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Reviewing the book "An Introduction to the Group Theory" By J.J.Rotman, I have found out that I have neglected the way $G'$, the commutator subgroup, becomes a subgroup of a given group $G$.

Unfortunately, I thought; under group operation of $G$: $$[a,b]\ast_G[c,d];\ a,b,c,d\in G$$ should be written as an element of $G'$. But according to problem 2.42:$$[a,b][c,d]=(aba^{-1}b^{-1})(cdc^{-1}d^{-1})=a(ba^{-1})b^{-1}c(dc^{-1})d^{-1}a^{-1}(ab^{-1})bc^{-1}(cd^{-1})d$$ which means that the product of two (or more) commutators need not be a commutator as well. So, how can I explain for a student that $G'$ is a subgroup of $G$ when he doesn't know a free group. In fact, should I only tell him to accept that $G’$ contains some forms like $$x_1*x_2*x_3*\ldots *x_n$$ , $n \geq1$ and each $x_i$ is a commutator in $G$? And tell him some basic facts about free groups and forms of combining their elements? Is there any simple example which I can use for a student, who learns basics of the Group theory? Thanks for any hints.

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The difficulty is present for any group given by generators, so I don't think $G'$ should be especially troublesome. $G'$ is the smallest subgroup containing the commutators. It need not be equal to the set of commutators. – Jack Schmidt Sep 4 '12 at 19:59
@JackSchmidt: Yes I see in exercise 2.43 that P.J.Cassidy gave a example about set of all commutators in a group need not be as subgroup. – Babak S. Sep 4 '12 at 20:17
up vote 6 down vote accepted

The set of commutators of a group is, in general, not a subgroup.

$G'$ is defined as the subgroup of $G$ generated by commutators, and then it is obviously a subgroup!

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See… for examples. – Mariano Suárez-Alvarez Sep 4 '12 at 20:00
Thanks. The link gave me more than I had expeted. – Babak S. Sep 4 '12 at 20:12
A very nice exposition is here:… and of course Isaacs's paper: – user641 Sep 4 '12 at 20:21
Mentioned implicitlym see also On commutators in groups, by Luise-Charlotte Kappe and Robert Fitzgerald Morse, – Nicky Hekster Sep 5 '12 at 22:13

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