Prove that the map $f: M \rightarrow \{1,r\}$ defined by $t_a \rho_{\theta} \mapsto 1$, $t_a \rho_{\theta}r \mapsto r$ is a homomorphism. M denotes the set of isometries of the plane; r the reflexion about the x-axis; $t_a$ the translation of vector $a$ and $\rho_{\theta}$ the rotation of angle $\theta$ with respect to the x-axis.
What I did so far: Let $m,m' \in M$ such that $\begin{cases}m=t_a \rho_{\theta}\\ m'=t_b \rho_{\phi} \end{cases}$
$f(mm')=f(t_a \rho_{\theta}t_b \rho_{\phi})=f(t_a t_{\rho_{\theta}(b)}\rho_{\theta}\rho_{\phi})=f(t_{a+{\rho_{\theta}(b)}}\rho_{{\theta}+{\phi}})=1 $
and $f(m)f(m')=f(t_a \rho_{\theta})f(t_b \rho_{\phi})=1\times 1=1$
Therefore the first map is an homomorphism.
I was planning on proceeding the same way for the second map just wanted to make sure wanted I did for the first one was correct.