# Determining commutator subgroups

Here's the full question:

Determine the commutator subgroups of the following groups:

a) $SO_2$

b) $O_2$

c) the group $M$ of isometries of the plane

d) $S_n$

e) $SO_3$

f) the group $G$ of $3 \times 3$ upper triangle matrices with 1's along the diagonal over the prime field $F_p$

I have little to no intuition for commutator subgroups. I imagine I need to find the generating set for each, but I suppose I can't just say "the set of all commutators". Then they want something explicit. But then where do you start? I was thinking I could start with the commutators of generators of each group. Would that be the right track, and where should I go from there if it is?

Any help much appreciated.

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@DonAntonio I just saw that too... I just registered with an email so I guess I neglected that. Going through old questions now, thanks for the note. – Benjamin Lu Nov 19 '12 at 4:19
Anyway: sometimes it is necessary to "do some dirty work", so I'd advice you to explicitly evaluate the commutator of two elements in each case and try to find out some pattern. In some cases you'll get a pretty easy result (e.g., $\,\mathcal O_2\,$ is abelian , so its commutator subgroup is...), in others some apparent pattern appears (triangular upper matrices, say). – DonAntonio Nov 19 '12 at 4:22
@DonAntonio Thanks. I will follow your advice. This business just seems like walking in the dark. – Benjamin Lu Nov 19 '12 at 4:32
@Benjamin: and why is that a bad thing? If you already knew what you were going to find then you wouldn't be learning anything. – Qiaochu Yuan Nov 19 '12 at 5:05
By the way, if you just answered "the set of all commutators", then it would not necessarily be correct. The commutator subgroup is the subgroup generated by all commutators, and sometimes contains elements that are not themselves commutators. – Derek Holt Nov 19 '12 at 13:23

c) Note that the commutator of two isometries is always orientation preserving because the commutator always consists of an even number of orientation-inverting factors. Thus $[M,M]$ can contain only rotations and translations. In fact $[M,M]$ contains at least all rotations and all translations: The rotation by half the angle or the translation by half the distance can be written as product of two reflections; the comuutator of these reflections is then the given arbitrary rotation or translation (because the reflections are their own inverse).
d) Note that $[a,b]$ is always an even permutation, i.e. $[S_n,S_n]\le A_n$. Do you know a set of generators for $A_n$? COuld you maybe write these as commutators?