# prove that a conjugate of a glide reflection is a glide reflection

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?

• "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? – Dietrich Burde Jan 29 at 9:42
• @ 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. – user35687 Jan 29 at 13:29