Example of $H$, $N$ two normal subgroups of a group G such that $G/N \cong G/H$ but $N \ncong H$

Question

I have tried to prove that if $H$, $N$ are two normal subgroups of a group G such that $G/N \cong G/H$, then $N \cong H$. I think that it is not possible, but I can't find a counterexample. What is an illustrative example for to prove that $N \ncong H$?

Answer 1

Let $G=\mathbb Z/4\mathbb Z\times \mathbb Z/\mathbb 2\mathbb Z$. Since $G$ is abelian, every subgroup is normal. Let $N_1=\mathbb Z/4\mathbb Z$ be the first direct factor in the definition of $G$, and let $N_2=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ come from taking the direct product of a subgroup of the first factor and the second. Clearly, $N_1\not\cong N_2$. However, $G/N_1$ and $G/N_2$ are both groups of order $2$, and thus isomorphic.

Answer 2

How about $$G = \mathbb Z \times \mathbb Z \times \mathbb Z \times \cdots$$

$$H = \mathbb Z \times \{0\} \times \{0\} \times \cdots$$

$$N = \mathbb Z \times \mathbb Z \times \{0\} \times \cdots$$

Then $H$ and $N$ are not isomorphic, but $G/N \cong G \cong G/H$.