How do you verify that $(1,2)$ is a generator of $\mathbb{Z}_2 \times \mathbb{Z}_5$? (Abstract Algebra Question) I know that for $(1,2)$, in $\mathbb{Z}_2$, $1$ is of order $2$, and in $\mathbb{Z}_5$, $2$ is of order $5$. Since the lcm$(2,5)=10$, is it correct to state that the element $(1,2)$ has order $10$? I don't know how to show that this is the first multiple where $(1,2)=(0,0)$ in $\mathbb{Z}_2 \times\mathbb{Z}_5$ (which I know has 10 elements). So is showing the order of the element as equal to the number of elements in the entire group a verification that $(1,2)$ is a generator? If not, what would strengthen this?
 A: Just do the sums:
\begin{align*}
(1,2)+(1,2) & = (0,4), \\
(0,4)+(1,2) & = (1,1), \\
(1,1)+(1,2) & = (0,3), \\
(0,3)+(1,2) & = (1,0), \\
(1,0)+(1,2) & = (0,2), \\
(0,2)+(1,2) & = (1,4), \\
(1,4)+(1,2) & = (0,1), \\
(0,1)+(1,2) & = (1,3), \\
(1,3)+(1,2) & = (0,0), \\
(0,0)+(1,2) & = (1,2).
\end{align*}
Look carefully at all the elements you got: they are all elements of $\mathbb Z_2 \times \mathbb Z_5$.
A: My preference is to notice that in $\mathbb Z_2 \times \mathbb Z_5$
$$ 5 \cdot (1,2) =(1,0)$$
and
$$ 8 \cdot (1,2)=(0,1)$$
so the subgroup generated by $(1,2)$ contains those two elements, which clearly generate the whole group.
If I wanted to be super explicit, I would use the notation $\langle A\rangle$ to denote the subgroup generated by $A$, for any $A$ that is a subset of a group $G$, and the fact that if $B \subset A$ then $\langle B \rangle \subset \langle A\rangle$. (I haven't written it all out as a detailed answer, I've just kind of given you all the tools necessary. Is it clear enough now, or do you think you need more explanation? You can always take the answers/suggestions you get and write them up yourself as an answer, and ask for people to critique it.)
A: We can do this more generally:

Lemma. Let $G$, $H$ be groups and $A\subseteq G$, $B\subseteq H$ generating sets, each containing elements of finite order only. Assume that $gcd(|x|,|y|)=1$ for any $x\in A$ and $y\in B$. Then $A\times B$ generates $G\times H$.

Proof. Of course $C=A\times\{e_H\}\cup\{e_G\}\times B$ generates $G\times H$, so it is enough to generate $C$ from $A\times B$. In particular it is enough to show that given $(x,y)\in A\times B$ we can generate $(x,e_H)$ and $(e_G,y)$ out of it. We will only focus on $(x,e_H)$ since the other case is symmetric.
As a first step we consider $(x,y)^{|y|}=(x^{|y|},e_H)$. Since $gcd(|x|,|y|)=1$, then by the Bézout's identity there are $\alpha,\beta\in\mathbb{Z}$ such that
$$\alpha\cdot |x|+\beta\cdot |y|=1$$
Therefore
$$x=x^{1}=x^{\alpha\cdot |x|+\beta\cdot |y|}=x^{\beta\cdot |y|}=(x^{|y|})^\beta$$
and so we can generate $x$ from $x^{|y|}$. Meaning we can generate $(x,e_H)$ from $(x^{|y|},e_H)$ and thus from $(x,y)$. This completes the proof. $\Box$

Corollary. $\mathbb{Z}_2\times\mathbb{Z}_5$ is generated by $(1,2)$ element.

which I leave as a simple application of the Lemma above.
A: Hint
In a direct product,  the order of any element $(x,y)$ is the $\rm lcm(\lvert x\rvert, \lvert y\rvert).$
