The derived series of a group is constructed iteratively, taking repeated commutator subgroups. A commutator subgroup is famously not only the set of commutators but the group they generate.

This raises the question of whether the second derived group is generated by commutators of commutators, as opposed to commutators of products of commutators. And so on iteratively.

Now the identity $$[x, zy] = [x, y] [x, z] [[x, z], y]$$ implies that commutators of products can be expressed as products of commutators. Unless I'm mistaken, a little thought now allows one to infer that you can move all products to the outermost level of commutator brackets, and conclude that in fact the $n$th derived group is generated by $n$th nested commutators.

But I find this argument a little messy and unsatisfying. Is there a nicer argument - and preferably an intuitive one, rather than one relying on non-obvious theorems? Thanks!

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    $\begingroup$ You can do better than the identity you write: $[x,yz]=[x,z][x,y]^z=[x,z][x^z,y^z]$ and $[xz,y]=[x,y]^z[z,y]=[x^z,y^z][z,y]$, so a commutator of a product is a product of commutators “at the same level”. $\endgroup$ Apr 11, 2021 at 18:41
  • $\begingroup$ @Arturo It took me a minute there, but yes, I see the argument you're making - conjugates of commutators at solved level are inherently commutators at the same level ($[x, y]^z=[x^z, y^z]$ applied repeatedly), so using the form of the identity which involves conjugates is preferable. Good. $\endgroup$ Apr 11, 2021 at 20:35

1 Answer 1


By the way, you are right to have some qualms, since in general, $[X,Y,Z]=[[X,Y],Z]$ is not generated by elements of the form $[[x,y],z]$ with $x\in X$, $y\in Y$, $z\in Z$, even though $[X,Y]$ is generated by elements $[x,y]$ with $x\in X$ and $y\in Y$. So there is something a bit special about the sets involved here.

I think the key is something like this:

Theorem. If $G=\langle Y\rangle$ and $N=\langle X\rangle^G$ for subsets $X$ and $Y$ of $G$, then $[N,G]=[X,Y]^G$.

(This is Proposition 1.2.8 in Susan McKay’s Finite $p$-groups, Queen Mary Maths Notes 18)

Here, $H^G$ for subgroup $H$ means the normal closure of $H$ in $G$, and $[X,Y] = \langle [x,y]\mid x\in X, y\in Y\rangle$.

Now, taking $N=G=X=Y$ gives that $[G,G]$ is the normal closure of the subgroup generated by the commutators; but since $[x,y]^g = [x^g,y^g]$, this subgroup is already normal.

Now, we can replace $G$ with $G’$, $N$ with $G’$, and let $X=Y=\{[x,y]\mid x,y\in G\}$. We note that $X$ and $Y$ are closed under conjugation. Then $[G’,G’] = [X,Y]^G = [X,Y]$ is generated by elements of the form $\bigl[ [x,y],[z,w]\bigr]$.

Then we apply the theorem taking $G$ and $N$ to be $G’’$, and taking $X$ and $Y$ to be the set of commutators-of-commutators, $[[x,y],[z,w]]$. Etc.

To prove the theorem: let’s state the commutator identities: $$\begin{align*} [y,x]^{-1} &= [x,y] \\ [xy,z] &= [x,z]^y[y,z]\\ [x,yz] &= [x,z][x,y]^z\\ [x,y^{-1}] &= [y,x]^{y^{-1}}\\ [x^{-1},y] &= [y,x]^{x^{-1}} \end{align*}$$ My commutator convention is that $[x,y]=x^{-1}y^{-1}xy$; if your convention is $[x,y]=xyx^{-1}y^{-1}$ you need to tweak the identities above.

Now, clearly each element of the generating set of $[X,Y]$ lies in $[N,G]$, which is normal, so $[X,Y]^G$ is contained in $[N,G]$. Thanks to the commutator identities, every element of $[N,G]$ is a product of conjugates of powers of elements of the form $[x^z,y]$ with $x\in X$, $y\in Y$, and $z\in G$. Since $[x^z,y] = [x,y^{z^{-1}}]^z$ and $y^{z^{-1}}\in\langle Y\rangle$, the commutator identities show that $[N,G]\subseteq [X,Y]^G$, giving the desired equality.

  • $\begingroup$ Okay, nice. I think for my desire for a more direct argument it's helpful to state and prove a special case of the theorem, i.e. if the sets $X, Y$ are closed under conjugation, then the subgroups their commutator generates is the commutator of the subgroups they generate, $[\left<X\right>, \left<Y\right>]=\left<[X, Y] \right>$. As you brought up previously in the comments, this stresses the property of the generator set being closed under conjugation. Does that seem like a good idea to you? $\endgroup$ Apr 11, 2021 at 21:41
  • $\begingroup$ @Sharkos: Ehr, by definition, $[X,Y]$ is already a subgroup, even when $X$ and $Y$ are just subsets. I think you mean normal closure: if $X$ and $Y$ are closed under conjugation, then $[X,Y]=[X,Y]^G$. $\endgroup$ Apr 11, 2021 at 21:45

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