# Thread: drivation of Q8 and D8=D2.4

1. ## drivation of Q8 and D8=D2.4

Hi.
please help me , I need drivation of Q8 and derivation of D8 ( 8 element ).

2. ## Re: drivation of Q8 and D8=D2.4

what do you mean by "derivation"? the derived series? the way they are defined?

3. ## Re: drivation of Q8 and D8=D2.4

Originally Posted by Deveno
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 >

4. ## 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).