$\bullet$ Find an element of order 2 in $G$ and an element of infinite order
$\bullet$ explain why there is only one-nontrivial element of finite order
$\bullet$ To which group is $G$ isomorphic
I looked up similar problems but I'm just getting more confused, If the question would find an element of order 2 in $ \mathbb{Z_n} \oplus \mathbb{Z_n}/ \langle(4,2)\rangle $ for some $n$ it's straight forward but now I'm not sure how an element of order 2 in $G$ looks like I think its $(3,1)$ , but I'm not sure.
Why is there only one element of finite order, doesn't any element $(k,k)$ with $k$ being a prime $>2$ in $ \mathbb{Z} \oplus \mathbb{Z}/ \langle(4,2)\rangle $ have infinite order in the Group.
Best Answer
$v_1=(2,1)$ has order $2$ in $G$.
$v_2=(1,0)$ has infinite order in $G$.
$n(a,b)$ is $0$ in $G$ iff $(na,nb)=m(4,2)=(4m,2m)$ iff $a=2b$.
$G \cong \mathbb Z \times \mathbb Z_2$ because $\mathbb Z \oplus \mathbb Z = v_1 \mathbb Z \oplus v_2\mathbb Z$.