Let $G=\langle x,y,z\mid z^2=1\rangle\cong \mathbb{Z}*\mathbb{Z}*\mathbb{Z}/2$.

I am interested to compute by hand (or by any other means) the quotient groups $\gamma_n/\gamma_{n+1}$ where $\gamma_n$ is the $n$th term of the lower central series. That is $\gamma_1=G$, $\gamma_2=[G,G]$, $\gamma_3=[\gamma_2,G]$, etc.

For $\gamma_1/\gamma_2$, it is essentially the abelianization of $G$, hence it is $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/2$.

I am facing some difficulties computing $\gamma_2/\gamma_3$.

I computed that $\gamma_2=[G,G]=\langle x^{-1}y^{-1}xy,x^{-1}z^{-1}xz,y^{-1}z^{-1}yz\mid z^2=1\rangle$. (Update: This is probably wrong.)

However things start to get complicated with $\gamma_3=[\gamma_2,G]$. Is there an "easy" way to find $\gamma_2/\gamma_3$ or is brute-force the way to go?


Update: I think my expression for $\gamma_2$ may be wrong, there should be way more generators than just the 3 commutators $x^{-1}y^{-1}xy,x^{-1}z^{-1}xz,y^{-1}z^{-1}yz$, in fact $\gamma_2$ may not even be finitely generated?


$[G,G] = \gamma_2$ is not finitely generated. But it is the normal closure of the set $\{ [x,y],[x,z],[y,z]\}$ of the commutators of the three generators.

In the quotient group $\gamma_2/\gamma_3$, all conjugates of these generators have the same images. For example $[x,y]^z\gamma_3 = [x,y]\gamma_3$, because $[x,y]^{-z}[x,y] = [z,[x,y]] \in \gamma_3$.

So $\gamma_2/\gamma_3 = \langle [x,y]\gamma_3,[x,z]\gamma_3,[y,z]\gamma_3 \rangle$.

The first of these three generators has infinite order, but $z^2=1$ implies that the second and third have order $2$, so $\gamma_2/\gamma_3 \cong Z \oplus Z/(2Z) \oplus Z/(2Z)$ (I have been lazy and written $Z$ instead of ${\mathbb Z}$.)

It is possible to compute further factors $\gamma_i/\gamma_{i+1}$ but it gets much more difficult as $i$ increases.

  • $\begingroup$ Thanks! How do we see that $[G,G]$ is the normal closure? $\endgroup$ – yoyostein Sep 9 '18 at 18:49
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    $\begingroup$ Let $N$ be the normal closure of the commutators of the generators. Then the generators of $G/N$ commute, so $G/N$ is abelian and hence $[G,G] \le N$. $\endgroup$ – Derek Holt Sep 9 '18 at 18:53
  • $\begingroup$ Lastly, how do we prove $N\leq [G,G]$? I know this commutator identity $[x,y]^g=[x,y][[x,y],g]\in [G,G]$. However elements in $N$ may be conjugate of product of commutators? Thanks! $\endgroup$ – yoyostein Sep 9 '18 at 19:10
  • $\begingroup$ $N \le [G,G]$ is obvious from the definition of $N$. Why have you accepted the answer if you don't understand it? $\endgroup$ – Derek Holt Sep 9 '18 at 19:36
  • $\begingroup$ Ok I get it now, normal closure is smallest normal subgroup containing the set of 3 commutators, while $[G,G]$ also contains that set clearly, and it is known that $[G,G]$ is also a normal subgroup. $\endgroup$ – yoyostein Sep 10 '18 at 3:21

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