Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it's infinite.

The four cosets are listed as $(0,0) + \langle (2,2) \rangle$, $(0,1) + \langle (2,2) \rangle$, $(1,0)+ \langle (2,2) \rangle$ and $(1,1) + \langle (2,2) \rangle$. However, $(2,0)$ doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong.

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    $\begingroup$ It is infinite, it contains a copy of $\mathbb{Z}$ per $(k,0)$, but it is not cyclic, as $(1,1)$ has order $2$. $(\mathbb{Z}\oplus\mathbb{Z})/(\langle 2\rangle \oplus\langle 2\rangle)$ on the other hand... $\endgroup$ – Daniel Fischer Oct 25 '13 at 11:58
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    $\begingroup$ Answer given to you would be correct for $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,0),(0,2) \rangle$ $\endgroup$ – Adam Oct 25 '13 at 12:02

You are right, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ is infinite. You can embed $\mathbb{Z}$ via $k\mapsto (k,0)$ (and in other ways) into it. The quotient is not cyclic, because it contains elements of finite order, $(1,1)$ for example.

Probably it was meant to be $(\mathbb{Z} \oplus \mathbb{Z})/ (\langle 2\rangle\oplus \langle 2 \rangle)$ which indeed is a group of order $4$ (a Klein $4$-group).

  • $\begingroup$ Just a quick question to see if I am "seeing" the groups in this question correctly: The group $\mathbb Z \oplus \mathbb Z$ contains $3$ copies of $\mathbb Z$ -- one looks like $(k,0)$, one like $(0,k)$ and one like $(k,k)$. Then taking the quotient by $\langle (2,2) \rangle$ eliminates half of the copy $(k,k)$ (namely the even numbers). So what remains is all elements of $\mathbb Z \oplus \mathbb Z$ minus all elements of the form $(2n,2n)$? $\endgroup$ – Rudy the Reindeer Feb 28 '14 at 19:27
  • $\begingroup$ $\mathbb{Z}\oplus\mathbb{Z}$ contains more copies of $\mathbb{Z}$; everything of the form $(k,m\cdot k)$ or $(m\cdot k,k)$ gives an embedding of $\mathbb{Z}$ and of course you have the subgroups of them. The "diagonal" $(k,k)$ is mapped to a subgroup $\cong C_2$ of the quotient. The entire quotient is $\cong \mathbb{Z}\oplus C_2$, that's probably easiest seen via the automorphism $(k,m) \mapsto (k-m,m)$, which transports it to $(\mathbb{Z}\oplus\mathbb{Z})/\langle(0,2)\rangle$. As is, you have $(k,2m+r) \equiv (k-2m,r)$ which gives you a nice set of representatives. $\endgroup$ – Daniel Fischer Feb 28 '14 at 19:45
  • $\begingroup$ Ooh, you are right, I missed lots of copies of $\mathbb Z$. Thank you very much for your patient comments. I do not fully understand your comment somehow but it gave me the idea on how to prove that the image is $\mathbb Z \oplus \mathbb Z_2$: define a map $\mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus\mathbb Z$ that maps $\langle (0,2) \rangle$ to $(0,0)$ and whose image is $\mathbb Z \oplus \mathbb Z_2$. Like e.g. $f((n,k)) = (n, k \mod 2)$. My only problem now is that I don't know how to figure out what the image looks like. How did you figure out it's $\mathbb Z \oplus \mathbb Z_2$? $\endgroup$ – Rudy the Reindeer Mar 1 '14 at 9:19
  • $\begingroup$ A change of basis. If we use the basis $\{(1,0),(1,1)\}$, i.e. write $(k,m) = (k-m)\cdot(1,0) + m\cdot (1,1)$, the subgroup factored out is entirely contained in the second factor, and $(\mathbb{Z}\alpha \oplus \mathbb{Z}\beta)/\langle 2\beta\rangle$ is easily recognised as (isomorphic to) $\mathbb{Z}\oplus \mathbb{Z}_2$. $\endgroup$ – Daniel Fischer Mar 1 '14 at 11:36

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