Assuming that I know that:
1) A plane isometry is either a rotation, a traslation, a reflection, or a glide reflection
2) Every isometry can be expressed as a product of reflections,
3) Rotations and translations are pair isometries (in the sense that the number of reflections at which they can be expressed is pair),
4) A a rotation is product of two reflections through two lines that intersect and a traslation is a product of two reflextions through two parallel lines
what's is wrong in the following proof of the question:
Given $\tau=\sigma_{a}\sigma_{b}$ where $a || b$, and $\rho=\sigma_{a'}\sigma_{b'}$ where $a'$ and $b'$ are not parallel, then $\tau\rho=\sigma_{b}\sigma_{a}\sigma_{a'}\sigma_{b'}$ is a pair isometry, so it can only be a rotation or a translation. But in order it to be a translation, $a,b,a',b'$ must be parallel, and we know that $a'$ and $b'$ are not parallel, so the product is not a translation, so it must be a rotation.
$\endgroup$ 72 Answers
$\begingroup$A translation $T$ of vector $\vec v$ is the product of the reflections about ANY two parallel lines, perpendicular to $\vec v$ and with distance $|\vec v|/2$ between them. A rotation $R$ of center $O$ and angle $\theta$ is the product of the reflections about ANY two lines, intersecting at $O$ and forming an angle $\theta/2$ between them.
If $r$ is the line passing through $O$ and perpendicular to $\vec v$, there exist then two other lines, $s$ (parallel to $r$) and $t$ (passing through $O$), such that $T=\sigma_r\circ\sigma_s$ and $R=\sigma_t\circ\sigma_r$. It follows that $$ R\circ T=\sigma_t\circ\sigma_r\circ\sigma_r\circ\sigma_s=\sigma_t\circ\sigma_s. $$ But $t$ and $s$ are not parallel and form the same angle as $t$ and $r$, hence this is a rotation $R'$ of angle $\theta$. The center $O'$ of this rotation is located at the intersection of $s$ and $t$.
$\endgroup$ $\begingroup$You reasoning is correct. I just dislike the sentence: "But in order it to be a translation, $a,b,a′,b′$ must be parallel. Actually just $a',b'$ must be parallel and they can interesct $a,b$, lines that are already parallel.
$\endgroup$ 2
