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 October 8th, 2009, 03:22 PM #1 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Is the set of all glide-reflections a group? Hi, Is the set of all glide-reflections a group? (Note: the group was never defined in the question) I assumed that the group consisted of all isometries. This was a question on a quiz I had two weeks ago. I said it was a group. Apparently I got it wrong though, and I wanted a clarification as to why I got it wrong. Or if my TA made a mistake and I got it right. I wanted to know if the 4 axioms that define a group (identity, inverse, closed under product, and associativity) are met for the set consisting of all glide-reflections. I can see how inverse and closed under multiplication are met (assuming the group consists of all isometries), but is there an identity and is it associative (remember the set consists of only all glide-reflections and I assumed the group consisted of all isometries)? Side Note: In our book, it says, "The set T of all transformations of the plane, along with transformation multiplication, is a group.' It goes on to cite 4 theorems (in relation to the 4 group axioms). I used this as the backbone to me showing the 4 axioms and why I thought the set of all glide reflections is a group. Any clarification is greatly appreciated!
 October 8th, 2009, 05:47 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Is the set of all glide-reflections a group? I assume the set in question is that of all ordered pairs (vector, theta) which sends a point (x, y) to (x, y) + vector then flips it about theta. With that interpretation I would say "not a group" because the identity isn't a member. As the identity is the product of two members, it's not even closed under the operation...
October 8th, 2009, 06:54 PM   #3
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Re: Is the set of all glide-reflections a group?

Quote:
 Originally Posted by CRGreathouse I assume the set in question is that of all ordered pairs (vector, theta) which sends a point (x, y) to (x, y) + vector then flips it about theta. With that interpretation I would say "not a group" because the identity isn't a member. As the identity is the product of two members, it's not even closed under the operation...
i'm not too sure that's the correct set. the set in question is all glide reflections. glide reflections are a product of 3 reflections (or 1 reflection and 1 translation (which is, itself, 2 reflections)). think of footprints that show someone's path of walking.

October 8th, 2009, 07:36 PM   #4
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Re: Is the set of all glide-reflections a group?

Quote:
 Originally Posted by envision (or 1 reflection and 1 translation (which is, itself, 2 reflections))
That's precisely what I was assuming.

If you reflect a "4", there's no way to translate it to make it a "4" again. A second reflection will put it back, though.

October 8th, 2009, 08:48 PM   #5
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Re: Is the set of all glide-reflections a group?

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by envision (or 1 reflection and 1 translation (which is, itself, 2 reflections))
That's precisely what I was assuming.

If you reflect a "4", there's no way to translate it to make it a "4" again. A second reflection will put it back, though.
this is what i am seeing to take the "4" back to it's original position. tell me if i am making a mistake.

let element a be the glide reflection (composed of three reflections) that takes a figure somewhere.
let element b be the glide reflection (composed of the same three reflections, but in the opposite order) that takes the figure back to it's original position (b=a^-1).

is that possible? does that show inverse?

the only thing i can't think of is an identity element (is that what you are trying to show me?), and i don't know if associativity holds or not.

 October 8th, 2009, 09:45 PM #6 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Is the set of all glide-reflections a group? I think, the set of all glide reflections is a subset of all isometries. I believe the reflection is forced-- there needs to be one. So you do not have closure, as the composition of two will be a translation. Otherwise, an isometry is a glide reflection. Anyway, you were onto the right idea, because the set of glide reflections generates the set of isometries. (This is my basis fol assuming the issue is one of closure) Edit: Definitely an algebra problem. Moved there.
 October 9th, 2009, 03:37 AM #7 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Is the set of all glide-reflections a group? I'm merging this withe the thread I moved to the algebra section. Cheers.
 October 9th, 2009, 05:06 AM #8 Senior Member   Joined: Jun 2009 Posts: 150 Thanks: 0 Re: Is the set of all glide-reflections a group? the group of all isometries of the plane... This is made up of 4 types of maps: translations, rotations, reflections, glide-reflections The set of translations is a group. The set of translations and rotations is a group. All four together form a group. But the composition of two glide-reflections is in general not a glide-reflection, so by themselves the glide-reflections do not form a group.

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