# 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

Let $A:\newcommand{\ZZ}{\mathbb{Z}}\ZZ^2\to\ZZ^4$ be the map sending $e_1$ to $(5,-2,-4,1)$ and $e_2$ to $(-5,4,4,1)$. Let's call the group you care about $G$. Then $G=\newcommand{\coker}{\operatorname{coker}}\coker A$. Then let $B:\ZZ^2\to\ZZ^2$ and $C:\ZZ^4\to\ZZ^4$ be automorphisms. Then the following diagram commutes, $$\require{AMScd}\newcommand{\inv}{^{-1}} \begin{CD} \ZZ^2 @>A>> \ZZ^4@>>>G@>>>0\\ @VBVV @VVCV @VVDV\\ \ZZ^2 @>CAB\inv>> \ZZ^4@>>>G'@>>>0,\\ \end{CD}$$ where $G'=\coker CAB\inv$, and $D$ is the induced map $\coker A \to \coker CAB\inv$. Then since $B$ and $C$ are isomorphisms, $D$ is too. You can either cite the five lemma or do the diagram chase yourself to prove it.

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.

$$A = \newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}} \bmat 5 & -5 \\ -2 & 4 \\ -4 & 4 \\ 1 & 1 \\ \emat.$$ First we'll swap the first and last rows, and subtract the first row from the rest to make all the rest of the entries in the first column 0: $$\newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}} \bmat 1 & 1 \\ 0 & 6 \\ 0 & 8 \\ 0 & -10 \\ \emat.$$ Now subtract the first column from the second, to get $$\newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}} \bmat 1 & 0 \\ 0 & 6 \\ 0 & 8 \\ 0 & -10 \\ \emat.$$ Subtract the second row from the third and swap them, to get $$\newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}} \bmat 1 & 0 \\ 0 & 2 \\ 0 & 6 \\ 0 & -10 \\ \emat.$$ Now zero the last two rows using the second row: $$\newcommand{\bmat}{\begin{pmatrix}}\newcommand{\emat}{\end{pmatrix}} \bmat 1 & 0 \\ 0 & 2 \\ 0 & 0 \\ 0 & 0 \\ \emat.$$ Thus $G\cong \ZZ^4/(e_1,2e_2)\cong \ZZ/2\ZZ\times \ZZ^2$.

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