Let $G\le S_n$ be a transitive subgroup with trivial centralizer in $S_n$.

Can we deduce any nontrivial lower bounds on the order of $G$?

I'd also be interested in asymptotic results as $n\rightarrow\infty$.

  • $\begingroup$ That depends on $c$. For $c=1$,all you can say is that $|G| \ge n-1$, and for $c=n!$ you have $|G|=1$. For $c=(n/2)!$ (with $n$ even) you get something like $n/2-1 \le |G| \le (n/2)!$, which is a large range. Did you have any particular values of $c$ in mind? $\endgroup$ – Derek Holt Jan 9 '17 at 19:15
  • $\begingroup$ @DerekHolt Okay I've edited the question. So, my interest comes from this - When $M$ is a finite nonabelian simple group, $F_2$ is a free group of rank 2, and $\varphi : F_2\twoheadrightarrow M$ is a surjection, then I've checked computationally (for the smallest 23 nonab finite simple groups) that the action of $Aut(F_2)$ on the $Aut(F_2)$-orbit of $\varphi$ has precisely $Out(M)$ as its centralizer. In particular, $Out(M)$ is always rather small. I'd like to show that the permutation image of $Aut(F_2)$ on the orbit of $\varphi$ is always "pretty large" $\endgroup$ – user355183 Jan 9 '17 at 19:26
  • $\begingroup$ @DerekHolt In particular, if $e$ is the exponent of $M$, then I'd like to show that the permutation image always has size at least $e^3$ (computationally this is true, by massive margins). I suppose I should write this as a separate question. $\endgroup$ – user355183 Jan 9 '17 at 19:28

I think the best possible general bound for your edited question is $|G| \ge 2n$. For $n$ odd, the dihedral group of order $2n$ has trivial centralizer.

  • $\begingroup$ How do you see that the centralizer of $D_{2n}$ is trivial? It certainly has trivial center, but I assume you're embedding $D_{2n}$ inside $S_n$? If $n$ is even you can consider the action of $D_{2n}$ on vertices, but if $n$ is odd this doesn't work, right? $\endgroup$ – user355183 Jan 9 '17 at 19:37
  • $\begingroup$ The order of the centralizer in $S_n$ of a transitive subgroup $G$ is equal to the number of fixed points of the stabilizer $G_a$ of a point $a$. This is $1$ for $D_{2n}$ when $n$ is odd, but $2$ for $n$ even. $\endgroup$ – Derek Holt Jan 9 '17 at 20:25
  • $\begingroup$ I agree with the first sentence, and perhaps this is a stupid question - but how are you having $D_{2n}$ act on a set of $n$ elements when $n$ is odd? $\endgroup$ – user355183 Jan 9 '17 at 21:30
  • $\begingroup$ There must be some misunderstanding here! The dihedral group of order $2n$ is often defined as the group of rotations and reflections of an $n$-sided regular polygon, so it is acting on the $n$ vertices. $\endgroup$ – Derek Holt Jan 9 '17 at 21:47
  • $\begingroup$ But if $n$ is odd, the reflections don't send a vertex to another of the $n$ vertices! $\endgroup$ – user355183 Jan 9 '17 at 21:51

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