Q : Prove that a conjugate of a glide reflection by any isometry of the plane is again a glide reflection, and show that the glide vectors have the same length.
I know that:
the conjugate of a reflection by a translation (or by any isometry, for that matter) is another reflection, by some explicitly calculation (haven't try).
And 3 reflections leads to glide.
So I was thinking write a glide as 3 reflections and using the same process as proving 1 to prove the statement here. Is this promising? Can someone tell me? Or, should I try other methods?
Feel like this question is worth thinking. Thanks in advance~
Best Answer
Since the Euclidean group is generated by reflections, it suffices to show that the conjugation of a glide reflection by a reflection is a glide reflection.
Let $\ell_{1}$ and $\ell_{2}$ be parallel lines, $\ell_{3}$ a line perpendicular to $\ell_{1}$, and $R_{i}$ the reflection across $\ell_{i}$, so that $$ G = R_{1} R_{2} R_{3} $$ is a glide reflection. (Conversely, every glide reflection can be represented in this way.)
If $R$ is an arbitrary reflection, then:
$R^{-1} = R$;
$RR_{i}R^{-1} = RR_{i}R$ is the reflection across the line $R\ell_{i}$. Denote this reflection by $R_{i}'$.
The lines $R\ell_{1}$ and $R\ell_{2}$ are parallel, and separated by the same distance as $\ell_{1}$ and $\ell_{2}$. The lines $R\ell_{1}$ and $R\ell_{3}$ are perpendicular.
Consequently, $$ RGR^{-1} = R(R_{1} R_{2} R_{3})R = (RR_{1}R)(RR_{2}R)(RR_{3}R) = R_{1}' R_{2}' R_{3}' $$ is a glide reflection having glide vector of the same length as $G$.