consider a regular $2n-gon$

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 $R_{180}$

Question

consider a regular $2n-gon$

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 $R_{180}$

zzr0ck3r · Answer

https://ysharifi.wordpress.com/2011/02/02/center-of-dihedral-groups/

zzr0ck3r · Answer

I can help with any questions.

zzr0ck3r · Answer

if needed....

ganeshie8 · Answer

Thank 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 · Answer

I believe reflections have order $2$ as two reflections does nothing 

and rotations have order $n$ as $n*\dfrac{360}{n}=360$

zzr0ck3r · Answer

often it is the defining relation.

ganeshie8 · Answer

so in that proof, i assume

$a$ = reflection
$b$ = rotation

zzr0ck3r · Answer

if it is not, you can get there from the rules of the group.

zzr0ck3r · Answer

yeah, and order is |2n| (some people say D_4 had 8 elements) booo

ganeshie8 · Answer

$ba=ab^{-1}$ works only for dihedral groups right ?

it shouldn't work for groups in general.. ?

zzr0ck3r · Answer

correct

zzr0ck3r · Answer

well, it might work for other, but it defined the dihedrals

zzr0ck3r · Answer

you can also use $b^{-1}a=ab$

ganeshie8 · Answer

I 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 · Answer

A 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 · Answer

sure we can think of rotation and reflection as functions right

zzr0ck3r · Answer

This 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 · Answer

yeah so its aa bijection in two space where every pair of points maintains distance

zzr0ck3r · Answer

and something way of saying that it stays in the same spot.

zzr0ck3r · Answer

sorry it is late :)

ganeshie8 · Answer

I'm liking group theory  but also feeling overwhelmed with all the different terminology and new stuff..

ganeshie8 · Answer

I think I understood the proof, thanks again! gnite !

zzr0ck3r · Answer

Yeah 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:)