# Let $K\lhd G$ be s.t. both $K$ and $G/K$ are simple. Show that either $K$ is the only proper normal subgroup of $G$, or $G \cong K \times (G / K)$.

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.

• I think that here a proper normal subgroup cannot be either of the trivial normal subgroups ($\{1\}$ or $G$). After all, the statement is to apply in the situation $G=S_5$, $K=A_5$ as well, when we don't have $G\simeq K\times (G/K)$, but still $K$ and $G/K$ are both simple. Commented Feb 4 at 4:44
• Just assume the case where $K$ is not the only non-trivial normal subgroup. If $N$ is another, what can you say about $K\cap N$, $KN$ etc? Commented Feb 4 at 4:46
• The trivial group is not simple, so we cannot have $K=\{1\}$. Commented Feb 4 at 8:58

## 1 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.