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I read this recent question on proving that the identity is never the product of an odd number of reflections. After googling a bit, I found a Lemma in J. Aarts Plane and Solid Geometry, but I didn't fully understand the proof. I'll post the relevant sections, and the google book can be found here.

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The Lemma that answers the linked question can be found on the google book, but I think I will understand it if I understand the proof of this above theorem.

I have two questions.

  1. Why is it the case that if the three reflection axes are concurrent or all parallel does it follow that $\mathcal{F}$ is a reflection? For instance, I see in the case that if the axes are all parallel, then lines perpendicular to them will be invariant, but I don't necessarily see why there should be just one new axis of reflection that can replace the other three. I also played around with reflecting points across three concurrent axes, but didn't see why there should be one new reflection axis that "fits all" in a sense.

  2. I'm uncomfortable with the line 'We can find $l_1^\prime$ and $l_2^\prime$ such that $S_{l_2^\prime}\circ S_{l_1^\prime}=S_{l_2}\circ S_{l_1}$ and $l^\prime_2$ is perpendicular to $l_3$. This seems like a lot to demand of these two lines, why should we be sure such lines exist?

Thanks for taking the time to read, it's been bugging me since I googled it when the other question was first posted.

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A composition of two reflections whose axis are parallel is a translation. Since translations commute with each other, this implies that if you conjuguate the reflections with a translation, their composition will be the same translation as before. This means that in the case where you are composing reflections with parallel axis, you can translate those two axis by the same vector without changing the result.

A composition of two reflections whose axis intersect at $P$ is a rotation of center $P$. Since rotations of center $P$ commute with each other, this again implies that you can rotate the axis of your reflections around $P$ both by the same angle without changing the result.

These two things answer your questions :

  1. if the 3 axis are parallel, translate the first two so that the axis of the 2nd reflection coincides with the axis of the 3rd. Then, since you are composing two reflections with the same axis, they cancel each other, and you are left with only one reflection.

  2. you simply need to find the rotation around $A$ which will bring $l_2$ perpendicular to $l_3$, and rotate the two axis $l_1$ and $l_2$ accordingly. Then you rotate $l'_2$ and $l_3$ around $C$ until $l'_3$ and $l'_1$ are parallel.

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Thanks for your clear answer chandok, much appreciated. –  yunone Mar 23 '11 at 8:30
    
Hi @chandok, it's been a while since I posted this, but I had a question I over looked when I was less familiar. Suppose I have a rotation $S_2S_1$ given by two reflections $S_1$ and $S_2$ around a center $P$. So you say I can rotate this axes around $P$ by the same angle without changing the result. Suppose this new angle is applied by some rotation $R$. Then rotating both axes would be a new motion $RS_2RS_1$? How does $RS_2RS_1=S_2S_1$? I could commute to get $RRS_2S_1$, but I don't believe $R^2=\text{id}$. What am I doing wrong? Thanks. –  yunone Apr 7 '11 at 3:43
    
@yunone : No, the new reflecction is not simply $RS_1$. Suppose the axis of $S_1$ is $D_1$, so $S'_1$ is the reflexion by $R(D_1)$. If you have a point $x$, and call $y$ its reflection by $R(D_1)$. Call $x' = R^{-1}(x)$ and $y' = R^{-1}(y)$. You will find that $y'$ is the reflected of $x'$ by $D_1$ : $y' = S_1(x')$. This means that $y = RS_1R^{-1}(x)$, so the "rotated" reflexion is $RS_1R^{-1}$, it is the original reflexion conjugated by the rotation you apply (and not simply composed). So you get $RS_2R^{-1}RS_1R^{-1} = R(S_2S_1)R^{-1} = RR^{-1}(S_2S_1) = S_2S_1$ –  mercio Apr 7 '11 at 8:06
    
thanks, I would upvote twice if I could. –  yunone Apr 7 '11 at 8:36
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