3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The keyword to look up is Smith normal form: en.wikipedia.org/wiki/Smith_normal_form $\endgroup$ – Qiaochu Yuan Jan 7 '18 at 6:26
  • $\begingroup$ There are a bunch of posts on this topic: 1, 2, 3, for example. $\endgroup$ – André 3000 Jan 7 '18 at 6:43
  • $\begingroup$ @Quasicoherent how to efficiently find those links? I tried approach0 out too but it seems I still could dig them up $\endgroup$ – frogeyedpeas Jan 7 '18 at 6:51
  • $\begingroup$ @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. $\endgroup$ – André 3000 Jan 7 '18 at 7:39
3
$\begingroup$

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: $$\ZZ^4/(e_1,2e_2) = (\ZZ/\ZZ)\oplus (\ZZ/2\ZZ) \oplus (\ZZ/0) \oplus (\ZZ/0) = \ZZ/2\ZZ \times \ZZ^2.$$

$\endgroup$
  • $\begingroup$ @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} $? $\endgroup$ – frogeyedpeas Jan 7 '18 at 6:56
  • $\begingroup$ 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) $\endgroup$ – frogeyedpeas Jan 7 '18 at 6:58
  • $\begingroup$ I apologize, your answer matches mine too, It seems I forgot that $3-1=2$ $\endgroup$ – frogeyedpeas Jan 7 '18 at 6:59
  • $\begingroup$ Ah cool, glad it got worked out. $\endgroup$ – jgon Jan 7 '18 at 7:00
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.