tchebytchev wrote: Let $G$ be a non-empty set of non-constant functions $f$ such that $f(x)=ax + b$ (where $a$ and $b$ are two reals) and satisfying the following conditions: 1) if $f \in G$ and $g \in G$ then $gof \in G$, 2) if $f \in G$ then $f^ {-1} \in G$, 3) for all $f \in G$ there exists $x_f \in \mathbb{R}$ such that $f(x_f)=x_f$. Prove that there is a real $k$ such that for all $f \in G$ we have $f(k)=k$ If $|G|=1$ all elements of $G$ trivially have the same fixed point. Q.E.D. If $f(x)=ax+b$ and $g(x)=cx+d$ are distinct, different from identity, and both belong to $G$, then : Fixed point of $f$ is $x_f=\frac b{1-a}$ and fixed point of $g$ is $x_g=\frac d{1-c}$ $h(x)=f(g(f^{-1}(g^{-1}(x))))\in G$ And since $h(x)=x+b(1-c)-d(1-a)$ we need, in order to have a fixed point for $h(x)$ : $b(1-c)=d(1-a)$ And so $x_f=x_g$ Q.E.D.