Show that $\mathbb{Z_3 x Z_4}$ is a cyclic group Q. Show that $\mathbb{Z_3 x Z_4}$ is a cyclic group.
So my question is there a faster way besides listing all the elements and besides knowing the theorem. 
Since the process I am doing is:
I know:
$\mathbb{Z_3 x Z_4}=\{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3)\}$
and I've been taking each element and figuring out the order such as:
$$(1,0): (1,0); (1,0)+(1,0)=(2,0);(2,0)+(1,0)=(0,0)$$ which is order 3.
and I believe I have to keep doing this until I find the element that gives me order 12.
but I realize that is kinda tedious. So my question is, is there a simpler way or would be it suggested to continue on the method I am doing.
 A: You can calculate the order of an element without listing every element. For example, you can show that $(1,0)$ has order $3$ and $(0,1)$ has order $4$. 
So $(1,1)$ has order divisible by $3$, since the first element has order $3$. Similarly its order must be divisible by $4$. Hence, it must have order $12$ since $3,4$ are coprime and the order of the group is $12$.
This is a group theoretic way of phrasing the Chinese Remainder theorem. 
A: To find an element of order $12$ don't check elements randomly: think about which elements might possibly be of order $12$.  For example, if for $(x,y)\in \mathbb{Z}_3\times\mathbb{Z}_4$ one of the components is not a generator for it's group, you won't possible generate all $12$ different elements.  So why don't you look at a pair $(x,y)$ where $x$ is a generator for $\mathbb{Z}_3$ and $y$ is a generator for $\mathbb{Z}_4$.  
A: Taking $(1,1)$ in $\mathbb{Z_3 x Z_4}$, we have $12(1,1)=0$ and $n(1,1)\neq0$ if $n<12$. That is, it is cyclic.
A: This is the abstract version of the Chinese Remainder Theorem::
If $m$  and $n$ are two coprime integers, then $\mathbf Z/m\mathbf Z\times \mathbf Z/n\mathbf Z\simeq \mathbf Z/mn\mathbf Z$.
In the present case, consider the canonical morphism:
\begin{align*}
\varphi \colon\mathbf Z\ &\rightarrow \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z\\
x&\mapsto (x\bmod 3,x\bmod 4)
\end{align*}
Bézout's identity for $3$  and $4$ is simply $4-3=1$. It is easy to see $\varphi$ is surjective. Indeed, if $(\bar a, \bar b)$ is an element of $\mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z$, we have: 
$$\varphi(4a-3b)=(4a-3b\bmod 3, 4a-3b\bmod 4)=(a\bmod 3, b\bmod 4)$$
since $4\equiv 1\mod 3$, $-3\equiv 1\mod 4$.
Now $x\in \ker\varphi$ if and only if $x$ is divisible by both $3$  and $4$, i.e. $x$  is divisible by $12$, whence an isomorphism;
$$\bar\varphi\colon \mathbf Z/12\mathbf Z\rightarrow \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z. $$
This isomorphism proves $ \mathbf Z/3\mathbf Z\times \mathbf Z/4\mathbf Z$ is cyclic, of order $12$.
A: The fastest "minimal computation" way:
$6(1,1) = (0,2) \neq (0,0)$
$4(1,1) = (1,0) \neq (0,0)$. Therefore, $(1,1)$ has order $12$ (thanks to Bill Dubuque).
