TITLE: Quotient Group of $\mathbb{Z}^4$ and a Lattice
QUESTION [5 upvotes]: I am working on problem 2 of the Rutgers 2017 Fall Algebra Qualifier where we are tasked with determining the structure of $\mathbb{Z}^4 /S$ where $S$ is the group generated by the vectors $(5,-2,-4,1)$, $(-5,4,4,1)$, $(0,6,0,6)$.
So the first thing I noted was that 
$$(5,-2,-4,1) + (-5,4,4,1) = (0,2,0,2).$$
So it follows the third vector given $(0,6,0,6)$ is in the span of first two, and so the question remains to show: 
Find $\mathbb{Z}^4 / \lbrace a (5,-2,-4,1)  + b (-5,4,4,1), a, b \in \mathbb{Z} \rbrace$
Now I tried to look for similar problems to this to make sense of it and came across the following: Is $\mathbb{Z}\times\mathbb{Z}/((6,5),(3,4))$ is finitely generated?
But I'm not sure how to use the matrix techniques there correctly and rigorously. 

So I now I'm working with equivalence classes but the ease with which one can declare $$ \mathbb{Z} / k\mathbb{Z} = \mathbb{Z}_k$$ seems to be lost when I move into the 2 basis vector situation.

REPLY [2 votes]: A slightly different approach that amounts to the same thing (row reduction)  using tietze transformations.
Let's rewrite the group in terms of its presentation (although I will suppress all the commutation relations such as $ab=ba$, $ac=ca$ etc.) Additionally, each $a,b,c,d$ are just standard basis vectors. We get:
$$\langle a,b,c,d: a^{-5}b^4c^4d=1, a^{-5}b^2c^4d=1 \rangle$$
The first thing to note is that we get $d^{-1}=c^{-4}b^{-4}a^5$ (and likewise for the second relation) so we in fact obtain that  $c^{-4}b^{-2}a^{-5}=c^{-4}b^{-4}a^{-5},$ and by cancellation $b^2=1$.
Hence, we can remove $d$ as a generator and replace the relations by $b^2=1$ so the new presentation is
$$\langle a,b,c : b^2=1 \rangle$$
along with the usual commutation relations, or $\mathbb Z^2 \times \mathbb Z_2$.