I was asked to find the order of the elements in $G/H$ where $G=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ and $H=<(0,2)>$.
I found $G/H$ to be $\{H,(0,1)+H,(1,0)+H,(1,1)+H\}$ which I think is right.
Then I proceeded to find the order of the elements in $G/H$.
$H$ is of order $1$.
$(0,1)+H$ is of order $4$ because $[(0,1)+H]+[(0,1)+H]+[(0,1)+H]+[(0,1)+H]=(0,4)+H=H=e$
$(1,0)+H$ is of order $2$ because $[(1,0)+H]+[(1,0)+H]=(2,0)+H=H=e$
$(1,1)+H$ is of order $4$ because $[(1,1)+H]+[(1,1)+H]+[(1,1)+H]+[(1,1)+H]=(4,4)+H=H=e$
According to the solutions though $(0,1)+H$ and $(1,1)+H$ are of order $2$. Can someone tell me where I have gone wrong?