Let $G$ be a group generated by $3$ elemets with the following relation
$$G=\langle a,b,c\text{ }|\text{ }a^2bcb^{-1}c^{-1}=1\rangle$$
Let $G'$ be the commutator of $G$, that is $$G'=\langle xyx^{-1}y^{-1}\text{ }|\text{ }x,y\in G\rangle$$
It is a standard fact that
$G'$ is normal in $G$ and $G/G'$ is abelian.
I have to describe $G/G'$.
I intuitively feel that $G/G'$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\bigoplus\mathbb{Z}\bigoplus\mathbb{Z}$. But I don't know how to formally prove it.
My idea:
(1) First of all, notice that $a^2G'=1$ in $G/G'$
(2) Since everything commutes in $G/G'$, any element of $G/G'$ is of the form $ab^ic^jG'$.
Explanation: Just collect all the $aG'$, $bG'$ and $cG'$ together by using commutativity.By the previous point the highest power of $aG'$ that can appear is $1$.
It is well known that if $G$ is generated by $x_1,\ldots,x_k$ then $G'$ is the normal closure of the group generated by $[x_i,x_j]=x_ix_jx_i^{-1}x_j^{-1}$.
In this case, this implies $G/G'$ is defined by the relations $$a^2bcb^{-1}c^{-1}=aba^{-1}b^{-1}=aca^{-1}c^{-1}=bcb^{-1}c^{-1}=1$$
This is easily seen to be equivalent to $$a^2=aba^{-1}b^{-1}=aca^{-1}c^{-1}=bcb^{-1}c^{-1}=1$$
These are precisely the relations for $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ with generators $a,b,c$, hence $$G/G'\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$$