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Let $H_1$ be the subgroup of $\mathbb{Z}^2$ generated by $\{(1,2),(4,1)\}$, let $H_2$ be the subgroup of $\mathbb{Z}^2$ generated by $\{(3,2),(1,3)\}$. Is it true that $\mathbb{Z}^2/H_1\cong \mathbb{Z}^2/H_2?$

In general, if $H_1$ is generated by $\{x,y\}$ and $H_2$ is generated by $\{u,v\}$, what is a sufficient and necessary condition that $\mathbb{Z}^2/H_1\cong \mathbb{Z}^2/H_2$?

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In this particular case, you need only calculate the two determinants: $-7$ for the first pair, $7$ for the second. So in both cases, the quotient groups are of order seven, thus necessarily isomorphic. – Lubin Jul 26 '14 at 19:38
But then one needs to show somehow that the order of the quotient is the absolute value of the determinant :-) (that's nicely done using the SNF) – Mariano Suárez-Alvarez Jul 26 '14 at 21:33
@MarianoSuárez-Alvarez One could do this using SNF, but one doesn't have to. Let $A : \mathbb{Z}^n \to \mathbb{Z}^n$ with $\det A \neq 0$. Any two fundamental domains for $\mathbb{R}^n/A \mathbb{Z}^n$ must have the same volume. One such domain is $A [0,1)^n$, with volume $|\det A|$. Another is $\bigcup_i \left( v_i + [0,1)^n \right)$, with $v_i$ ranging over a set of coset representatives for $\mathbb{Z}^n/A \mathbb{Z}^n$. So $|\det A| = |\mathbb{Z}^n/A \mathbb{Z}^n|$. – David Speyer Aug 21 '14 at 14:42
That's a really nice argument! Do yoy know an elementary way to show that fundamental domains must have the same volume? – Mariano Suárez-Alvarez Aug 21 '14 at 15:19
@MarianoSuárez-Alvarez I'm going to assume that, in whatever way you define volume, it is clearly additive. Let $G$ act freely on $X$, preserving volume. Let $D_1$ and $D_2$ be two fundamental domains. Then $\mathrm{Vol}(D_1) = \sum_{g \in G} \mathrm{Vol}(D_1 \cap g D_2) = \sum_{g \in G} \mathrm{Vol}(g^{-1} D_1 \cap D_2) = \mathrm{Vol}(D_2)$. In our particular case, one can prove that there are only finitely many nonzero terms in the union, so you only need finite additivity, which makes life a little easier. – David Speyer Aug 21 '14 at 19:59
up vote 1 down vote accepted

Let $R$ be a ring and $M$ an $R$-module. Suppose $u,v\in R^\times$ are units. Then for any $\pi\in R$,

$$\frac{M}{u\pi\color{Green}{vM}}=\frac{M}{u\pi\color{Green}{M}}\cong \frac{\color{Blue}{u^{-1}M}}{\pi M}=\frac{\color{Blue}{M}}{\pi M} \implies \frac{M}{u\pi v M}\cong\frac{M}{\pi M}.$$

The isomorphism is given by multiplication-by-$u^{-1}$. Let $M=\Bbb Z^n$ and $R=M_n(\Bbb Z)$ and take any integer matrix $A\in M_n(\Bbb Z)$, then $A$ has a smith normal form $A=UDV$ where $U,V\in{\rm GL}_n(\Bbb Z)$ are invertible and $D={\rm diag}(d_1,\cdots,d_n)$ is a diagonal integer matrix. Since $D\Bbb Z^n=d_1\Bbb Z\times\cdots\times d_n\Bbb Z$,

$$\frac{\Bbb Z^n}{A\Bbb Z^n}\cong\frac{\Bbb Z^n}{D\Bbb Z^n}=\frac{\Bbb Z\times\cdots\times\Bbb Z}{d_1\Bbb Z\times\cdots\times d_n\Bbb Z}\cong\frac{\Bbb Z}{d_1\Bbb Z}\times\cdots\times\frac{\Bbb Z}{d_n\Bbb Z}.$$

This gives the invariant factor decomposition of $\Bbb Z^n/A\Bbb Z^n$. Lastly, using column vectors,

$$\left\langle\begin{pmatrix}a_{11}\\ \vdots\\ a_{n1}\end{pmatrix},\cdots,\begin{pmatrix}a_{n1}\\ \vdots \\ a_{nn}\end{pmatrix}\right\rangle =\left\{b_1\begin{pmatrix}a_{11}\\ \vdots\\ a_{n1}\end{pmatrix}+\cdots+b_n\begin{pmatrix}a_{n1}\\ \vdots \\ a_{nn}\end{pmatrix}:b_1,\cdots,b_n\in\Bbb Z\right\} $$

$$=\left\{\begin{pmatrix}a_{11} & \cdots & a_{nn} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}\begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}:~\begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}\in\Bbb Z^n \right\}=A\Bbb Z^n.$$

This justifies the use of matrices. If the number of vectors you're using to generate a subgroup of $\Bbb Z^n$ is less than $n$, then one can pad the list using zero vectors to get a list of $n$ vectors. If one is using a list of more than $n$ vectors, then one will have to use the more general SNF for nonsquare matrices, and justify its use in much the same way. Note that as a corollary, since $|\det A|=|\det D|=d_1\cdots d_n$ is equal to the size of $|\Bbb Z/d_1\Bbb Z\times\cdots\times\Bbb Z/d_n\Bbb Z|$ if all $d_i\ne0$, we know $|\Bbb Z^n/A\Bbb Z^n|=|\det A|$ if $\det A\ne0$.

Here one can compute $\det(\begin{smallmatrix}1&4\\2&1\end{smallmatrix})=1-8=-7$ and $\det(\begin{smallmatrix}3&1\\2&3\end{smallmatrix})=9-2=7$. Since there is only one abelian group of order $7$, the quotients must be isomorphic. In general, a sufficient condition for the quotients $\Bbb Z^n/A\Bbb Z^n\cong\Bbb Z^m/B\Bbb Z^m$ to be isomorphic is $|\det A|=|\det B|$ being a squarefree integer (forcing there to be only one abelian group of that order, up to isomorphism), but this is far from a necessary condition.

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Answer in the form of a keyword which will answer your question and its generalization to $\mathbb Z^n$: Smith normal form.

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Can you please explain why the smith normal form gives the structure of the quotient? – mez Jul 28 '14 at 12:11
Let $A = UDV$ where $U$ and $V$ are invertible over $\mathbb{Z}$. I claim that $\mathbb{Z}^n / A \mathbb{Z}^n \cong \mathbb{Z}^n/D \mathbb{Z}^n$. Proof: Since $V$ is invertible, $V \mathbb{Z}^n = \mathbb{Z}^n$ and so $UDV \mathbb{Z}^n = UD \mathbb{Z}^n$. Then $z \mapsto U z$ is an automorphism of $\mathbb{Z}^n$ carrying $D \mathbb{Z}^n$ to $UD \mathbb{Z}^n$, so the quotients $\mathbb{Z}^n/UD \mathbb{Z}^n$ and $\mathbb{Z}^n/D \mathbb{Z}^n$ are isomorphic. Putting it all together, $\mathbb{Z}^n / UDV \mathbb{Z}^n \cong \mathbb{Z}^n/D \mathbb{Z}^n$ as claimed. $\square$. – David Speyer Aug 25 '14 at 15:52
The structure of $\mathbb{Z}^n/D \mathbb{Z}^n$ is obvious, and we have just shown the structure of $\mathbb{Z}^n/A \mathbb{Z}^n$ is the same. – David Speyer Aug 25 '14 at 15:53

Both determinants are 7 and therefore the quotient is a group of order 7. Since such a group is unique, the quotients are isomorphic.

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In general, the quotient is isomorphic to $\prod_{k=1}^{n}\frac{\mathbb Z}{d_k\mathbb Z}$ where the $d_k$ are the "elementary divisors", i.e. the diagonal coefficients in the Smith Normal Form.

To expand on Mariano’s answer in your example : the Smith normal form for $((1,2),(4,1))$ is ${\sf diag}(1,7)$ and $H_1={\sf span}((7,0),(-3,1))$. Similarly, the Smith normal form for $((3,2),(1,3))$ is ${\sf diag}(1,7)$ and $H_2={\sf span}((7,0),(-2,1))$. Denote by $p_i$ the natural projection ${\mathbb Z}^2 \to \frac{{\mathbb Z}^2}{H_i}$. For $(x,y)\in{\mathbb Z}^2$, we have $(x,y)=y(-3,1)+(x+3y,0)$, so $p_1(x,y)=p_1(x+3y,0)$ and hence $$H_1=p_1({\mathbb Z}^2)=p_1({\mathbb Z}\times\lbrace 0 \rbrace) \approx \frac{\mathbb Z}{7{\mathbb Z}}.$$

A similar argument shows that $H_2\approx \frac{\mathbb Z}{7{\mathbb Z}}$.

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