I'm trying to prove that the group $\mathbb{Z}_{3} \times \mathbb{Z}_{4}$ is isomorphic to the generated subgroup by $(2,20)$ and $(9,10)$ of $\mathbb{Z}_{12}\times\mathbb{Z}_{40}$ .
Here is my solution so far :
$(9,10)-4(2,20)=(9,10)-(8,80 \bmod 40)=(9,10)-(8,0)=(1,10)$
$(9,10)=9(1,10)=(9,90 \bmod 40)=(9,10)$
$(2,20)=2(1,10)$
Meaning , using the element $(1,10)$ we can generate the requested subgroup . Now , what is left is to prove that the order of $(1,10)$ is 12 (meaning $|(1,10)|=12$) :
Using the theorem : a group of order $N$ is cyclic iff it has an element of order $N$. Then I wrote the following WRONG solution :
$12(1,10)=(12,120)=(12 \bmod 12 ,120 \bmod 40)=(0,0)$ ......we mark it as $(**)$
So $(**)$ is incorrect .
Can someone please explain how I can prove that the order of $|(1,10)|=12$ ?
After that, the rest is easy since: $\gcd(4,3)=1\longrightarrow\mathbb{Z}_{12}\cong\mathbb{Z}_{3}\times\mathbb{Z}_{4}$
Regards