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ganeshie8

  • one year ago

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 \(R_{180}\)

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  1. zzr0ck3r
    • one year ago
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    https://ysharifi.wordpress.com/2011/02/02/center-of-dihedral-groups/

  2. zzr0ck3r
    • one year ago
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    I can help with any questions.

  3. zzr0ck3r
    • one year ago
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    if needed....

  4. ganeshie8
    • one year ago
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    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 ?

  5. ganeshie8
    • one year ago
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    I believe reflections have order \(2\) as two reflections does nothing and rotations have order \(n\) as \(n*\dfrac{360}{n}=360\)

  6. zzr0ck3r
    • one year ago
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    often it is the defining relation.

  7. ganeshie8
    • one year ago
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    so in that proof, i assume \(a\) = reflection \(b\) = rotation

  8. zzr0ck3r
    • one year ago
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    if it is not, you can get there from the rules of the group.

  9. zzr0ck3r
    • one year ago
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    yeah, and order is |2n| (some people say D_4 had 8 elements) booo

  10. ganeshie8
    • one year ago
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    \(ba=ab^{-1}\) works only for dihedral groups right ? it shouldn't work for groups in general.. ?

  11. zzr0ck3r
    • one year ago
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    correct

  12. zzr0ck3r
    • one year ago
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    well, it might work for other, but it defined the dihedrals

  13. zzr0ck3r
    • one year ago
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    you can also use \(b^{-1}a=ab\)

  14. ganeshie8
    • one year ago
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    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.. .

  15. zzr0ck3r
    • one year ago
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    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.

  16. ganeshie8
    • one year ago
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    sure we can think of rotation and reflection as functions right

  17. zzr0ck3r
    • one year ago
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    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

  18. zzr0ck3r
    • one year ago
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    yeah so its aa bijection in two space where every pair of points maintains distance

  19. zzr0ck3r
    • one year ago
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    and something way of saying that it stays in the same spot.

  20. zzr0ck3r
    • one year ago
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    sorry it is late :)

  21. ganeshie8
    • one year ago
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    I'm liking group theory but also feeling overwhelmed with all the different terminology and new stuff..

  22. ganeshie8
    • one year ago
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    I think I understood the proof, thanks again! gnite !

  23. zzr0ck3r
    • one year ago
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    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:)

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