How is $\Bbb{Z}_2\otimes \Bbb{Z}_3$ isomorphic to $\Bbb{Z}_6$? I am currently reading the book  Group Theory in a Nutshell for Physicists by Anthony Zee and he wrote something about isomorphism I could not quite understand.

Now we come in for a bit of a surprise. Consider $\Bbb{Z}_2 \otimes \Bbb{Z}_3$ consisting of $(n, m)$, which we compose by $(n + n' \mod 2, m + m' \mod3)$. Again, we start with $(0, 0)$ and add $(1, 1)$ repeatedly:
$(0,0)\{+(1,1)\}\rightarrow (1,1) \{+(1,1)\}\rightarrow(0,2)\{+(1,1)\}\rightarrow(1,0)\{+(1,1)\}\rightarrow(0,1)\{+(1,1)\}\rightarrow(1,2)\{+(1,1)\}\rightarrow (0,0)$
We are back where we started! In the process, we cycled through all
six elements of $\Bbb{Z}_2 \otimes \Bbb{Z}_3$. We conclude that the
six elements $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$ } describe
$\Bbb{Z}_6$

Further he writes-

Thus, $\Bbb Z_2 \otimes \Bbb Z_3$ and $\Bbb Z_6$ are isomorphic; they are literally the same group. Note that this
phenomenon, of a possible isomorphism between $\Bbb Z_p \otimes \Bbb Z_q$ and $\Bbb{Z}_{pq}$, does not require p
and q to be prime, only relatively prime. (Consider the example of $\Bbb Z_4 \otimes \Bbb Z_9$.)

And hence my question: How did the author establish the one-to-one mapping between the groups claimed to be isomorphic? Any help is appreciated.
 A: You can write his mapping by literally taking the elements of $\mathbb{Z}_6$ mod 2 and 3 respectively:




$a \in \mathbb{Z}_6$
$a \mod 2$
$a \mod 3$
$(n, m) \in \mathbb{Z}_2 \times \mathbb{Z}_3$




$0$
$0$
$0$
$(0, 0)$


$1$
$1$
$1$
$(1, 1)$


$2$
$0$
$2$
$(0, 2)$


$3$
$1$
$0$
$(1, 0)$


$4$
$0$
$1$
$(0, 1)$


$5$
$1$
$2$
$(1, 2)$




Note that since $\mathbb{Z}_6$ under addition is the cyclic group of order 6, you can form its elements by repeated addition of $1$, and likewise in $\mathbb{Z}_2 \times \mathbb{Z}_3$ the element $(1, 1)$ is a generator that behaves identically.
A: Lemma: If a group $G$ has order $2p$ for prime $p>2$, then $G$ is isomorphic to either $\Bbb Z_{2p}$ or $D_p$.
Since $\Bbb Z_2\times \Bbb Z_3$ is abelian and $D_3$ is not, we must have that $\Bbb Z_2\times\Bbb Z_3$ is isomorphic to $\Bbb Z_6$.
Here $D_p$ is the dihedral group of order $2p$.
A: If you read the text, you will see that the correspondence $[1]\in \mathbb{Z}_6 \mapsto ([1],[1]) \in \mathbb{Z}_2\times\mathbb{Z}_3$ is one possible isomorphism.
