Group isomorphisms don't preserve everything?

Question

Consider the following claim:

Suppose $G$ and $H$ are groups such that $G \simeq H.$

If $N \triangleleft G$ and $ K \triangleleft H $ are normal subgroups of $G,H$ such that $ N \simeq K $, then $G/N \simeq H/K$.

it seems to me that it should be true, since it looks like a pretty basic property of groups, but $G=H=N=\mathbb{Z}$ and $K=2\mathbb{Z} $ is a counterexample.

Doesn't it show that isomorphism is somewhat a bad tool for researching groups?

Answer 1

You have already found a nice counterexample, so the claim cannot be true. Why is that?

A way to answer this could be by saying that: $G\cong H$ is not enough, we need to know what the isomorphism is.

If you have a specific isomorphism $f:G\to H$, then any normal subgroup $N$ of $G$ is sent via $f$ to a normal subgroup $K$ of $G$, and then we do have $G/N\cong H/K$. Without knowing this relationship between $H$ and $K$, nothing much can be said.

So an idea to keep in mind would be that existence of an isomorphism between two groups is good, but a specific isomorphism is better.

Nevertheless, isomorphisms are still a good tool to study groups, since many properties (abelian-ness, solvability, representation theory, etc.) only depend on the isomorphism class of a group.

Answer 2

I will transform my comment into an answer. You have two inclusion morphisms $\iota_N:N\hookrightarrow G$ and $\iota_K:K\hookrightarrow H$ and two isomorphisms $\phi: G\to H$ and $\psi: N\to K$. To conclude that $G/N\cong H/K$, you need to have $\phi\circ\iota_N=\iota_K\circ\psi$.

This is certainly reasonable, since $G/N$ is the cokernel of the morphism $\iota_N$, so it is essentially an object associated to this morphism. In order to have compatibility between the two objects, you need the stated compatibility of the morphisms.