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Prove that every finite group whose order is square free is soluble.

I think it is enough to show that every sylow subgroup of this is cyclic.

Please tell me if my idea is right and if it is wrong please give me a little help,thanks.

user1729
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kpax
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  • It's clear that every Sylow subgroup is cyclic, but the result doesn't follow immediately from that step. Induct on the order of the group and use the fact that $G$ is solvable if $N$ and $G/N$ are solvable for some normal $N\subset G$. – anomaly Oct 29 '14 at 18:36
  • That they're cyclic is immediate. What's your rationale from there? – zibadawa timmy Oct 29 '14 at 18:36
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    It is actually enough that all Sylow subgroups are cyclic, but you need some non-trivial group theory to prove that (transfer theorems or normal $p$-complement theorems). – Geoff Robinson Oct 29 '14 at 19:12

1 Answers1

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Let $G$ be a finite group of square-free order and let $p$ be the smallest prime divisor of $|G|$, with $P$ being a Sylow $p$-subgroup of G. Then $P$ is cyclic, solvable, and has a normal complement $K$ in $G$ (Burnside 1900). Then $|K|$ is square-free, and, by induction on the order of the group, $K$ is solvable. Also, $G/K \cong P$ is solvable, so $G$ is, as a solvable extension also solvable.

Reference: Theorem 17 here.

Dietrich Burde
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