# Proof that commutator subgroups are always non-abelian?

I can't seem to come up with a proof that the commutator subgroup is necessarily non abelian, is this even true?

Let $aba^{-1}b^{-1}$ and $cdc^{-1}d^{-1}$ be elements of the commutator subgroup $C$ of a group $G$. I want to show that $aba^{-1}b^{-1}cdc^{-1}d^{-1} \neq cdc^{-1}d^{-1}aba^{-1}b^{-1}$. I don't see why this couldn't be equality? Where am I going wrong?

• It's not true. In an abelian group the commutator subgroup is just the identity element, which forms a one-element abelian group. – vadim123 Aug 10 '16 at 18:46
• Gotcha, are there nontrivial examples? – user308716 Aug 10 '16 at 18:47
• Vadim's example is better than mine! but try $S_3$, which is non abelian, has an abelian non-trivial quotient, and all of its proper non-trivial subgroups are abelian – peter a g Aug 10 '16 at 18:48
• Yes. The commutator of $A_4$ is the Klein 4 group, which is abelian. See the examples here. – vadim123 Aug 10 '16 at 18:50
• The Heisenberg group is another (interesting) counterexample → en.wikipedia.org/wiki/Heisenberg_group. By the way, please note that you cannot write a general element of the commutator subgroup as a commutator $aba^{-1}b^{-1}$. The commutator subgroup is only generated by commutators, so a general element in it is only a finite product of commutators $a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}$. – PseudoNeo Aug 10 '16 at 18:58

Not true. Take $G=S_3$: $G'=A_3$, cyclic of order $3$.
The class of counterexamples is precisely the class of metabelian groups: those groups $G$ for which $G'$ is abelian.
Take $G$ any abelian group, then the commutator subgroup is a subgroup of an abelian group so it is abelian.