Non trivial counterexample: $H_1\cap H_2$ is normal subgroup $\not\Rightarrow H_1$ and $ H_2$ are normal.

Question

I know that if $H_1, H_2$ are normal subgroups then $H_1 \cap H_2$ is normal.

The converse is not true, e.g. $H_1 = \langle(1, 2)\rangle, H_2 = \langle(3, 4)\rangle $ (permutations) are not normal, but $H_1 \cap H_2 = \{ e \}$ is normal.

Are they any non trivial ($|H_1\cap H_2| > 1$) counterexamples?

Answer 1

Take $H_1, H_2 \subset S_4$ as in your example. Now, define $G := S_4 \times H$, where $H$ is your favorite non-trivial group. Set now $H_1' := H_1 \times H$, $H_2' := H_2 \times H$. Then $H_1' \cap H_2' = \{e\} \times H$ is normal in $G$, but $H_1'$ and $H_2'$ are not normal in $G$, since $H_1, H_2$ are not normal in $S_4$.