I have been asked to prove that, if I have a movement $f$ in the plane that can be written as the product of three reflections , then it can be written as the product of three reflections, where one of their axis is perpendicular to the others.
I thought that , if $ L_f = \emptyset $ ( $ L_f $ being the set of fixed points by $f$) then I can take two reflections with their axis paralel to each other , and then another one perpendicular to them, and if $ L_f = \{P\}$ for some point $P$ , then I can take two of the axis to be the same , and the other one perpendicular to them.
I'm not sure that this proves it. If it isn't right , how could I prove it?
Let $a$, $b$, $c$ three given lines and $P_c\circ P_b\circ P_a$ the given isometry.
• If $a$ and $b$ intersect at $O$, then $R=P_b\circ P_a$ is a rotation of center $O$ and angle equal to twice the angle $\angle{ab}$. If $b'$ is the line through $O$ perpendicular to $c$ and $a'$ is a line through $O$ such that $\angle{a'b'}=\angle{ab}$, then $$ R=P_{b'}\circ P_{a'}\quad\text{and}\quad P_c\circ P_b\circ P_a=P_c\circ P_{b'}\circ P_{a'}. $$ But $c$ and $b'$ intersect at a certain point $O'$, hence we can repeat the above argument and $$ P_c\circ P_{b'}\circ P_{a'}=P_{c'}\circ P_{b''}\circ P_{a'}, $$ where $b''$ passes through $O'$ and is perpendicular to $a'$, while $c'$ passes through $O'$ and $\angle{c'b''}=\angle{cb'}=90°$. Notice that $c'$ and $a'$ are both perpendicular to $b''$, as requested.
• If $a$ and $b$ are parallel but $c$ intersects them, one can repeat the same reasoning, starting with $R=P_c\circ P_b$.
• If $a\parallel b\parallel c$ then $P_b\circ P_a$ is a translation and one can argue as above, proving that $P_c\circ P_b\circ P_a=P_{a'}=P_{a'}P_{r}P_{r}$, where $a'$ is a suitable line and $r\perp a'$.