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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 (?)

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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 $H$.

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  • $\begingroup$ Sorry, a bit of elaboration would be helpful $\endgroup$ – S.D. May 11 '20 at 14:21
  • $\begingroup$ Which one is the simple and solvable group here? $\endgroup$ – S.D. May 11 '20 at 14:29
  • $\begingroup$ I know that a simple and solvable group is cyclic, but I'm not sure how this is relevant here $\endgroup$ – S.D. May 11 '20 at 14:38
  • $\begingroup$ $K$ is certainly not maximal in $G$ so $G/K$ can't be simple $\endgroup$ – S.D. May 11 '20 at 14:40
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    $\begingroup$ Saw that you posted this in a new question. Well done. Apologies if I did not take your level of group theory knowledge into account. Never stop asking! $\endgroup$ – Nicky Hekster May 11 '20 at 17:40

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