TITLE: Writing $G/[G, G]$ as a direct product of cyclic groups
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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.