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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

  • 0
    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.2012-03-22
  • 1
    You have $(1,10)+\ldots+(1,10)=(k,...)$ if $k$ is the number of addends. Can this be $(0,0)$ if $k\in\{1,...,11\}$?2012-03-22

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