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I am trying to solve the following exercise from Dixon and Mortimer:

Let $G$ be a finite primitive permutation group with abelian point stabilizers. Show that $G$ has a regular normal elementary abelian subgroup.

As hints the authors suggest to use Frobenius' theorem and the Frattini argument.

My attempts: Let $H$ be some of those abelian stabilizers. By primitivity $H$ is maximal. If there is some element $g \in G \setminus H$ with $H \cap H^g \neq 1$ it is easy to show that this intersection contains such a regular normal subgroup we are looking for. So let us suppose there is no such element, i.e. $G$ is a Frobenius group with Frobenius kernel $K$, say. It is easy to see that $K$ is regular and characteristically simple, hence a direct product of isomorphic simple groups. So it remains to show that $K$ is abelian.

Of course by a well known theorem we know that Frobenius kernels are nilpotent in general, and hence $K$ must be abelian in this case. But do we really need such a strong theorem here or is there an easier argumentation? I did not use the Frattini argument yet.

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  • $\begingroup$ The idea I tried before is rubbish; thanks for pointing out the error! The (dihedral) group of degree $5$ (acting naturally on a regular pentagon) has point stabilisers of order $2$, providing a counter-example. It might still be possible to show that $\operatorname{Aut}(K)$ is transitive on $K\setminus 1$ by some other method but, if there is, I'm not seeing it right now. $\endgroup$ – James Aug 15 '14 at 18:01
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I didn't understand the first part of your argument. If $H \cap H^g \ne 1$, then $C_G(H \cap H^g)$ is a proper subgroup of $G$ that strictly contains $H$, contradicting primitivity.

(To see that $C_G(H \cap H^g)$ is a proper subgroup of $G$, note that $H \cap H^g$ fixes some but not all of the permuted points, and its centralizer permutes the set of its fixed points. But $G$ is transitive, so $G$ cannot permute the fixed point set of $H \cap H^g$, and hence $C_G(H \cap H^g) \ne G$.)

Suppose that $G$ is a Frobenius group with kernel $K$ that is isomorphic to a diretc product of nonabelian simple groups. Let $P \in {\rm Syl}_p(K)$ for some prime $p$ dividing $|K|$. Then by the Frattini argument $G = N_G(P)K$, where $N_G(P)$ is a proper subgroup of $G$.

(So, by the 2nd Isomorphism Theorem, $H \cong G/K \cong N_G(P)/N_K(P)$, and hence we can apply Schur-Zassenhaus to $N_G(P)$ and deduce that $N_K(P)$ has a complement that is isomorphic to $H$.)

By Schur-Zassenhaus, $N_G(P)$ has a subgroup of order $|H|$ (complementing $N_K(P)$) which, by Schur-Zassenhaus again, must be conjugate to $H$, and hence is a point stabilizer, so again we contradict the primitivity of $G$.

The conjugacy part of the Schur-Zassenhaus Theorem requires that at least one of the normal subgroup and the complement is solvable. In this case, $H$ is abelian and hence solvable.

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  • $\begingroup$ Thank you very much or your answer, Derek! Unfortunately, it is hard for me to follow your argumentation. Why is $C_G(H \cap H^g)$ a proper subgroup? How do you know the order of $N_G(P)$ to apply Schur-Zassenhaus? Which argument do you use to avoid the necessity of the odd order theorem for the conjugacy statement of Schur-Zassenhaus? $\endgroup$ – Dune Aug 15 '14 at 20:28
  • $\begingroup$ I've added some more explanation. We don't need the odd order theorem, because we know that $H$ is abelian, and hence solvable. $\endgroup$ – Derek Holt Aug 16 '14 at 19:32

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