# How can this isomorphism be valid?

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?

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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

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 $nx=(0,0)$.

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A stupid question, but how did you come up with (1,1) to counter example? –  Hawk 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

I presume you mean ring isomorphism. If so, you should define what is the 0, the 1, the sum and the multiplication. Of course, those are rather implicit. But there lies your doubts. As @KReiser points out, the unit in $\mathbb{Z}_4 \times \mathbb{Z}_6 / \langle(2,3)\rangle$ is (the class of) (1,1).

In order to prove the isomorphism to $\mathbb{Z}_{12}$, you can define the bijective function and show that it is an isomorphism of rings. That is, let $\Phi$ be the function that maps $1\in\mathbb{Z}_{12}$ to the class of $(1,1)\in \mathbb{Z}_4 \times \mathbb{Z}_6 / \langle(2,3)\rangle$. First, show that it is a morphism, that it is injective and surjective. Since both sets have the same amount of equivalence classes, they thus turn out to be isomorphic.

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