Why $\mathbb{Z}_m \times \mathbb{Z}_n, \, \text{gcd}(m,n) >1$ isn't cyclic? I'm aware of the theorem which states that $\mathbb{Z}_m \times \mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_{mn}$ (and thus cyclic) if and only if $\text{gcd}(m,n)=1$.
However, in $G=\mathbb{Z}_2 \times \mathbb{Z}_4$, where $\text{gcd}(2,4)=2$, it seems to me that if we take $(0,0)$ and add to it $(1,1)$ $4$ times (least common multiple of $2,4$) then we return to $(0,0)$. That's what misled me into thinking that $G$ is cyclic.
Could you explain to me why this is a faulty syllogism and provide me with a more practical way to visualize cyclic direct products?
 A: Suppose that $m$ and $n$ are not coprime. Let $d=\text{gcd}(m,n)$ and 
$$
a=\frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) \in \mathbb{Z}_m \times \mathbb{Z}_n$. We have 
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $\mathbb{Z}_m \times \mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $\mathbb{Z}_m \times \mathbb{Z}_n$ is not cyclic. 
A: In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H=\{(0,0),(1,1),(0,2),(1,3)\}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
A: By the definition of the direct product of groups $Z_m \times Z_n$, the order of any element  $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m \times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m \times Z_n$ does not have any element of order $mn$. Since $Z_m \times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
