0

I couldn't figure out how to solve this. I found another explanation for the same questions here and I didn't understand the hints.

  • For the first coordinate I see that it has to be 0 or 1 being mod 2, so this will create $\mathbb{Z}_2$
  • For the second coordinate we could also create a $\mathbb{Z}_2$ , but the answer is that $\mathbb{Z}\times\mathbb{Z}/\langle(2,2)\rangle$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}$

So I couldn't see how we get that the second coordinate is from $\mathbb{Z}$.

Any help is greatly appreciated.

Classifying the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$

Community
  • 1
Impermanence
  • 161
  • In the answers two explicit isomorphisms are shown. The quotient can't be torsion, because it has rank $1=2-1$. – egreg Nov 07 '14 at 23:19
  • @egreg: I get the rank part but lost on the prove part. Can you also help explain how to show mu+nv is a subgroup? – Impermanence Nov 08 '14 at 01:43

2 Answers2

2

$a=(1,1)$ and $b=(1,0)$ form a free generating set of ${\mathbb Z}^2$. Since the subgroup you are factoring out is $\langle 2a \rangle$, the quotient group is isomorphic to ${\mathbb Z}_2 \oplus {\mathbb Z}$.

Derek Holt
  • 96,726
1

You're right that we could make the first coordinate be an element of $\mathbb{Z}_2$, and we could make the second coordinate an element of $\mathbb{Z}_2$. However, we can't do both of these things at the same time.

For a concrete example, let's looks at $(5,17)$. The coset of $\mathbb{Z} \times \mathbb{Z} / \langle(2,2)\rangle$ containing $(5,17)$ also contains $(1, 13)$ and $(-11, 1)$. More generally, if $(x,y) \in \mathbb{Z} \times \mathbb{Z}$, the coset containing $(x,y)$ will have elements of the form $(x+2n,y+2n)$ for some integer $n$. We could certainly pick $n$ to make first component be either $0$ or $1$, but at that point the second component $y + 2n$ could be any integer.

(I'll point out that we could have just as well chosen to make the second component $0$ or $1$, so that the first component ranged over the integers. This is because $\mathbb{Z}_2 \times \mathbb{Z} \cong \mathbb{Z} \times \mathbb{Z}_2$.)

  • My confusion: y+2n can be anything but once it has been mod 2 then that would result 1 or 2 also. When you said we cant do both at the same time, is there some rules or logic to this? – Impermanence Nov 08 '14 at 06:07
  • 1
    Well, if you have some element $(x,y)$ in a coset of $\langle(2,2)\rangle$, you can form an element of that coset by adding a multiple of $2$ (the same multiple of $2$) to both components.

    Can you add the same multiple of $2$ to $5$ and $17$ to make them both be either $1$ or $2$?

     –  Nov 08 '14 at 19:12