# Proving that something equals the commutator subgroup and conjugacy classes/normal subgroups

I've learned that the commutator subgroup is generated by the commutators. Now this says little about its elements (to me) because I don't see how they need to be commutators themselves. I'm interested in what these commutators look like in general (maybe an intuitive idea). And what some easy ways of proving that something equals a commutator subgroup.

I'm also quite confused about the difference of normal subgroups and conjugacy classes. I've read that normal subgroups are usually the 'union' of conjugacy classes, however.. Can't we construct normal subgroups then by taking individual conjugacy classes?

• "because I don't see how they need to be commutators themselves" — That's right, they need not be commutators themselves. But for any group of order less than $96$, the commutator subgroup is the same as the set of all commutators! See math.stackexchange.com/questions/7811/… and mathoverflow.net/questions/44269/…. – M. Vinay May 28 '16 at 1:19
• I'm confused by your question about normal subgroups. Yes, we can construct normal subgroups by taking appropriate conjugacy classes and finding their union — but note that if you take any arbitrary collection of conjugacy classes, their union need not be a subgroup at all! – M. Vinay May 28 '16 at 1:26
• Have you had a look at my answer? Do you need any help with it? – Gerry Myerson May 30 '16 at 13:22
• Sorry for my late reaction, I did @gerry and it's sattisfying – Bahbi May 30 '16 at 20:02
• Glad to hear it. You have the option of clicking in the check mark next to my answer to indicate that you "accept" it. – Gerry Myerson May 30 '16 at 22:45

## 1 Answer

The commutator subgroup of a group $G$ enjoys the property of being the smallest normal subgroup of $G$ with abelian quotient. This often helps when you are trying to find the commutator subgroup of a given group. For example, if you can get your hands on all the normal subgroups, you can go calculating quotient groups, starting with the smallest normal subgroup, and as soon as you hit one with abelian quotient, you win. Or, you can calculate a few commutators, work out the subgroup $H$ they generate, find the smallest normal subgroup containing $H$, and see whether $G/H$ is abelian. What exactly works best will depend on $G$.

If $N$ is normal in $G$, then $N$ must be a union of conjugacy classes. But not every union of conjugacy classes is a normal subgroup, simply because not every union of conjugacy classes is a subgroup. If a union of conjugacy classes is a subgroup, then it is a normal subgroup.