# Identifying the quotient class of infinite group under the classification of finitely generated abelian groups.

The exact problem asks to Identify the isomorphism class of the quotient group $$\mathbb{Z}\times\mathbb{Z}/\langle(2,2)\rangle$$ within the classification of finitely generated abelian groups.

I'm trying to find the cosets, but I can't understand what they might look like outside a finite group. I know that $$\langle(2,2)\rangle = \{..., (-2,-2), (0,0), (2,2), (4,4), ...\}$$ in $$\mathbb{Z}\times\mathbb{Z}$$, so I started with $$(1,n)$$ and $$(0,m)$$ for any $$n,m$$ as coset representatives.

Using those two, I tried to solve for the finite order cosets $$(xi, ni)=(2j,2j)$$ where $$x=0,1$$ and I only found the identity coset for $$x=1$$ and the coset $$(1,1)+\langle(2,2)\rangle$$ for $$x=1$$. However it appears to me that this has order $$2$$.

My intuitive guess is the quotient group is isomorphic to $$\mathbb{Z}\times\mathbb{Z_2}$$, especially since the other finite coset seems to have order $$2$$. I really can't grasp these concepts around infinite groups. I don't know where I'm going wrong, but ultimately a general explanation of what is going on would be preferred so I can attempt to apply it myself in this context.

• Try thinking like you would for vector spaces, and write down a different "basis" for $\mathbb{Z} \times \mathbb{Z}$, which plays well with the element $(2, 2)$. A strong hint: every element can be uniquely written as $a(1, 1) + b(1, 0)$ for $a, b \in \mathbb{Z}$. – Joppy Oct 1 '20 at 22:52
• $2$ is prime... – Qiaochu Yuan Oct 1 '20 at 22:53
• @QiaochuYuan oh yea. so $\mathbb{Z}\times\mathbb{Z_2}$ does work then, so would that be the answer or am I still doing something wrong. – Jack Oct 1 '20 at 22:56

A presentation for $$\Bbb Z\times \Bbb Z$$ is
$$\langle a,b\mid ab=ba\rangle,\tag{1}$$
where $$a\mapsto (1,0)$$, say, and $$b\mapsto (0,1)$$; in which case $$(2,2)$$ corresponds so $$(ab)^2$$. The quotient by $$\langle (2,2)\rangle$$ is, in effect, the same as killing $$(ab)^2$$ in $$(1)$$, like so: let $$c=ab$$; then:
\begin{align} \Bbb Z\times \Bbb Z/\langle (2,2)\rangle&\cong \langle a,b, c\mid (ab)^2, c=ab=ba\rangle\\ &\cong \langle a,b, c\mid c^2, c=ba\rangle\\ &\cong \langle a,b,c\mid c^2, cb=bab\rangle\\ &\cong \langle b,c\mid c^2, cb=bc\rangle\\ &\cong \Bbb Z\times \Bbb Z_2, \end{align}
• In the right hand sides of your isomorphisms, the generators should be $a,b,c$ for the first three, and $b,c$ for the final one. – Derek Holt Oct 2 '20 at 7:42