Problem about isomorphic groups Show that the group $(Z_6,+_6)$ is isomorphic to the direct group of the groups $(Z_3,+_3)$ and $(Z_2,+_2)$
Approach:
$Z_6=\{0,1,2,3,4,5\}$
$z_3=\{0,1,2\}$
$z_2=\{0,1\}$
$Z_3 \times Z_2=\{(0,0),(0,1),(1,0),(1,1),(2,0),(2,1)\}$
As you may notice, the cardinality of each group is the same, so we can easily create a bijection , but the problem is how we label each element in $Z_3 \times Z_2$, so $f$ is homomorphic. Basically we have to do something like
$f(0)=(0,0)$,$f(1)=(0,1)$,$f(2)=(1,0)$,$f(3)=(1,1)$,$f(4)=(2,0)$, $f(5)=(2,1)$. Obviously the previous one doesn't work, but I just can't find the right labeling. Is there a discernible pattern to do this?
 A: Yes, there is a general approach here (it is important that $2$ and $3$ be coprime for this to work!). For any class representative $[x]_{6}$ in $\mathbb{Z}/6\mathbb{Z}$, you can further reduce $x$ mod $3$ and $x$ mod $2$ to obtain a pair $([x]_{3}, [x]_{2}) \in \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Try showing that the map $\varphi \colon \mathbb{Z}/6\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ given by $[x]_{6} \mapsto ([x]_{3}, [x]_{2})$ is a well-defined group homomorphism. Since the sets have the same cardinality, as you observe, it is enough to show that $\varphi$ is surjective in order to show it is an isomorphism. Feel free to comment if you need more assistance!
A: There are $6!=720$ possible bijections between the two sets, most of which are not homomorphisms. So you're right, it's no good to try random bijections.
On the other hand, there are only $6$ possible homomorphisms from $Z_6$ to $Z_3\times Z_2$, because $Z_6$ is generated by $1$, so once you choose $f(1)$, that fixes the rest of $f$. So you should try random homomorphisms until you find one that's a bijection. It won't take too long.
A: $\mathbb{Z}_{3}\times\mathbb{Z}_{2}$ is cyclic generated by $(1,1)$ because the order of $(1,1)$ is 6. The map $$\begin{matrix}\mathbb{Z}_{6}&\longrightarrow&\mathbb{Z}_{3}\times\mathbb{Z}_{2}\\1&\longmapsto&(1,1)\end{matrix}$$ is an isomorphism.
