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?
Thanks.
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?
Best Answer
$[G,G] = \gamma_2$ 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.