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ganeshie8
 one year ago
consider a regular \(2ngon\)
show that \(R_{180}\) commutes with all other rotations and flips.
for example, below two sequence of transformations give the same final effect :
1) a. rotate 180
b. flip over the symmetry line passing through vertex A
2) a. flip over the symmetry line passing through vertex A
b. rotate 180
in other words, show that the order doesn't matter if one transformation is \(R_{180}\)
ganeshie8
 one year ago
consider a regular \(2ngon\) show that \(R_{180}\) commutes with all other rotations and flips. for example, below two sequence of transformations give the same final effect : 1) a. rotate 180 b. flip over the symmetry line passing through vertex A 2) a. flip over the symmetry line passing through vertex A b. rotate 180 in other words, show that the order doesn't matter if one transformation is \(R_{180}\)

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1https://ysharifi.wordpress.com/2011/02/02/centerofdihedralgroups/

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I can help with any questions.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Thank you! that looks like a neat proof :) Im still going through the proof... Is there any intuitive way to convince ourselves that \(ba = ab^{1}\) for dihedral groups ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I believe reflections have order \(2\) as two reflections does nothing and rotations have order \(n\) as \(n*\dfrac{360}{n}=360\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1often it is the defining relation.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so in that proof, i assume \(a\) = reflection \(b\) = rotation

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1if it is not, you can get there from the rules of the group.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1yeah, and order is 2n (some people say D_4 had 8 elements) booo

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(ba=ab^{1}\) works only for dihedral groups right ? it shouldn't work for groups in general.. ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1well, it might work for other, but it defined the dihedrals

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1you can also use \(b^{1}a=ab\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I see that rotation composed with reflection changes the orientation, so it is essentially a reflection. Since reflection is its own inverse, we have : \[(rotation)(reflection) \\= ((rotation)(reflection))^{1} \\= (reflection)^{1}(rotation)^{1} \\=(reflection)(rotation)^{1}\] that convinces me but not sure if it is a valid proof.. .

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1A valid proof (in my eyes) would need to be a proof in terms of functions, and then you must define what a symmetry is in terms of function, which is strange enough and often skipped, and essentially what you will be doing is creating group theory.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sure we can think of rotation and reflection as functions right

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1This is exactly how we start our class in group theory. 1) hand them triangles. 2) give them hints untill they can sort of define what a symmetry is. 3) let them play around with names and notation 4) give them hints until they figure out that relation hehe

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1yeah so its aa bijection in two space where every pair of points maintains distance

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1and something way of saying that it stays in the same spot.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I'm liking group theory but also feeling overwhelmed with all the different terminology and new stuff..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think I understood the proof, thanks again! gnite !

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Yeah it gets better :) So much notation and definitions. But like with the rest of this stuff, once you start to figure out why they do it to begin with, that other stuff will seem more natural:)
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