Prove that $G$ has at most $2$ normal subgroups

Question

Problem
Let $G$ be a non-abelian group satisfying the following condition.

"$N_1\nsubseteq N_2$ for all non-trivial normal subgroups $N_1\neq N_2\subset G$"

Show that $G$ has at most $2$ normal subgroups.

I found that if $G$ has two non-trivial normal subgroups $N_1\neq N_2, G=N_1\times N_2$ because $N_1\cap N_2=1, N_1N_2=G$ from the condition. But I don't know the next step.

Answer 1

We may assume $N_1$ is non-abelian, since $G$ is non-abelian.

Assume $N_3$ is the third non-trivial normal subgroup of $G$, then as you did, $G = N_1\times N_3$ and $G = N_2\times N_3$. In particular, $N_3$ is commutative with both $N_1$ and $N_2$.

Now suppose $(g_1,g_2)\in N_3\le G$ with $g_1\in N_1$ and $g_2\in N_2$. Since $N_3$ is commutative with $N_1$, we have $g_1\in Z(N_1)$. But $Z(N_1)$ is a characteristic subgroup of $N_1$, and thus a normal subgroup of $G$, it must be trivial or the whole $N_1$. By assumption, $N_1$ is non-abelian and so $Z(N_1) =1$. Hence, $g_1 = 1$. This gives $N_3 = 1$ or $N_3 = N_2$, a contradiction.