Sorry about the title, I couldn't fit the whole exercise (Exercise 8.1.6, Nicholson Introduction to Abstract Algebra 4th edition):
Let $K \triangleleft G$ be such that both $K$ and $G/K$ are simple. Show that either $K$ is the only proper normal subgroup of $G$, or $G \cong K \times (G / K)$.
I first noticed that if $K$ is the only proper normal subgroup of $G$, then $G$ is simple and $K = \{1\}$ (assuming $G$ isn't the trivial group, for which we can't even have such a $K$). Furthermore, since $G / K$ is simple, $K$ must be a maximal normal subgroup.
To start off I assumed that $K$ is not the only normal proper subgroup, so there exists another proper normal subgroup $N$. But I don't know where to from here. All I know is that I have to show that $G \cong K \times (G / K)$. The section is about products (not direct products) and one of the more significant theorems covered was the correspondence theorem, but I don't see any connections.
I've made some progress:
Suppose that $K$ isn't the only proper normal subgroup of $G$. Then there exists another proper normal subgroup $N \triangleleft G$. By Theorem 3, $N K / (N \cap K) \cong N / (N \cap K) \times K / (N \cap K)$. Since $K$ is simple, we have $N \cap K = K$ or $N \cap K = 1$. If $N \cap K = K$, then $K \subseteq N$. Since $K$ is maximal by Theorem 6, we have $N = K$ or $N=G$. Each case contradicts our hypothesis. Hence, $H \cap K = 1$, so $NK \cong N \times K$.
Theorem 3 states that if $N$ and $K$ are simple, then $N K / (N \cap K) \cong N / (N \cap K) \times K / (N \cap K)$.
Theorem 6 states that $K$ is a maximal normal subgroup if and only if $G/K$ is simple.
Best Answer
Yes the first step is to suppose that there is a proper normal subgroup $N\neq K$.
From the mere fact that $K$ is simple, as you have said, it automatically follows that $N\cap K$ is trivial.
From the fact that $G/K$ is simple, what does the image of $N$ look like under the obvious mapping from $G$ to $G/K$?
Once you understand the answer to this, all that is left is to show that every element of $N$ commutes with every element of $K$, but you have done that in your progress.