Is group defined by action on $\mathbb{R}$ a free group?

Question

Assume that I have two bijective functions $f,h:\mathbb{R}\to\mathbb{R}$ such that $f(h(x))\neq h(f(x))$ as well as $f(x)\neq x \neq h(x)$ for all $x\in\mathbb{R}$. Additionally, I consider the group $G$ generated by a set $S:=\{a,b\}$ with an action of $G$ on $\mathbb{R}$ given for $x\in\mathbb{R}$ by $e(x)=x$, $a(x) = f(x)$, $b(x)=h(x)$ and $ab(x) = f(h(x))$, $ba(x) = h(f(x))$.

I am wondering whether the group $G$ is simply (up to homomorphism) the non-abelian free group $F_2$ generated by the two elements $a,b$. This confuses me, because that would allow me to define uncountably many actions of $G$ on $\mathbb{R}$ without any implications on either $f,h$ or $G$. Does $G$ bear any information on this action with respect to $f$ and $h$ or can I use $G$ in any way to say something about concationations of arbitrary length of $f$ and $h$ or does the previous observation prevent everythig in that regard?

Answer 1

This doesn't necessarily show that your group is free, just that for some presentation $$ H \cong \langle S \mid R \rangle := F(S)/ N $$ that $aba^{-1}b^{-1} \not \in N$. (With $N = \langle\langle R \rangle\rangle$ the smallest normal subgroup containing the relations $R$ and in this case $S = \{a,b\}$).

So for example, taking $f(x) = x+1$, $h(x) = -x$, you get your group acting on $\mathbb R$ via $$ H = \langle a,b \mid a = ba^{-1}b, b^2 = 1 \rangle $$ since $x-1 = -(-x+1)$ and $-(-x)=x$, but you still have non-commutativity $fh(x) = -x+1 \not = -x-1=hf(x)$ for all $x$.

But on the other hand, you can also still act via a free group: for $G = F_2 = \langle a,b \rangle$ and $H$ a group generated by two elements $h,g \in H$, then there's always a homomorphism

\begin{align} F_2 &\to H \\ a &\mapsto g \\ b &\mapsto h \end{align}

which you can then see as acting on $\mathbb R$: $$ F_2 \to H \to \textrm{Bijections}( \mathbb R) $$ and more generally for any group $H = \langle S \rangle$, for some set $S \subseteq H$, there's a homomorhpism from the free group on $S$ (as symbols with no relations) by extending the inclusion $S \hookrightarrow H$, so any $H$ action extends to an $F(S)$ action.