Determine the invariant factors of $\mathbb{Z^2}$ I want to show that $(G,.)$ is a finitely generated abelian group and determine its invariant factor, where $G=\mathbb{Z}^2$ and the binary operation is $(x_1,y_1).(x_2,y_2):=(x_1+x_2,y_1+y_2+x_1x_2)$.
It is easy to show that “$.$” is associative and commutative. The identity element is $(0,0)$, and the inverse is $(-x,x^2-y)$. Thus, $G$ is an abelian group.
Furthermore, $G=\langle (1,0),(0,1) \rangle$, so it is finitely generated.
What I am struggling with is the invariant factors part. I know that it would be something like
$\mathbb{Z}/n_1 \times … \times \mathbb{Z}/n_s \times \mathbb{Z}^r$
Where $n_1,…,n_s$ are the invariant factors and $r$ is the free rank. However, I am not sure how to do that. Any help would be appreciated.
 A: We have a short exact sequence
$$0 \to \mathbb{Z} \overset{y \mapsto (0, y)}{\longrightarrow} G \overset{(x, y) \mapsto x}{\longrightarrow} \mathbb{Z} \to 0.$$
Moreover, since $\mathbb{Z}$ is a free abelian group, this sequence automatically splits.  Therefore, $G \simeq \mathbb{Z}^2$.
A: For a response that doesn't use exact sequences (although this is the simplest way to get the answer if you have learned them):
First, note that there is no torsion, since the first coordinate of $n(x,y)$ is $nx$, and thus any torsion point must have $x = 0$. But if $x = 0$, then the second coordinate of $n(x,y)$ is $ny$, as can be seen by induction. Thus $y = 0$ as well, so the identity is the only point of finite order.
Next, check that $n(0,1) = (0,n)$ and $n(1,0) = (n, (n(n-1))/2)$ for any $n \in \mathbb{Z}$. Thus if $n(1,0) = (0,m)$ for any $n$, we must have that $n = 0$, so $m = 0$, and the subgroups $\langle (0,1) \rangle$ and $\langle (1,0) \rangle$, intersect trivially. Since, as you noted, $(0,1)$ and $(1,0)$ generate $G$, we get that $G \cong \langle (0,1) \rangle \times \langle (1,0) \rangle \cong \mathbb{Z} \times \mathbb{Z}$.
