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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 $\mathbb{Z} \times \mathbb{Z} \big / \langle (1,2) \rangle \cong \mathbb{Z}$
Shaun
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Shauna
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    $\langle (1,2) \rangle$ is the group generated by $(1,2)$.. it is isomorphic to $\mathbb{Z}$, so you seem to have a confusion on that judging by what you have written in the question you have posed .. one way to come up with explicit isomorphisms of factor groups is to construct a homomorphism which is surjective with kernel, the subgroup you want to quotient out .. like here $\mathbb{Z}\times \mathbb{Z} \rightarrow \mathbb{Z}, ,\ (a,b) \mapsto b -2a$ – skeptic Jun 01 '21 at 07:32

2 Answers2

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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.

Richard Jensen
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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 $(a,b)\mapsto(a, 2a-b)$.

user26857
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