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I'm self studying with Artins Algebra and reached chapter 5, section 2. There, he defines four types of symmetry (orientation-preserving: rotation and translation; orientation-reversing: reflection and glide-reflection) and goes on to show that they can all be represented as the composition of a reflection on the x1-axis (if it's orientation-reversing), a rotation around the origin and a translation. In one of the exercises, you are then asked to "prove that a conjugate of a reflection or a glide-reflection is a motion of the same type", but I don't understand how I can tell reflection and glide reflection apart on an algebraic level since a reflection around a line that doesn't intersect the origin also includes translations, and I haven't found anything in the chapter about it.

Edit: using $f(f(x))=x$ for a reflection $f(x)=t_a(\rho_\theta(r(x)))$, I got the result that, for a translation vector $a$ and a rotation angle $\theta$, we would have

$$\begin{bmatrix}a_1\\a_2 \end{bmatrix}=\begin{bmatrix}-\cos(\theta)a_1-\sin(\theta)a_2\\ -\sin(\theta)a_1+\cos(\theta)a_2 \end{bmatrix}$$ if and only if $f$ is a reflection. Is this correct?

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    $\begingroup$ A reflection $f$ will satisfy $f(f(x)) = x$. For a glide reflection, $f(f(x))$ is a translation. $\endgroup$ Aug 15 '20 at 9:24
  • $\begingroup$ Ah okay that helps a lot. Thanks! $\endgroup$ Aug 15 '20 at 9:29
  • $\begingroup$ I edited my post with what I got as an answer for an equation a reflection has to satisfy. Is this correct? (Also sorry if this breaks any etiquette I'm not aware of) $\endgroup$ Aug 15 '20 at 9:56
  • $\begingroup$ I don't understand your notation. What are $\rho_{\theta}$ and $r(x)$? $\endgroup$ Aug 15 '20 at 10:14
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    $\begingroup$ Note that there is a proof strategy that avoids these explicit coordinates: first, prove that the conugate of a translation is also a translation. Then, using the fact that $f$ is a glide reflection iff $f^2$ is a translation, we see that any element $gfg^{-1}$ conjugate to $f$ satisfies $$ (gfg^{-1})^{2} = gf^{2}g^{-1}, $$ which means that $(gfg^{-1})^{2}$ is the conjugate of a translation, which means that $(gfg^{-1})^{2}$ is a translation, which means that $gfg^{-1}$ is a glide reflection. $\endgroup$ Aug 15 '20 at 10:32
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Given that $f(x) = t_a\circ \rho_\theta\circ r$, we see that $f$ is a reflection if and only if $f(f(0)) = 0$. We note that $$ \begin{align} f(f(0)) & = f(a) = a + \rho_\theta(r(a)) = a + \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta } \pmatrix{a_1\\ -a_2} \\ & = \pmatrix{a_1\\a_2} + \pmatrix{\cos \theta a_1 + \sin \theta a_2\\ \sin \theta a_1 - \cos \theta a_2}. \end{align} $$ So indeed, your statement is correct.

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