# $G$ is a group with a normal subgroup $K$ such that $G/K$ is soluble, and $H$ is a nonabelian simple subgroup of $G$, then $H \leq K$

I'm trying to see why the following theorem is true:

If $$G$$ is a group with a normal subgroup $$K$$ such that $$G/K$$ is solvable, and $$H$$ is a nonabelian simple subgroup of $$G$$, then $$H \leq K$$.

My attempt:

As $$K \lhd G$$ we can construct the normal series: $$\{e\} \lhd K \lhd G.$$

We also know that $$G/K$$ has a finite composition series with all factors prime cyclic (simple abelian) as it is solvable.

Since any finite group has a composition series, we can write the composition series for $$G$$ as:

$$\{e\} = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G.$$

Suppose $$G_k = K$$ but since it's not mentioned that $$K$$ is solvable, we don't know whether a composition series of $$K$$ will have prime cyclic (simple abelian) factors. So we cannot suppose $$G_k = K$$.

I'm not sure how to proceed from here (?)

Hint: look at the image of $$H$$ in $$G/K$$. What can be said about a group that is both simple and solvable?
The image of $$H$$ in the quotient $$G/K$$ is $$HK/K \cong H/(H \cap K)$$. But $$H$$ is simple and $$H \cap K \unlhd H$$. So either $$H \cap K=1$$ or $$H \cap K=H$$ and the latter is equivalent to $$H \subseteq K$$. If $$H \cap K=1$$ then $$HK/K \cong H$$ and is solvable since $$G/K$$ is. So $$H$$ is both simple and solvable, that $$H$$ is cyclic of prime order, contradicting the assumption on $$H$$.
• $K$ is certainly not maximal in $G$ so $G/K$ can't be simple – S.D. May 11 '20 at 14:40