Let $G$ be a finite group such that all subgroups of it (except $G$) are nilpotent. Then prove that $G$ is solvable.
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Here is an outline for a proof. To prove this, we can apply the following
I will not prove the lemma here. Let $G$ be a non-nilpotent finite group such that every proper subgroup of $G$ is nilpotent. A group $G$ with this property is often called a minimal non-nilpotent group.
To prove that $G$ is solvable, it suffices to show that $G$ is not simple (prove this). Hoping to find a contradiction, we assume that $G$ is simple. We can see that there are at least two maximal subgroups of $G$, let $D = K \cap L$ be an intersection of two maximal subgroups $K \neq L$ such that $D$ has largest possible order.
Next you should show that $D$ is normal in $G$. Since $G$ is simple, this implies that $D$ is trivial. But this is in contradiction with the lemma, so $G$ cannot be simple.
To prove that $D$ is normal, show $K$ is the only maximal subgroup containing $N_K(D)$ and $L$ is the only maximal subgroup containing $N_L(D)$. Hence $N_G(D)$ is not contained in a maximal subgroup and $N_G(D) = G$ since $G$ is finite.
This proof is the one given in Derek Robinson's group theory book, which contains a detailed proof (the lemma is also proven) and a couple of more facts about "minimal non-nilpotent groups". For example, with a little more work you can show that the order of $G$ has exactly two prime divisors and that exactly one Sylow subgroup of $G$ is normal. These results were first proven by O. J. Schmidt in 1929.