How is this a counter example to " $A$ and $B$ are isomorphic but $G/A \ncong G/B"$? If $G = \mathbb{Z}_2+ \mathbb{Z}_4$, and $A=\langle(1, 0)\rangle$ and $B=\langle(0, 2) \rangle$, how does it follow $G/A \ncong G/B$. How do we know $A$ and $B$ are normal subgroups of $G$? 

Note that $G$ is an abelian group so $A$ and $B$ are normal in $G$.Further note that 
$G/A\cong\mathbb{Z} / 4 \mathbb{Z} $ and $G/B\cong\mathbb{Z} / 2 \mathbb{Z} +\mathbb{Z} / 2 \mathbb{Z} $. Hence they are not isomorphic!
