# If $H$ and $K$ are normal subgroups of $G$ and $H \cong K$, prove that $G/H \cong G/K$.

It is natural. But I want to prove exactly.

My idea is to use isomorphism theorem.

Let $f: G \to G/H$ be the quotient map, then $f$ is a homomorphism and obviously surjective.

Hence, I just need to show $\ker(f) = K$.

However, I couldn't find the way. Definitely, $\ker(f) = H$ and $H \cong K$

But I can say $\ker (f) = K$.

How to prove it?

This is not true. Consider $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, $H=\langle (1,0)\rangle$, and $K=\langle (0,2)\rangle$.
• Indeed, a classical example. Another one is $G = \mathbb{Z}, H = 2\mathbb{Z},K = 3\mathbb{Z}$. Jun 7, 2017 at 7:14