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Automorphism group of the Alternating Group - a proof

In the above question, Derek Holt asserts that a primitive permutation groups has trivial centraliser in the symmetric group. Since I could not understand why, I thought to open a new question (maybe someone like me could be interested in this). Is there any simple proof of this?

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  • $\begingroup$ I do not know a lot about groups of permutations. If $G$ has order prime we cannot saying anything, right? The orbit of $X$ are blocks for $G$ and so $X$ must be transitive (since $G$ is primitive). Why $X$ permutes the fixed points of $g$ for any $g\in G$? and how it follows that no non-identity element of $G$ fixes any point? $\endgroup$ – W4cc0 Dec 9 '14 at 7:48
  • $\begingroup$ You should be able to prove yourself that if two permutations $g$ and $h$ commute, then $g$ permutes the fixed points of $h$, and vice versa. $\endgroup$ – Derek Holt Dec 9 '14 at 8:41
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    $\begingroup$ Just a minor point: you don't need to assume that $n$ is not prime, but just that $|G|$ is not prime, which is a weaker assumption. $\endgroup$ – Derek Holt Dec 9 '14 at 10:02
  • $\begingroup$ @Geoff Robinson: Maybe you can update your solution as the answer. $\endgroup$ – W4cc0 Dec 9 '14 at 17:07
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I'm not sure whether clarifying an answer to an earlier question merits a formal answer ( and I have incorporated a minor point made by Derek in this answer anyway), but here goes : First of all, groups of prime order $p$ in $S_{p}$ are an exception, as Derek pointed out in his answer to the previous question. Suppose then that $G$ is a primitive subgroup of $S_n$ which is not of prime order, and let $X$ be the centralizer of $G.$ Then $G$ permutes the orbits of $X$, so if $X≠1$, then $X$ is transitive. Now $X$ permutes the fixed points of $g$ for any $g∈G$, so no non-identity element of $G$ fixes any point. Hence $G$ is regular (since it is transitive). Hence the trivial subgroup of $G$ is maximal, so $G$ has prime order, contrary to assumption.

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