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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.