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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

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What you have proved is that $12$ is a multiple of the order of $(1,10)$. You still need to prove that the order is not some proper divisor of 12 -- which, however, should be easy. –  Henning Makholm Mar 22 '12 at 18:31
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You have $(1,10)+\ldots+(1,10)=(k,...)$ if $k$ is the number of addends. Can this be $(0,0)$ if $k\in\{1,...,11\}$? –  Andrea Mori Mar 22 '12 at 18:35
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1 Answer

The order of $(a,b) \in G \times H$ is the least common multiple of the orders of $a$ and $b$. [Why? If $(a,b)^k=(a^k,b^k)$ If $(a,b)^k=(e_G,e_H)$ (the identity), then $a^k=e_G$ so $k$ must be a multiple of the order of $a$ and likewise a multiple of the order of $b$. Thus $k$ is a common multiple of the orders. Conversely if $k$ is a common multiple of the orders $(a,b)^k=(a^k,b^k)=(e_G,e_H)$. Thus the order of $(a,b)$ is the smallest (i.e. least) common multiple of the orders of $a$ and $b$.]

So $|(1,10)| = \mathrm{lcm}(|1|$ in $\mathbb{Z}_{12},|10|$ in $\mathbb{Z}_{40})=\dots$

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