IMO 1973 Problem 5

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 36m23s

Problem

$G$ is a set of non-constant functions of the real variable $x$ of the form $$ f(x) = ax + b, a \text{ and } b \text{ are real numbers,} $$ and $G$ has the following properties:

(a) If $f$ and $g$ are in $G$, then $g \circ f$ is in $G$; here $(g \circ f)(x) = g[f(x)]$.

(b) If $f$ is in $G$, then its inverse $f^{-1}$ is in $G$; here the inverse of $f(x) = ax + b$ is $f^{-1}(x) = (x - b) / a$.

Prove that there exists a real number $k$ such that $f(k) = k$ for all $f$ in $G$.

Problem Understanding

Let $G$ be a set of non-constant affine functions of the real variable $x$ of the form

$$f(x) = ax + b, \quad a \neq 0,$$

satisfying the following properties:

If $f,g \in G$, then $g \circ f \in G$.
If $f \in G$, then $f^{-1} \in G$.
For every $f \in G$, there exists $x_f \in \mathbb{R}$ such that $f(x_f) = x_f$.

The goal is to show the existence of a real number $k$ such that

$$f(k) = k \quad \text{for all } f \in G.$$

The previous proof correctly identified that the essential issue is whether two functions in $G$ can have different fixed points. The invalid approach tried to compare fixed points of a function and its inverse, which yields no useful information. A more robust strategy is to rewrite affine maps relative to their fixed points and construct a function whose fixed-point property enforces equality of the fixed points.

Key Observations

Affine maps relative to a fixed point.

If $f(x) = ax + b$ and $f(p) = p$, then

$$ap + b = p \quad\implies\quad b = (1-a)p,$$

$$f(x) = a(x-p) + p.$$

If $a = 1$, then $b = 0$ by property (3), so $f(x) = x$ is the identity.

Slope 1 implies identity.

Any map $f(x) = x+b \in G$ must satisfy $b=0$, so the only slope-1 map is the identity. The identity fixes every point, so it will automatically fix the eventual common fixed point. 2. Construction of a translation via the commutator.

For non-identity maps $f(x) = a(x-p) + p$ and $g(x) = c(x-q) + q$ with $a \neq 1$ and $c \neq 1$, consider

$$h = f^{-1} \circ g^{-1} \circ f \circ g.$$

Then $h \in G$ by closure under composition and inversion. Computing $h$ shows that it is a translation

$$h(x) = x + \frac{(a-1)(c-1)}{ac}(p-q).$$

Since property (3) requires $h$ to have a fixed point, the translation must be trivial, forcing $p = q$.

Handling identity maps.

If $f$ is the identity, it automatically fixes any point. Therefore, once the common fixed point of non-identity maps is found, identity maps present no obstruction. 2. Degenerate case.

If $G$ contains only the identity map (possible if all functions of slope $1$ are identity), any real number can serve as the common fixed point. This case was previously omitted.

Solution

Let $f, g \in G$.

Step 1: Rewrite in terms of fixed points.

By property (3), write

$$f(x) = a(x-p) + p, \quad g(x) = c(x-q) + q,$$

where $p$ and $q$ are fixed points of $f$ and $g$, respectively.

Step 2: Handle slope 1.

If $a = 1$, then $f(x) = x+b$. Property (3) implies there exists $x$ with $x+b=x$, so $b=0$. Hence $f$ is the identity. Similarly for $g$ if $c=1$.

Step 3: Non-identity maps.

Assume $a \neq 1$ and $c \neq 1$. Consider

$$h = f^{-1} \circ g^{-1} \circ f \circ g \in G.$$

Step 4: Compute the commutator.

$$\begin{aligned} g(x) &= c x + (1-c) q, \ f(g(x)) &= a(g(x)-p)+p = a(c x + (1-c) q - p) + p = ac x + a(1-c)q + (1-a)p, \ g^{-1}(f(g(x))) &= \frac{f(g(x)) - q}{c} + q = \frac{ac x + a(1-c)q + (1-a)p - q}{c} + q \ &= a x + \frac{(1-a)p}{c} + \frac{a(1-c)}{c} q + \left(1-\frac1c\right) q, \ f^{-1}(g^{-1}(f(g(x)))) &= \frac{g^{-1}(f(g(x))) - p}{a} + p = x + \frac{(a-1)(c-1)}{ac} (p-q). \end{aligned}$$

Hence

$$h(x) = x + \frac{(a-1)(c-1)}{ac} (p-q).$$

Step 5: Apply property (3).

Since $h \in G$ and every element of $G$ has a fixed point, we require

$$x + \frac{(a-1)(c-1)}{ac} (p-q) = x \quad\implies\quad \frac{(a-1)(c-1)}{ac} (p-q) = 0.$$

Because $a \neq 0$, $a \neq 1$, $c \neq 0$, $c \neq 1$, it follows that $p-q=0$.

Step 6: Conclude common fixed point.

Thus any two non-identity maps in $G$ share the same fixed point. Let this point be $k$.

If $f$ is the identity, then $f(k) = k$ automatically. Hence every function in $G$ fixes the same real number $k$.

Step 7: Handle degenerate case.

If $G$ contains no non-identity elements, choose any $k \in \mathbb{R}$. The conclusion still holds.

∎

Verification of Key Steps

Step 1: Writing $f(x) = a(x-p) + p$ is correct because solving $f(p) = p$ yields $b = (1-a)p$.
Step 2: Slope 1 forces the identity since $x+b=x$ has unique solution $b=0$.
Step 4: Computation of the commutator was explicitly expanded above and verified against concrete examples ($a=2, c=3, p=2, q=3$ gives $h(x) = x-1/3$).
Step 5: The translation has a fixed point iff it is trivial; thus $p=q$ is rigorously justified.
Step 7: The degenerate case $G = {\mathrm{id}}$ is now explicitly addressed.

All slope configurations are covered:

Both non-identity ($a,c\neq1$) → commutator argument applies.
One identity ($a=1$ or $c=1$) → identity fixes $k$ automatically.
Both identity ($a=c=1$) → any $k$ works.
Only identity in $G$ → any $k$ works.

Alternative Approaches

An alternative is to use conjugation:

$$f(x) = a(x-p) + p, \quad g(x) = c(x-q) + q.$$

Then

$$g^{-1} \circ f \circ g(x) = a(x - g^{-1}(p)) + g^{-1}(p).$$

Since $g^{-1} \circ f \circ g \in G$, its fixed point must coincide with that of $f$ if uniqueness is required, yielding

$$g^{-1}(p) = p \quad \implies \quad g(p) = p.$$

Repeating this for arbitrary $f, g$ gives a universal fixed point.

While more abstract, this conjugation argument handles all slopes simultaneously and leads to the same conclusion as the commutator method.