Compute factor group $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle$.

Question

Compute the factor group: $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle$

I don't understand why the answer is $\mathbb{Z}$. Isn't $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle \cong \mathbb{Z} / \langle 1 \rangle \times \mathbb{Z} / \langle 2 \rangle$?

1. What is the order of the cyclic groups $\langle 1 \rangle$ $\langle 2 \rangle$?
2. Why is $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle \cong \mathbb{Z}$

Answer 1

Your proposed identity $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle \cong \mathbb{Z} / \langle 1 \rangle \times \mathbb{Z} / \langle 2 \rangle$ is not true. The group on the right has order 2, while the group on the left has an infinite subgroup generated by $(1,0)$.

As usual when proving an identity of factor groups, it is easiest to find a surjective homomorphism with the group factored out as the kernel. We want elements of the form $n(1,2) = (n, 2n)$ to be mapped to $0$, so how about we try a function

$\phi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$

defined by

$\phi(a,b) = 2a-b$

I leave it to you to show that this is a surjective homomorphism with the right kernel.

Answer 2

The subgroup generated by $(1,2)$ can be "moved" by an automorphism of $\mathbb Z\times\mathbb Z$ into a simpler one: the subgroup generated by $(1,0)$. Why is this simpler? Since it can be written as a direct product, that is, $\langle(1,0)\rangle=\mathbb Z\times\{0\}$, and we can use the natural isomorphism of groups $$G_1\times G_2/H_1\times H_2\simeq G_1/H_1\times G_2/H_2.$$ This way we find out that our quotient group is isomorphic to $\mathbb Z/\mathbb Z\times\mathbb Z/\{0\}$ which, as its turn is isomorphic to $\mathbb Z$.

The last (or the first) step to figure out is the automorphism of $\mathbb Z\times\mathbb Z$ that does the job, and this is $(a,b)\mapsto(a, 2a-b)$.