Let $n = p^k$ be a power of a prime. Let $G$ be a transitive permutation group on $n$ letters. Show that there exists a fixed-point free element of order $p$.

I can show that there must be a fixed point: The transitivity of the group implies that there is only $1$ orbit. Burnside's lemma then implies that the average number of fixed points for an element in the group is $1$. Since the identity fixes everything, there is a fixed point free element.

I can show that there exists an element of order $p$: Again we use that the group is transitive and that there is therefore only one orbit. Since there is only one orbit, the order of the orbit of $x$ is $p^k$. By the orbit-stabilizer theorem, the order of the stabilizer of some element $x$ is equal to the order of the group divided by $p^k$. Since the order of a stabilizer is an integer, $p^k$ divides the order of the group. By Cauchy's theorem, there is an element of order $p$ in the group.

I cannot, however, show that there is an element of order $p$ with no fixed points. Does anyone have any hints?


1 Answer 1


As you ask only for hints, here are two:

Step 1: Show that some/any $p$-Sylow subgroup $P$ of $G$ acts transitively on the $n$ letters.

Step 2: Then consider the action of a central element of $P$ of order $p$.

  • $\begingroup$ SPOILER (solution of step 1): For a $p$-Sylow $P$ of $G$ the one-point stabilizer $P_x = P \cap G_x$ has index $p^k$ in $P$, hence $P$ is transitive. $\endgroup$
    – j.p.
    Oct 4, 2010 at 9:44
  • $\begingroup$ SPOILER (solution of step 2): Let $G$ acting on a set $\Omega$ have a normal subgroup $N$. Then $G$ acts on the set of fixed points of $N$: If $x$ is fixed by $N$, given $n\in N$, $g\in G$, as $N$ normal, there is an $n' \in N$ such that $n'g = gn$. Now $x^g = x^{n'g} = x^{g n} shows that $n$ fixes also $x^g$. $\endgroup$
    – j.p.
    Oct 4, 2010 at 10:57
  • $\begingroup$ Sorry I accidentally hit enter: I figured out step 1, but step 2, even with your hint eludes me. What do you by central element? Also: $x$ is fixed by the entire subgroup $N$ by assumption, and your spoiler proves that $n$ also fixes any image of $x$ by a group element. Since the group is transitive $N$ fixes the whole set? And then we have a contradiction? $\endgroup$
    – Abelsh
    Oct 4, 2010 at 21:02
  • $\begingroup$ As $P$ is a $p$-group, it has a nontrivial center $Z(P) \ne 1$. Take an element $1 \ne z \in Z(P)$ (central element = element of the center). $\endgroup$
    – j.p.
    Oct 5, 2010 at 7:42
  • $\begingroup$ Yes, as the group is transitive, $N$ either doesn't fix any element or it fixes all elements (since the set of fixed points of $N$ is union of orbits of $G$). $\endgroup$
    – j.p.
    Oct 5, 2010 at 7:44

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