Let $G$ be a finite group, $p$ a prime dividing the order of $G$, $X$ the set of $p$-Sylow subgroups of $G$. $G$ acts on $X$ by conjugation, and by the Sylow theorems the action is transitive and $\lvert X \rvert \equiv 1\pmod p$. Can we say anything more about the action? An answer to any single one of the following questions would be very welcome.

  1. If the action is imprimitive, so the normalizers are not maximal, what do the blocks of imprimitivity look like?

  2. If it is primitive, when is it $2$-transitive? $k$-transitive?

  3. Is the action always faithful? How do two normalizers (point stabilizers) intersect?

  4. What does the action of the normalizer $G_x$ on $X \, \backslash \, \{ x \}$ look like? What about the action of the $p$-Sylow alone?

  5. We can also look at the action of a subgroup $H \leq G$ on $X$: when is it transitive? Otherwise, what do the orbits look like?

  6. What can we say about a subgroup $H \leq G$ that acts regularly on $X$?

I know that's a lot of questions, but I'm just trying to get a better grasp of the action of $G$ on $X$, because I didn't find much more than transitivity itself in the litterature. Thanks in advance!

  • 2
    $\begingroup$ Re 3: of course the action is not always faithful. The kernel of the action is the largest normal subgroup contained in any $p$-Sylow subgroup $P$. This is not trivial for example if G is abelian and $P$ is non-trivial. $\endgroup$ – Konstantin Ardakov Oct 21 '17 at 15:31
  • $\begingroup$ Actually the kernel of the action is $C_G(O^{p'}(G)/O_p(G))$, where $O_p(G)$ is the largest normal $p$-subgroup, and $O^{p'}(G)$ is the group generated by the $p$-Sylow subgroups. $\endgroup$ – Colin Oct 24 '17 at 5:31

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