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Dihedral Group ##D_4##

  1. Oct 30, 2014 #1
    In my class, we have denoted the elements of the dihedral group ##D_4## as ##\{R_0, R_{90}, R_{180}, R_{270}, F_{\nearrow}, F_{\nwarrow}, F_{\leftrightarrow}, F_{\updownarrow} \}##. Not surprising, I was rather bewildered when I searched the internet for information on this group and encountered other denotations of the elements of ##D_4##.

    The standard denotation appears to be using ##r##'s and ##s##'s. I figured that ##r = R_{90}##, ##r^2 = R_{180}##, and ##R_{270} = r^3##. However, I was unsure of the correspondence between the other elements. Could someone possibly help me with this?
     
  2. jcsd
  3. Oct 31, 2014 #2

    jbunniii

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    Assuming ##R_{90}## denotes counterclockwise rotation by ##90## degrees, and we use the convention that ##ab## means "do ##b## first, then ##a##", you can check that
    ##R_{90} F_{\updownarrow} = F_{\nwarrow}##
    ##R_{180} F_{\updownarrow} = F_{\leftrightarrow}##
    ##R_{270} F_{\updownarrow} = F_{\nearrow}##
    so if we put ##r = R_{90}## and ##f = F_{\nwarrow}## then the group is generated by ##r## and ##f##. Moreover, ##r## has order ##4## and ##f## has order ##2##, and you can verify that ##f r f = r^{-1}##; indeed, the relations ##r^4 = f^2 = e## and ##frf = r^{-1}## suffice to define the group.
     
  4. Nov 2, 2014 #3
    Okay, but what is the correspondence between ##s## and the flips (or reflections)?
     
  5. Nov 2, 2014 #4

    jbunniii

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    You can take ##s## to be any one of the four flips/reflections, because all four of them satisfy ##s^2 = e## and ##srs = r^{-1}##. Try it with a square sheet of paper to verify this.

    To verify algebraically that it doesn't matter which flip we assign to ##s##, first choose ##s## to be one of the flips and verify geometrically that it satisfies ##s^2 = e## and ##srs = r^{-1}##. Then note that the other three flips are ##rs##, ##r^2s##, and ##r^3 s##. Then if we set ##f_k = r^k s## for ##k=0,1,2,3##, we compute
    $$f_k^2 = f_k f_k = r^k s r^k s = r^k (s r^k s) = r^k r^{-k} = e$$
    where the third inequality holds by repeatedly applying ##s r s = r^{-1}##. So ##f_k## is its own inverse. Then
    $$f_k r f_k = (r^k s) r (r^k s) = r^k (s r^{k+1} s) = r^k (r^{-(k+1)}) = r^{-1}$$
    We have therefore shown that ##f_k## satisfies ##f_k^2 = e## and ##f_k r f_k = r^{-1}## for ##k=0,1,2,3##, so any of the ##f_k## obeys the defining properties of ##s##.
     
    Last edited: Nov 2, 2014
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