This question has an answer elsewhere, a conjugate of a glide reflection by any isometry of the plane is again a glide reflection, but they use results which were not mentioned in the book. The question comes from Artin's Algebra 6.3:

"Prove that a conjugate of a glide reflection is a glide reflection, and that the glide vectors have the same length."

The answers given use the fact that 3 reflections lead to a glide. Assuming I can't use this result, what to do?

Here's some notation and rules given:
$t_a$: translation by a vector a
$p_\theta:$ rotation around the origin by angle $\theta$
r: reflection about the $e_1- axis$

$p_{\theta}t_v = t_{p_{\theta}(v)}p_{\theta}$
$rt_v = t_{r(v)}r$
$rp_{\theta}= p_{-\theta}r$

All isometries are of the form $t_{v}p_{\theta}$ (orientation preserving) or $t_{v}p_{\theta}r$ (orientation reversing). I tried conjugating a glide, $g = t_{a}p_{\theta}r$, with a pure translation $t_{b}$ and I got:
$$t_{b}({t_{a}p_{\theta}r})t_{-b} = t_{b+a}p_{\theta}rt_{-b}=t_{b+a}t_{p_{\theta}(r(-b))}p_{\theta}r = t_{b+a+p_{\theta}(r(-b))}p_{\theta}r$$
The last expression is a glide (it's of the right form), but I don't see how the "glide vectors" have the same length. Any ideas?

  • $\begingroup$ "But they use results which were not mentioned in the book". I think that the answer at the duplicate is very clear and does not assume more theory than what is necessary. What exactly do you mean? $\endgroup$ – Dietrich Burde Jan 29 at 9:42
  • $\begingroup$ @ Dietrich burde like I said in the question, they use the fact that a glide is 3 reflections, or that reflections generate the entire group, neither of which is discussed in the book. $\endgroup$ – user35687 Jan 29 at 13:29

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