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

Results 1 to 4 of 4

- August 30th 2012, 05:28 AM #1

- August 30th 2012, 06:30 PM #2

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,463
- Thanks
- 795

- August 30th 2012, 09:59 PM #3

- August 31st 2012, 05:01 AM #4

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,463
- Thanks
- 795

## 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 r^{k}, and reflections r^{k}s.

since the rotations all commute with each other, [r^{k},r^{m}] = e.

[r^{k},r^{m}s] = r^{k}r^{m}sr^{-k}r^{m}s (since reflections are their own inverses)

= r^{k+m}sr^{m-k}s = r^{k+m}r^{k-m}s^{2}= (r^{k})^{2}.

thus the commutator of a rotation and a reflection is a rotation that is a square, so either e, or r^{2}.

[r^{k}s,r^{m}] = r^{k}sr^{m}r^{k}sr^{-m}

= r^{k-m}sr^{k+m}s = r^{k-m}r^{-k-m}s^{2}= (r^{-m})^{2},

so in this case, too, the only commutators are {e,r^{2}}.

finally, we consider the commutators of two reflections:

[r^{k}s,r^{m}s] = r^{k}sr^{m}sr^{k}sr^{m}s

= r^{k-m}s^{2}r^{k-m}s^{2}= (r^{k-m})^{2}.

so the only commutators are {e,r^{2}}, and <e,r^{2}> = <r^{2}> = {e,r^{2}}

so [D4,D4] = Z(D4) = {e,r^{2}}.

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

[i,j] = ij(-i)(-j) = (ij)^{2}= k^{2}= -1

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

[j,k] = jk(-j)(-k) = i^{2}= -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).