show $\Bbb Z_6 \cong \Bbb Z_3 \times \Bbb Z_2 $. I am trying to determine if $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $. I noticed that $\Bbb Z_6$ has a generator $1$ and $\Bbb Z_3 \times \Bbb Z_2$ has generator $(1,1)$. Now I set up the bijection $f :\Bbb Z_6 \to   \Bbb Z_3 \times \Bbb Z_2 $ where
\begin{eqnarray}
f(1)&=&(1,1)\\
f(1+1)&=&(1,1)+(1,1) = (2,0)\\
f(2+1)&=&(2,0)+(1,1) = (0,1)\\
f(3+1)&=&(0,1)+(1,1) = (1,0)\\
f(4+1)&=&(1,0)+(1,1) = (2,1)\\
f(0)&=&(1,1)+(2,1) = (0,0)\\
\end{eqnarray}
So basically $f(1^n) = (1,1)^n$ where $1^n$ means we add the $1$ to itself $n$ times and take the result mod 6, so as per usual $1^n$ means applying the group operation on 1 to itself for $n$ times. So this is a bijection but I am stuck on showing that $f(ab)=f(a)f(b)$ for all $a,b \in \Bbb Z_6$. I could check all possibilities but there must be a better way to show this. I was thinking something like $f(1^n1^m) = f(1^{n+m})=(1,1)^{n+m}=(1,1)^n(1,1)^m = f(a)f(b) $ because every element in $\Bbb Z_6$ can be expressed as some power of $1$.
Is this right? And is it enough to show $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $?
 A: Usually the way to prove an isomorphism of this type is not to define a map for every element and then try to show that the map respects the group structure, but rather to define a homomorphism and then use the appropriate isomorphism theorem to prove that the resulting homomorphism is what you want. So here you would define the map on the generator by setting $f(1)=(1,1)$, then say "extend to a homomorphism" between the groups, and then prove that it's onto (or injective) which is enough to show isomorphism.
A: Consider the map $\mathbb Z \to \mathbb Z_3 \times \mathbb Z_2$ given by $x\mapsto (x \bmod 3, x\bmod 2)$. It is easy to prove that this map is a surjective homomorphism whose kernel is $6\mathbb Z$.
A: It might be rather instructive for you to work out this exercise by defining $f:\mathbb{Z}/3\times\mathbb{Z}/2\to\mathbb{Z}/6$ via $f(a,b) = 3b - 2a$, and verify that this is a ring homomorphism.
Next, define $g:\mathbb{Z}/6\to\mathbb{Z}/3\times\mathbb{Z}/2$ via $g(x) = (x\pmod{3},x\pmod{2})$ and show that $f\circ g$ and $g\circ f$ are the identity maps. In other words, $f$ is inverse to $g$ and $g$ is inverse to $f$.
This is a special case of a more general phenomenon that you will likely see under the guise of the Chinese remainder theorem.
