Let $G$ be a group and $H\unlhd G$. Prove that if $G/H$ is a simple group, then $G\neq H$ and $\not\exists K$ s.t. $H\leq K\leq G,K \unlhd G.$

Question

Let $G$ be a group and let $H \trianglelefteq G$.

Prove that

$G/H$ is a simple group $\Longrightarrow$ $G\neq H$ and there doesn't exist $K$ such that $H\leq K \leq G, K \trianglelefteq G.$

If I suppose $G=H,$ then $G/H=H/H=\{H\}$ and this contradicts the fact that $G/H$ is a simple group.

But I cannot prove there doesn't exist $K$ such that $H\leq K \leq G, K \trianglelefteq G.$

If I suppose such $K$ exists, how can I lead a contradiction ?

Answer 1

I think you meant $H < K < G,,$ since of course $K=H$ and $K=G$ are both normal.

Anyways, $G/H$ is a simple group means that it has no non-trivial normal subgroups. If $K$ has $H < K < G$ and $K$ normal in $G,$ then $K/H$ is a nontrivial normal subgroup of $G/H,$ which is absurd (it is normal by the second isomorphism theorem).

Answer 2

Suppose $K$ exists. Then show that $K/H = \{kH \in G/H \ | \ k \in K\}$ violates the condition that $G/H$ is simple.