When is $G/N \cong H/K$?

Suppose the following:

1. $G \cong H$
2. $N \trianglelefteq G$
3. $K \trianglelefteq H$
4. $N \cong K$

What is the requirement that $\alpha: G/N \to H/K$ is an isomorphism?

I found a counterexample like for $G = \mathbb{Z} = H$ and $N = 2\mathbb{Z}$ and $K = 3\mathbb{Z}$, but are there such groups that work?

EDIT:

I just practiced with Overleaf, and I want to know if it's correct. I made the detailed proof with what you guys helped me in the comments. I'm a beginner doing Overleaf papers, so please let me know if you see any mistakes. Here is the link.

Solution 1:

Since $H$ and $ G $ are isomorphic, there exist an isomorphism $\varphi $ between them. Consider the homomorphism $ \pi:G\to H/_{K} $ defined by $$ \pi\left(g\right)=\varphi\left(g\right)K $$

Now,a sufficient condition for what you desire would be that $ \text{ker}\left(\pi\right)\subseteq N $. Because then we can define $ \psi:G/_{N}\to H/_{K}$ by: $$ \psi\left(gN\right)=\pi\left(g\right) $$This homomorphism is obviously surjective, and if $ \psi\left(g_{1}N\right)=\psi\left(g_{2}N\right) $, then $ \pi\left(g_{1}^{-1}g_{2}\right)\in K $ so that $ g_{1}^{-1}g_{2}\in\ker\pi\subseteq N $.

Now, this implies that $ g_{1}N=g_{2}N $ so that $\psi $ is injective and surjective and thus isomorphism