Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)> \cong \mathbb{Z}_{12} = \mathbb{Z}_{4} \times \mathbb{Z}_{3}$?

I am not convinced at the least that $\mathbb{Z}_{12}$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$

For instance, doesn't $<1>$ have an order of 12 in $\mathbb{Z}_{12}$? And no element of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$ can even have an order of $12$ no?

What is the maximum order of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$? I know if the denominator isn't there, it is $lcm(4,6) = 12$, but even if it weren't there, I don't see how either component can produce an element of order 12.

What I mean is that the first component is in $\mathbb{Z}_4$, so all elements have max order 6 and likewise $\mathbb{Z}_6$, have order 6, so how can any element exceed the order of their group?

share|improve this question
I corrected in your other post, but please note that you should use \langle and \rangle to produce $\langle$ and $\rangle$ rather than $<$ and $>$. I will leave it to you to edit the post. –  JavaMan Dec 11 '12 at 6:50

1 Answer 1

up vote 2 down vote accepted

HINT: Look at $x=(1,1)\in\mathbb{Z}_4\times\mathbb{Z}_6$. What is its order? If necessary, write out $nx$ for all $x$ until $nx=(0,0)$.

share|improve this answer
A stupid question, but how did you come up with (1,1) to counter example? –  sidht Dec 11 '12 at 7:15
@sizz: To answer the question in your comment. What's the order of $x=(a,b)$ here? To answer that question you must find the smallest integer $n$ such that $n(a,b)=(na,nb)=(0,0)$. In other words you want $na$ to be divisible by four and $nb$ to be divisible by six. Now we're trying to find an element $x$ of as large an order as possible. In other words we want to choose $x$ in such a way that it is as difficult as possible for both these divisibility conditions to be simultaneously true. Then the number $a$ and $b$ should not assist $n$ in getting the desired divisibility. Ergo, pick $a=b=1.$ –  Jyrki Lahtonen Dec 11 '12 at 8:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.