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Math Help - drivation of Q8 and D8=D2.4

  1. #1
    Member vernal's Avatar
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    drivation of Q8 and D8=D2.4

    Hi.
    please help me , I need drivation of Q8 and derivation of D8 ( 8 element ).
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  2. #2
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    Re: drivation of Q8 and D8=D2.4

    what do you mean by "derivation"? the derived series? the way they are defined?
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    Member vernal's Avatar
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    Re: drivation of Q8 and D8=D2.4

    Quote Originally Posted by Deveno View Post
    what do you mean by "derivation"? the derived series? the way they are defined?
    excuse me

    [Q8,Q8]=?

    [D4,D4]=?

    [G,G]=<[x,y] | x,y \in G >
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  4. #4
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    Re: drivation of Q8 and D8=D2.4

    ok, the first group in the derived series, the commutator subgroup. let's tackle D4 first:

    elements of D4 come in two flavors, rotations rk, and reflections rks.

    since the rotations all commute with each other, [rk,rm] = e.

    [rk,rms] = rkrmsr-krms (since reflections are their own inverses)

    = rk+msrm-ks = rk+mrk-ms2 = (rk)2.

    thus the commutator of a rotation and a reflection is a rotation that is a square, so either e, or r2.

    [rks,rm] = rksrmrksr-m

    = rk-msrk+ms = rk-mr-k-ms2 = (r-m)2,

    so in this case, too, the only commutators are {e,r2}.

    finally, we consider the commutators of two reflections:

    [rks,rms] = rksrmsrksrms

    = rk-ms2rk-ms2 = (rk-m)2.

    so the only commutators are {e,r2}, and <e,r2> = <r2> = {e,r2}

    so [D4,D4] = Z(D4) = {e,r2}.

    Q8 is a little more interesting. it suffices to consider commutators involving just i,j and k:

    [i,j] = ij(-i)(-j) = (ij)2 = k2 = -1

    [i,k] = (ik)(-i)(-k) = (-j)2 = -1

    [j,k] = jk(-j)(-k) = i2 = -1

    it is easy to see that [j,i] = [i,j] (since ji(-j)(-i) = (-k)2 = -1) and similarly [k,i] = [i,k], and [k,j] = [j,k].

    furthermore [-i,j] = [i,-j] = [j,-i] = [-j,i] (since -1 is in the center), and similarly for the other combinations of i, j, k.

    (since i and -i commute, we need not check things like [i,-i], or [-j,j], etc.).

    thus, in this case, too, [Q8,Q8] = Z(Q8) = {1,-1}, although it is not usually true that [G,G] = Z(G) ([G,G] need not even be abelian, whereas Z(G) has to be).
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