Quotient group with $\mathbb{Z}_2$ and $\mathbb{Z}_4$

Question

I recently published another question about quotient groups. In this case, I was practising and trying to find the elements of the group $\mathbb{Z}_2 \times \mathbb{Z}_4 / \langle(1,1)\rangle$. The progress made is the following:
Using Lagrange's theorem, since the group $\mathbb{Z}_2 \times \mathbb{Z}_4$ has 8 elements and $\langle (1,1)\rangle = \{ (1,1), (0,2), (1,3), (0,0) \}$ has 4 elements, there must be two elements inside the quotient group. One element is the equivalence class given by $[0]$, which is $\langle(1,1)\rangle$ itself. However, when adding $(1,2)$, which is not in the cyclic group $\langle(1,1)\rangle$ and $(0,1)$, which is neither in that group, one obtains $(1,3)$, which is in the cyclic group. Is there something wrong in my reasoning or the two elements of the quotient group are just the subgroup itself one and the other the rest of the elements? Thanks for your attention.