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I will assume that $G$ is finite.

If all proper subgroups of $G$ are nilpotent, believe that it can be proven quite easily that $G$ is soluble. But since I don't want to write the details of the proof I will assume it.(See Prove that if $G$ is a finite group in which every proper subgroup is nilpotent, then $G$ is solvable.)

I believe that the following statement is false but I don't know where the argument goes wrong:

If all proper subgroups of $G$ are nilpotent and $G$ is solvable, then $G$ is nilpotent.

In order to show that $G$ is nilpotent, I want to show that all Sylow subgroups are normal.

Let $M$ be a maximal subgroup of $G$, since $G$ is soluble,we have that $|G:M|=p^a$ for some $p$ prime and some $a\in \mathbb{N}$.

So for any prime $q\not=p$, we have that $Syl_q(G)=Syl_q(M)$. Since $M$ is nilpotent there is only one Sylow-subgroup $Q$ in $M$, and so in $G$. We have that $Q$ is the unique Sylow subgroup of $G$, so $Q$ is normal in $G$. This holds for all primes $q\not=p$.

Let $P\in Syl_p(G)$. Then $P$ is contained in some maximal subgroup of $G$. Again since $G$ is soluble we have that $|G:M|=q^b$ for some $q$ prime and some $b\in \mathbb{N}$. As before $Syl_p(G)=Syl_p(M)=\{P\}$. Which implies that $P$ is normal in $G$.

Please let me know where I am wrong, thanks!

Arturo Magidin
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Gillyweeds
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    If your statement were true, all finite soluble groups would be nilpotent. – David A. Craven Oct 22 '21 at 13:45
  • I know my statement is false. My question is where does the proof fail? – Gillyweeds Oct 22 '21 at 13:51
  • $S_3$ is soluble but not nilpotent. Try seeing what happens for $S_3$. – Sean Eberhard Oct 22 '21 at 14:02
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    So for any prime $q\neq p$, we have that $Syl_q(G)=Syl_q(M)$. This is false. – David A. Craven Oct 22 '21 at 14:03
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    (It's only true if $M$ is normal in $G$.) – David A. Craven Oct 22 '21 at 14:11
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    The smallest counterexample is, of course, $S_3$, and it is easy to run your argument through it and see where it goes wrong. Taking $M$ to be a $2$-subgroup, you claim the $2$-Sylows of $S_3$ are the same as the $2$-Sylows of $M$... which is plainly incorrect. If $M$ is not normal (as I now see David Craven has noted), then any conjugate of $M$ will contains Sylows that are not contained in $M$ (pick an element of $gMg^{-1}$ that is not in $M$ of prime order, and take a Sylow of $gMg^{-1}$ that contains int). – Arturo Magidin Oct 22 '21 at 14:17
  • @DavidA.Craven And this could then be proven by induction on the order of the group. – Geoffrey Trang Oct 22 '21 at 14:42

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