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Suppose $G \leqslant S_p$ acts transitively on $\{1,...,p\}$ for prime $p$. Let $P \leqslant G$ be a Sylow p-subgroup. Is it true that $G$ is soluble <=> $P \triangleleft G$?

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  • $\begingroup$ mathoverflow.net/questions/24081/… may be of some interest? $\endgroup$ – hmmmm Mar 16 '14 at 22:03
  • $\begingroup$ @hmmmm well, that is nice. But is there such an argument without quoting the $AGL$? $\endgroup$ – user112072 Mar 16 '14 at 22:49
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This result is due to Galois who was quite familiar with AGL. Here is a rephrasing of Robin Chapman's answer.

(1) The normalizer of $P$ in $S_p$ is solvable. Hence if $P \unlhd G \leq S_p$, then $G \leq N_{S_p}(P)$ is solvable.

(2) If $G \leq S_p$ is solvable and transitive, then let $N$ be the last non-identity term of the derived series (or any non-identity abelian normal subgroup of $G$). Since $G$ is primitive and $N$ is non-identity normal, $N$ is transitive. Hence $N$ has order divisible by $p$, and so contains some Sylow $p$-subgroup $P$ of $S_p$. Since $N$ is abelian, $P \unlhd N$ is even characteristic in $N$, and since $N$ is normal in $G$, we get $P \unlhd G$.

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  • $\begingroup$ It must be $P \trianglelefteq G$. $\endgroup$ – user112072 Mar 16 '14 at 23:49
  • $\begingroup$ Thanks, fixed. :-) $\endgroup$ – Jack Schmidt Mar 17 '14 at 0:07

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