Let $G$ be a group of order $p_1...p_s$ where $p_1,...,p_s$ are distinct primes. If $G$ has a normal $p$-Sylow subgroup, then $G$ is solvable.

We proceed by induction on $s$. If $s = 1$, $G$ is cyclic, so $G$ is abelian and therefore it is solvable. If $H$ is a normal $p_i$-Sylow of $G$, then $G/H$ has order $p_1...p_{i-1}p_{i+1}...p_s$. Using the induction hypothesis, $G/H$ is solvable. Now $H$ is abelian, so it is solvable and so $G$ is solvable.

Is this correct? I'm not entirely sure about the induction step; shouldn't we need a normal $p$-Sylow subgroup for $G/H$?

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    $\begingroup$ Yes, the assumptions are not strong enough for an induction proof. The conclusion is true, but that is a more advanced statement. $\endgroup$ – Tobias Kildetoft Mar 5 at 18:28
  • $\begingroup$ @TobiasKildetoft Thank you. Is there a way I can complete the proof using the induction? $\endgroup$ – Michele Mar 5 at 18:34
  • $\begingroup$ I don't really see one, unless you were to assume that any group of squarefree order had a normal Sylow subgroup (this is true, but non-trivial to prove). $\endgroup$ – Tobias Kildetoft Mar 5 at 18:38
  • $\begingroup$ As Tobias said, once you quotient by the normal subgroup, you are just left with an arbitrary group of squarefree order, with no additional hypothesis, which is weaker than your original hypothesis. So that extra hypothesis is not helpful. $\endgroup$ – verret Mar 5 at 19:02
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    $\begingroup$ If there were a non-solvable group $N$ of square-free order, then there would also be one with square-free order and a normal $p$-Sylow subgroup. Just choose a prime $p$ not dividing the order of $N$ and consider the group $N\times(\mathbb Z/p)$. $\endgroup$ – Andreas Blass Mar 5 at 21:33

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