\bigl(A^{n}(x)\bigr)\bigr)$.

Thus

$$X_0 = A,\qquad X_1 = B,\qquad X_2 = A^{-1} B A,\qquad X_3 = A^{-2} B A^{2},$$

and in general $X_i = A^{-(i-1)} B A^{\,i-1}$ for every $i \ge 1$.

The word attached to a list of exponents

Let $c = (c_0, c_1, \dots, c_{n-1})$ be a finite list of natural numbers (so each $c_j \ge 0$). Its associated element of $G$ is

$$W(c) \;=\; X_0^{\,c_0} \, X_1^{\,c_1} \, X_2^{\,c_2} \cdots X_{n-1}^{\,c_{n-1}} ,$$

that is: the index into the family $X$ starts at $0$ and advances by one for each successive entry of the list, the entry in position $j$ being used as the exponent of $X_j$. The factors are multiplied in the order displayed, so — with the composition convention above — the factor $X_{n-1}^{\,c_{n-1}}$ coming from the last entry of the list acts on a point first, and $X_0^{\,c_0}$ acts last. The empty list gives $W(\,) = 1$. Entries equal to $0$ contribute the factor $1$.

Total indexing of lists

For a finite list $c$ of natural numbers and a natural number $k$, write

$$c_k \;=\; \begin{cases} \text{the entry of } c \text{ in position } k \ (\text{positions counted from } 0), & k < \operatorname{len}(c),\\[2pt] 0, & k \ge \operatorname{len}(c). \end{cases}$$

The indexing used in the hypothesis below is this total one: reading past the end of a list returns the default value $0$ rather than being undefined.

The statement

Claim. For all finite lists $a$ and $b$ of natural numbers satisfying the four conditions (1)–(4) below, one has

$$W(b) \cdot W(a)^{-1} \;\ne\; 1 \quad\text{in } G .$$

The conditions on the pair $(a,b)$ are, with $n = \operatorname{len}(a)$ and $m = n - 1$ (natural-number subtraction):

1. $a$ is not the empty list; hence $n \ge 1$ and $m = n-1$ is a legitimate position in $a$.
2. $\operatorname{len}(a) = \operatorname{len}(b)$; hence $b$ is not empty either, and $m$ is the last position of both lists.
3. At that last position, either $a_m = 0$ and $b_m > 0$, or $a_m > 0$ and $b_m = 0$. (The two alternatives are mutually exclusive, so this says: exactly one of the two final entries $a_m$, $b_m$ is zero and the other is strictly positive.) Both entries here are taken at the single index $\operatorname{len}(a) - 1$ computed from $a$; by (2) this is also the last index of $b$.
4. For every natural number $k$ with $k + 1 < n$: if $a_k > 0$ and $b_k > 0$, then $a_{k+1} > 0$ or $b_{k+1} > 0$.

Written out, the asserted inequality is

$$\Bigl(X_0^{\,b_0} X_1^{\,b_1} \cdots X_{n-1}^{\,b_{n-1}}\Bigr) \cdot \Bigl(X_0^{\,a_0} X_1^{\,a_1} \cdots X_{n-1}^{\,a_{n-1}}\Bigr)^{-1} \;\ne\; 1 .$$

Note which list plays which role: the list $b$ supplies the non-inverted word, and the list $a$ supplies the word that is inverted. The inverse is taken of the whole product $W(a)$, not of its individual factors, and (by the composition convention) the map $W(a)^{-1}$ acts on a point first, then $W(b)$. Since $G$ is a group, the assertion is literally equivalent to $W(b) \ne W(a)$, i.e. the two words determine different order isomorphisms of $[0,1]$.

Degenerate and edge cases silently covered

• The two lists $a$, $b$ are quantified universally and only implicitly; they are otherwise unconstrained apart from (1)–(4). There is no upper bound on their common length $n$ and no bound on the size of the entries.
• Entries are natural numbers, so all exponents in both words are $\ge 0$: no factor $X_j$ ever occurs with a negative exponent inside $W(a)$ or $W(b)$.
• Condition (4) is a conditional requirement that only takes effect at positions $k$ where both $a_k > 0$ and $b_k > 0$. At a position where at least one of the two entries is $0$ it imposes nothing. In particular interior positions where $a_k = b_k = 0$ are permitted, and such a position imposes no condition on position $k+1$.
• Condition (4) is vacuous when $n = 1$ (there is no $k$ with $k + 1 < 1$).
• Condition (4) never constrains the pair of entries at the last position $m$; only condition (3) does.
• Nothing requires the entries of $a$ or $b$ to be positive except as forced at the last position by (3), and nothing requires $a \ne b$ as lists (though (3) does force $a_m \ne b_m$, hence $a \ne b$).
• The total indexing convention means that the expressions $a_k$, $b_k$ would still be defined for $k \ge n$, but conditions (3) and (4) only ever evaluate them at indices $k \le n-1$, so the out-of-range default value $0$ is never actually reached.
• The hypothesis is satisfiable, not vacuous. For instance $a = (0)$, $b = (1)$ satisfies (1)–(4): the lists are nonempty of equal length $1$, at the last position $a_0 = 0 < 1 = b_0$, and (4) is vacuous. Further witnesses: $a = (1,0)$, $b = (1,2)$; $a = (1,0,0)$, $b = (0,0,3)$; and $a = (2,5,0)$, $b = (3,0,7)$, the last of which does exercise condition (4) nontrivially at $k = 0$.
• Both $W(a)$ and $W(b)$ are continuous piecewise-linear order isomorphisms of $[0,1]$ fixing $0$ and $1$, being products of powers of $A$, $B$ and their inverses; but the inequality is asserted in the full group of order isomorphisms of $[0,1]$, whose identity element is the identity map of $[0,1]$.