Here we give a proof of a proposition which is a particular case of a theorem of Huppert on solvable doubly transitive permutation groups (see, for example, [HB], Chapter XII, §7).

Proposition 1.Let D be a group of odd order which acts faithfully on an elementary abelian q-group E (q prime) and which is transitive on E#. Then F(D) is cyclic and acts without fixed points on E, and D/F(D) is abelian.

(Note that, under the hypotheses of this proposition, E × D acts doubly transitively on E.)

Lemma. Let p be a prime number, p ≠ 2, and let P be a p-group acting faithfully on the elementary abelian q-group E. Assume that |Pa| is independent of a for a ∈ E#. Then P is cyclic and acts without fixed points on E.

Proof. We will denote the operation in E additively and consider E as an Fq[P]-module.

(1) Preliminary steps.

Assume first of all that E = E1 ⊕ … ⊕ Er where r ≥ 2 and the Ei are subspaces of E permuted by P (i.e., (Ei)g is one of the subspaces Ej for g ∈ P and 1 ≤ i ≤ r).