Hello everyone I have a hard time trying to resolve this problem if anyone could help it would be a lot appreciated.
Let $$f_1\colon\mathbb R\rightarrow \mathbb R,\,f_1(x)=-x,\quad f_2\colon\mathbb R\rightarrow \mathbb R,\,f_2(x) = -x +1$$
We define $D_\infty = \langle f_1,f_2 \rangle$ , the infinite dihedral group and $A = \langle f_2 \circ f_1 \rangle$ and $B=\langle f_2\rangle$, then we have :
$$D_\infty = A \rtimes B \simeq \mathbb Z \rtimes \mathbb Z/2\mathbb Z$$
How can we prove that statement ?
Should we prove that $f_1 ^2=f_2 ^2=1$ , so I can use the following answer ? I also found here exactly what I want but how can I prove it using $f_1$ and $f_2$.
( How can we also prove infinite order of $\langle f_2 \circ f_1\rangle$, I thought maybe we have to compute $(f_2 \circ f_1)^k(x)$?)
$B$ and $\frac{\Bbb Z}{2\Bbb Z}$ act faithfully on $A$ and $\Bbb Z$, respectively, via $A$ automorphisms and $\Bbb Z$ automorphisms. The map $$\Psi:\langle f_2\circ f_1\rangle\rtimes \langle f_2\rangle\to\Bbb Z\rtimes\frac{\Bbb Z}{2\Bbb Z}\;\;\;\;\;\;\;\;\;\Psi:((f_2\circ f_1)^j,f_2^k)\mapsto (j,\mathbf k)\;\;\;\;$$ is obviously a bijection where $j\cdot(-1)^k$ is the action of the residue class $k+2\Bbb Z$ on the integer $j$ while$$f_2^k\circ(f_2\circ f_1)^j\circ f_2^k=(f_2\circ f_1)^{j\cdot(-1)^k}$$ is the action of $f_2^k$ on $(f_2\circ f_1)^j$ for any $j,k\in\Bbb Z$. Consequently,
$$\begin{align} \Psi(((f_2\circ f_1)^j,f_2^k)\cdot((f_2\circ f_1)^s,f_2^t))&=\Psi((f_2\circ f_1)^{j+s\cdot(-1)^k},f_2^{k+t}))\\ &=(j+s\cdot(-1)^k,k+t+2\Bbb Z)\\ &=(j,k+2\Bbb Z)\cdot(s,t+\Bbb Z)\\ &=\Psi(((f_2\circ f_1)^j,f_2^k))\cdot\Psi((f_2\circ f_1)^s,f_2^t)) \end{align}$$ $$\therefore\;\langle f_2\circ f_1\rangle\rtimes \langle f_2\rangle\approx\Bbb Z\rtimes\frac{\Bbb Z}{2\Bbb Z}$$