4
$\begingroup$

I'm working on some practice problems and would like to get a few solutions checked (more coming!). Let $H$ be the subgroup of $\mathbb{Z} \times \mathbb{Z}$ generated by the elements $(2, 2)$ and $(3, 4)$. First give a minimal set of elements which generate $(\mathbb{Z} \times \mathbb{Z}) ~ / ~ H$ (as their images by the natural homomorphism $g \mapsto g + H$) and list all the elements in this quotient group, and then describe this quotient group.

First we simplify the presentation of $H$ a bit by simplifying the generators: if $H$ is generated by $(2, 2)$ and $(3, 4)$, then it is generated by $(2, 2)$ and $(3, 4) - (2, 2) = (1, 2)$, and so it is generated by $(2, 2) - (1, 2) = (1, 0)$ and $(1, 2)$, and so it is generated by $(1, 0)$ and $(1, 2) - 2 (1, 0) = (0, 2)$ (this could be formalized into a theorem since it's essentially the Euclidean algorithm). So geometrically $H$ spans the set of lattice points of the form $(m, 2n)$ where $m$ and $n$ are integers. Hence it partitions $(\mathbb{Z} \times \mathbb{Z})$ into exactly two cosets $H$ and $(0, 1) + H$. Since $(0, 2) + H = H$, the quotient group is generated by a single element $(0, 1) + H$, and contains two elements $H$ and $(0, 1) + H$. As such, it is isomorphic to $C_2$.

Is this a (correct and) rigorous answer and are there things that could be improved? One thing I am a bit concerned about is that the problem asks me to find a set of generators for the quotient group before asking for its elements and its description, whereas I've done the opposite - was there a better approach to finding its generators?

$\endgroup$
  • 1
    $\begingroup$ Looks perfectly rigorous and well written to me. Is there a different perspective you're looking for or parts you are looking to sharpen? There are slight adjustments that could be made if you are interested, but I like it a lot as written. $\endgroup$ – Alex Wertheim Jun 13 '14 at 4:22
  • $\begingroup$ @AWertheim Well for instance if there was a better way to find the generators of the quotient group than completely working it out and then finding the elements which generate it? $\endgroup$ – Thomas Jun 13 '14 at 4:48
  • 1
    $\begingroup$ @Thomas, have you studied already the Smith Normal Form for an integer matrix? $\endgroup$ – DonAntonio Jun 13 '14 at 5:57
  • 1
    $\begingroup$ Yes @Thomas...and it is a rather easy, albeit sometimes slightly messy, algorithm for doing that. I shall try later to find you a good document about this. $\endgroup$ – DonAntonio Jun 13 '14 at 13:41
  • 1
    $\begingroup$ @Thomas, the following document is, I think, right on the money. If your background in abstract algebra is not too wide try to assimilate the algorithmic part of it with the matrices, which must be known to you: sierra.nmsu.edu/morandi/notes/SmithNormalForm.pdf $\endgroup$ – DonAntonio Jun 13 '14 at 13:44
2
$\begingroup$

Applying the already mentioned method of Smith Normal form: the generators matrix is

$$A=\begin{pmatrix}2&2\\3&4\end{pmatrix}\stackrel{C_2-C_1}\longrightarrow\begin{pmatrix}2&0\\3&1\end{pmatrix}\stackrel{C_1-3C_2}\longrightarrow\begin{pmatrix}2&0\\0&1\end{pmatrix}\stackrel{\begin{align*}C_1\leftrightarrow C_2\\R_1\leftrightarrow R_2\end{align*}}\longrightarrow\begin{pmatrix}1&0\\0&2\end{pmatrix}$$

Thus, $\;H\cong \Bbb Z\oplus 2\Bbb Z\;$ , and thus $\;\left(\Bbb Z\otimes\Bbb Z\right)/K\cong\Bbb Z/2\Bbb Z=:\Bbb Z_2\;$

Note the above gave precisely the same generators you got. The advantage the method of Smith Forms offers is that you can methodically apply it doing very basic row/column operations on some given relations matrix, very much like elementary operations on matrices what we know from linear algebra.

$\endgroup$
  • $\begingroup$ Thanks, that makes sense, the Smith normal form seems very handy. Just one question, to conclude that $(\mathbb{Z} \oplus \mathbb{Z}) / (\mathbb{Z} \oplus 2 \mathbb{Z}) = \mathbb{Z} / 2 \mathbb{Z}$ can you just cancel out $\mathbb{Z}$? I thought of the third isomorphism theorem but it isn't that and I don't think I've seen arithmetic done on direct products like this before. $\endgroup$ – Thomas Jun 13 '14 at 22:09
  • $\begingroup$ If you know/remember about finitely generated abelian groups, cyclic ones and etc., there's one part in which you actually use/prove that $\;(\Bbb Z\oplus\Bbb Z)/(\Bbb Z\oplus 2\Bbb Z)\cong\Bbb Z/\Bbb Z\oplus \Bbb Z/2\Bbb Z\;$ ... $\endgroup$ – DonAntonio Jun 13 '14 at 22:16
  • $\begingroup$ Ah! got it, it's clear now. Thanks again for your help! $\endgroup$ – Thomas Jun 13 '14 at 22:36
1
$\begingroup$

Let $\alpha=(2,2)$ and $\beta=(3,4)$. In general any member of $\mathbb{Z} \times \mathbb{Z}/H$ will be of the form $(a,b)+H$. We will distinguish between two cases $b$ is even and $b$ is odd. $$ (a,b)= \begin{cases} & 2a(\alpha -\beta) + \frac{b}{2}(2\beta-3\alpha), \text{ if } b \text{ is even}\\ & 2a(\alpha -\beta) + \frac{b-1}{2}(2\beta-3\alpha)+(0,1), \text{ if } b \text{ is odd} \end{cases} $$ Since $\alpha, \beta \in H$ and $H \leq \mathbb{Z} \times \mathbb{Z}$, therefore $$ (a,b)+H= \begin{cases} & H, \text{ if } b \text{ is even}\\ & (0,1)+H, \text{ if } b \text{ is odd} \end{cases} $$ This gives us the minimal set of generators as well as a complete description of the quotient group.

$\endgroup$
  • $\begingroup$ Nice approach for showing $(a, b)$ can only fall in two equivalence classes, I know about using algebra to rewrite things that way but occasionally get muddled up! Thanks! $\endgroup$ – Thomas Jun 13 '14 at 4:50

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.