Find $|\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle|$ and determine if its cyclic.

$\mathbb{Z} \oplus \mathbb{Z} = \{(1,1), (1, 2),... (2, 1),...(3,1),...\}$
$\langle(3,6)\rangle = \{(0,0),(3,6),(6,12),...\}$
$\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle = \{(a,b) + \langle(3,6)\rangle : a,b \in \mathbb{Z}\}$

How would I go by finding the order of the quotient group? Would it be infinite? Since both sets that make up the quotient group are infinite, does that mean that the order of the quotient group is also infinite? What are some things I'm not noticing that could be helpful?


Here's a hint at a general method (see Smith Normal Form).

The subgroup $\langle (3,6)\rangle$ can be thought of as the image of the homomorphism $$ \mathbb{Z} \xrightarrow{n\mapsto n(3,6)} \mathbb{Z}\oplus\mathbb{Z} $$ where we can think of $(3,6)$ as being a matrix. Then, the group in question is the codomain modulo the image of this homomorphism; such a quotient group is known as a cokernel.

Without changing the quotient group, we can multiply this matrix on either side by an invertible matrix with integer entries. The only thing we can multiply by on the left is $(1)$ or $(-1)$, which does not really change anything. On the right, we can multiply by the elementary matrix $$\begin{pmatrix} 1 & -2\\ 0 & 1 \end{pmatrix}$$ which has the effect of subtracting two of the first column from the second, so the resulting matrix is $(3,0)$. This sort of matrix multiplication corresponds to a change in generating set for $\mathbb{Z}\oplus \mathbb{Z}$.

So, the quotient group is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}/\langle(3,0)\rangle\cong (\mathbb{Z}/3\mathbb{Z})\oplus \mathbb{Z}$.

Put a different way, let $a,b$ be the generators of $\mathbb{Z}\oplus\mathbb{Z}$, and so the quotient group is given by the relation $3a+6b\equiv 0$. This relation is equivalently $3(a+2b)\equiv 0$. Let $a'=a+2b$ and $b'=b$, which is also a generating set since $a=a'-2b'$ and $b=b'$. Then the relation becomes $3a'\equiv 0$ with no relation on $b'$. It follows that the quotient group is $(\mathbb{Z}/3\mathbb{Z})\oplus \mathbb{Z}$ with the first summand generated by $a'$ and the second summand generated by $b'$.

The group $\mathbb{Z}/3\mathbb{Z}$ is an example of a finite group with both $\mathbb{Z}$ and $3\mathbb{Z}$ being infinite groups.


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