2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ It's not true. In an abelian group the commutator subgroup is just the identity element, which forms a one-element abelian group. $\endgroup$ – vadim123 Aug 10 '16 at 18:46
  • $\begingroup$ Gotcha, are there nontrivial examples? $\endgroup$ – user308716 Aug 10 '16 at 18:47
  • $\begingroup$ 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 $\endgroup$ – peter a g Aug 10 '16 at 18:48
  • $\begingroup$ Yes. The commutator of $A_4$ is the Klein 4 group, which is abelian. See the examples here. $\endgroup$ – vadim123 Aug 10 '16 at 18:50
  • 2
    $\begingroup$ 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 $a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}$. $\endgroup$ – PseudoNeo Aug 10 '16 at 18:58
1
$\begingroup$

A counterexample of this assertion is a non commutative nilpotent group of class 2.

http://groupprops.subwiki.org/wiki/Group_of_nilpotency_class_two

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Not true. Take $G=S_3$: $G'=A_3$, cyclic of order $3$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The class of counterexamples is precisely the class of metabelian groups: those groups $G$ for which $G'$ is abelian.

Examples abound- see the wikipedia page.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Take $G$ any abelian group, then the commutator subgroup is a subgroup of an abelian group so it is abelian.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.