Writing $G/[G, G]$ as a direct product of cyclic groups Let $G$ be the group given by the presentation $\langle x , y , z : x^2 , y^3 , (xyz)^4 \rangle$. I would like to write $G/[G , G]$ as a direct product of cyclic groups, where $[G , G]$ is the commutator subgroup of $G$.
I am familiar with writing (finitely generated) abelian groups as a product of cyclic groups using Smith normal forms, but the free group aspect of this problem is throwing me off (which I believe is the point). I am aware that the free group on a set $S$ modulo its commutator subgroup is isomorphic to $\mathbb{Z}^{|S|}$, but I am unsure of how the additional relations (such as those in the problem in question) impact things.
Obviously $G/[G , G]$ is a finitely generated abelian group, but is $G/[G , G]$ isomorphic to the abelian group, $A$, (written additively) generated by elements $a , b, c \in A$ subject to the relations $2a = 3b = 4(a + b + c) = 0$? If this is the case, then I am completely comfortable finishing off the problem and actually writing $G/[G , G]$ as a direct product of cyclic groups, but I would like some more justification as to why. If this is not the case, then what is the correct way to go about this problem?
 A: Standard general method:
Write the presentation in abelianized form in a matrix (rows for relators, columns correspond to generators):
$$
\left(\begin{array}{rrr}%
2&0&0\\%
0&3&0\\%
4&4&4\\%
\end{array}\right).
$$
Calculate the Smith Normal form (also called elementary divisor normal form):
$$
\left(\begin{array}{rrr}%
1&0&0\\%
0&2&0\\%
0&0&12\\%
\end{array}\right)%
$$
Thus the structure of $G/[G,G]$ is $C_2\times C_{12}$ (which is the same as the other andwer, using that $12=3\cdot 4$.
Justification for why Smith normal form works: Row transformations correspond to relator changes (replace relator by product with other), column operations to generator changes.
A: The presentation of $G/[G,G]$ is achieved by adding the relations
$$xyx^{-1}y^{-1}=1,xzx^{-1}z^{-1}=1,yzy^{-1}z^{-1}=1.$$
The commutators $xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}$ and their inverses are the generators of the normal subgroup $[G,G]$.
Then
$$G/[G,G]\cong\langle x,y,z\ : x^2,y^3,(xyz)^4, xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}\rangle.$$
This is also because the natural homomorphism
$
\langle x,y,z\ : x^2,y^3,(xyz)^4\rangle$ to
$\langle x,y,z\ : x^2,y^3,(xyz)^4, xyx^{-1}y^{-1},xzx^{-1}z^{-1},yzy^{-1}z^{-1}\rangle$ is surjective and has kernel the normal closure of $[G,G]$, which is $[G,G]$ itself.
The "abelian" presentation would be
$$\langle x,y,z:2x=0,3y=0, 4x+4y+4z=0,\\ x+y-x-y=0,x+z-x-z=0,y+z-y-z=0\rangle,$$ which can be simplified with
$t=x+y+z$ into
$$\langle x,y,t\ :\ 2x=0,3y=0, 4t=0\rangle,$$
and which is $\mathbb Z_2\oplus \mathbb Z_3\oplus \mathbb Z_4$
