# Quotient Group of $\mathbb{Z}^4$ and a Lattice

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.

• The keyword to look up is Smith normal form: en.wikipedia.org/wiki/Smith_normal_form – Qiaochu Yuan Jan 7 '18 at 6:26
• There are a bunch of posts on this topic: 1, 2, 3, for example. – André 3000 Jan 7 '18 at 6:43
• @Quasicoherent how to efficiently find those links? I tried approach0 out too but it seems I still could dig them up – frogeyedpeas Jan 7 '18 at 6:51
• @frogeyedpeas I'm not sure how to find them. I just had answered these sorts of questions several times, so I bookmarked them. Googling site:math.stackexchange.com smith normal form brought up this relevant post, but you'd have to know the Smith normal form is the right thing to search. – André 3000 Jan 7 '18 at 7:39


The point is to determine the structure of $G$, we can freely compose automorphisms with $A$ to get a nicer matrix, for which the cokernel is obvious. In particular, we can apply row and column operations to $A$.

At this point, we'll put $A$ in Smith Normal Form as is suggested in the linked question.


Edit: to elaborate, since we made $A$ essentially diagonal, the cokernel splits as a direct sum: $$\ZZ^4/(e_1,2e_2) = (\ZZ/\ZZ)\oplus (\ZZ/2\ZZ) \oplus (\ZZ/0) \oplus (\ZZ/0) = \ZZ/2\ZZ \times \ZZ^2.$$

• @jgon this made sense up until the last point do you mean to say $\mathbb{Z}^4/ {( \mathbb{Z} \times 2 \mathbb{Z } ) }$ which is $\mathbb{Z}^3 \times \mathbb{Z}/ 2 \mathbb{Z}$? – frogeyedpeas Jan 7 '18 at 6:56
• I was inspired by math.stackexchange.com/questions/1243829/… where they seem to take $k\mathbb{Z}$ into the denominator of the quotient for each $k$ along the diagonal of the smith normal form (I'm going to review the proof and exact mechanics in a bit so I should be able to prove it myself but wanted to clarify here first) – frogeyedpeas Jan 7 '18 at 6:58
• I apologize, your answer matches mine too, It seems I forgot that $3-1=2$ – frogeyedpeas Jan 7 '18 at 6:59
• Ah cool, glad it got worked out. – jgon Jan 7 '18 at 7:00

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 $\mathbb Z^2 \times \mathbb Z_2$.