5
$\begingroup$

Let $G$ be a multiplicative group. The commutator of $a,b\in G$ is the element $aba^{-1}b^{-1}$. (Now $a,b\in G$ commute iff their commutator is the identity.) The commutator subgroup of $G$, denoted $[G,G]$, is the subgroup generated by all commutators of all elements.

Example. If $G$ is abelian, then every commutator is the identity, so the commutator subgroup $[G,G]$ is the trivial group.

(The commutator subgroup is used to construct the abelianization of $G$ as the quotient $G/[G,G]$, which I've encountered in learning homology.)

Now, an arbitrary element of $[G,G]$ need not be itself a commutator. So, is the commutator subgroup of $[G,G]$ again all of $[G,G]$? If not, is there an easy characterizations of groups $G$ for which the commutator subgroup of $[G,G]$ is again $[G,G]$?

$\endgroup$
1
9
$\begingroup$

Let $G=S_3$. Then $[G,G]=A_3$ (see answers to this question) which is Abelian so $[A_3,A_3]=\langle e\rangle\ne[G,G]$.

$\endgroup$
7
$\begingroup$

A group for which $G=G'$ is called perfect. You might want to read this. All non-abelian simple groups belong to this category.

$\endgroup$
2
  • $\begingroup$ In particular, this answer led to an answer to my question (from the Wikipedia page on perfect groups): a group $G$ satisfying $G' = G''$ is called quasi-perfect, ie. its commutator subgroup $G'$ is perfect, ie. the commutator subgroup of the commutator subgroup is the commutator subgroup. $\endgroup$ May 30 '17 at 20:24
  • 1
    $\begingroup$ Yes that is correct, well done! Glad you learned something. $\endgroup$ May 30 '17 at 21:30
5
$\begingroup$

The commutator subgroup is often called the derived subgroup and denoted $G'$. This makes it easy to write its derived subgroup as $G''$. An example of a group with $G''\ne G'$ is $G=A_4$.

If instead we consider groups of upper triangular matrices over a finite field, we can get sequences $G\supset G'\supset G''\supset\cdots$ which finally stabilise after any given finite number of steps.

$\endgroup$

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.