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In a previous problem I asked about the notation used here.. I'm still not sure how to show it even though I now get what it's asking.

The following is an exercise from The Four Pillars of Geometry.

3.6.5. Show that the reflections in lines $L$, $M$, and $N$ (in that order) have the same outcome as reflections in lines $L'$, $M'$, and $N$, where $M'$ is perpendicular to $N$

I've had to ask around to clairify parts of the problem; it should be noted that the lines $L'$ and $M'$ are essentially arbitrary with no relation to $L$ and $M$ except that they appear in the same order when you do the reflections. Also, no picture is given as the lines are general though I've drawn a few pictures to help reason it out.

My main issue might be that I'm not sure what tools to use, so please be as specific as possible in your answer.

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    What does this have to do with a glide?2011-07-24
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    I just added a link to the related question. (I was usure what the problem was stating.) There are 2 other problems, the last asks you to prove that any 3 reflections results in a glide.2011-07-24
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    $L'$ and $M'$ are not arbitrary. GIven $L, M, N$ they are to be determined so that the condition $M'$ perpendicular to N is satisfied and so that the reflections in the new lines define the same isometry as the reflections in the old ones.2011-07-24
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    I don't think that this is true in general. The simplest case for me to imagine is the following. Assume that all 6 lines intersect at point $O$. Then the composition of the two first reflections is a rotation about $O$ by an angle that's twice the angle between the two lines. Therefore the orthogonality relation $M'\perp N$ can be basically ignored, because we are free to choose $L'$. Then the third reflection makes the total composite mapping into an arbitrary reflection about a line thru $O$. So you have left out something essential. -1 for asking the same question again?2011-07-24
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    @Jyrki: It’s not really the same question. The first time was to find out exactly what the question was; now that he understands that, he’d like some help in answering it.2011-07-24
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    @Brian - it isn't exactly the same question, but it does leave a trail of references to follow, and if you track back to the original book you'll find a huge hint - the exercise is part of a sequence.2011-07-24
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    @Gerry - if you track back the references to the original book you'll see that this is one of a sequence of exercises which progress from three reflections to a glide reflection.2011-07-24
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    @Mark: I know, and so, clearly, does Tony. I was arguing against the suggested $-1$, not denying the close relationship to the earlier question.2011-07-24
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    As you see I didn't really downvote in the end. I had gotten the impression that the discussion related to a single question should be kept in one place here, but apparently I was mistaken. IOW I plead guilty to being a bit trigger-happy here. Sorry.2011-07-24
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    (see comments to answer as well) Yeah, I tried to trackback a bit to show that I did look in the book and was in fact, regrettably, still confused. :P It's possible I'm over thinking these things.2011-07-25

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I understand the question as in Mark's comment: $L$, $M$ and $N$ are given lines, and we want to show that there are lines $L'$ and $M'$ with $M'\perp N$ such that reflections in $L'$, $M'$ and $N$ (in that order) have the same effect as reflections in $L$, $M$ and $N$ (in that order).

First, since the reflection in $N$ is last in both cases, we can cancel it, so the task is to show that there are lines $L'$ and $M'$ such that reflections in $L'$ and $M'$ have the same effect as reflections in $L$ and $M$, subject to $M'\perp N$.

This is false. If $L$ and $M$ are parallel, reflections in $L$ and $M$ amount to a translation perpendicular to them by twice their distance, and this can only be reproduced by $L'$ and $M'$ if they, too, are parallel to $L$ and $M$; hence we cannot choose $M'$ perpendicular to $N$ unless $M$ already happens to be.

It's true, however, if $L$ and $M$ aren't parallel. In this case, it follows almost immediately from the hint given in the book right above the exercise:

Reflections in any two lines meeting at the same angle $\theta/2$ at the same point $P$ give the same outcome.

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    And if Tony looks carefully at the notation for the next exercise in the book (3.6.6), he'll see that he has to apply the same observation to a different pair of reflections to prove the final theorem. It's a pity the special case doesn't get picked out. The primes in the book aren't standard notation, but I think they may be supposed to give a hint that the mirror lines are being transformed in some way.2011-07-24
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    I'm learning to trackback more often, as here honestly I knew about the hint, and the next exercise, and what all three are trying to say and I still had questions!2011-07-25
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    Thanks for pointing out the exception; does this reasoning work for the case where $L$ and $M$ intersect? Choose $M'$ to be the perpendicular to $N$ through the point reflected in $L$ then $M$. Choose $L'$ so that $L'$ and $M'$ intersect so that the initial, final, and intersection define a circle - that is, initial and final points are only $\theta$ apart. Have $L'$ bisect this angle. Done.2011-07-25
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    <-- I'm curious if the previous comment worked but it doesn't use the hint. Better to say initial and final point are at angle $\theta$ with the intersection of $L$ and $M$. Place $M'$ through the intersection. Place $L'$ at angle $\theta/2$. Done with hint!2011-07-25
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    I don't understand what you're referring to by "the point reflected in $L$ then $M$" and by "the initial, final, and intersection", but it's less complicated than that -- any two lines meeting where $L$ and $M$ meet, at the same angle, give the same rotation, so just turn $L$ and $M$ together such that $M$ becomes perpendicular to $N$.2011-07-25
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    I don't understand your second comment either -- don't know what "initial and final point" refer to, or what it means for two points to be at an angle with an intersection.2011-07-25
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    I wanted "initial point" to be some point in the plane. Then "final" to be the point arrived at after the reflections. Then, I claimed that you could place the lines so that the "intersection" was the center of a circle with "final" and "initial" at some radius. The angle was between the rays coming out of the center going towards "final" and "initial" You're right it's simpler but now I'm wondering if I got it - even if there is a little more justification to be done.2011-07-25