Prove or give a counterexample: If $H, K$ are normal subgroups to $G$ and $G/H$ is isomorphic to $G/K$, then $H$ is isomorphic to $K$.
proof:
Let $G$ be the Klein-4 Group ($V$), $H = \langle a\rangle$, and $K = \langle b\rangle$. $H$ and $K$ are normal subgroups to $G$ since $G$ is abelian.
Then $|G/H| = |V/\langle a\rangle| = 4/2 = 2$ and $|G/K| = |V/\langle b\rangle| = 4/2 = 2$. So $G/H$ is isomorphic to $G/K$.
But $H$ is not isomorphic to $K$ since $a$ does not equal $b$.
Consider a countably infinite product of copies of $\Bbb Z/2\Bbb Z$, and take $\Bbb Z/2\Bbb Z$ and $(\Bbb Z/2\Bbb Z)^2$. Another example is that $S^1/\langle e^{2\pi i/n}\rangle \simeq S^1$ for any $n$.